Calculus 1 Final Exam Review

now this video is for those of you who are taking the calculus one final exam that is if you&#39;re in college or if you&#39;re in high school you can also use this if you&#39;re taking the ap calculus exam you could benefit from this video as well so this video is going to cover continuity limits derivatives integration and all the subtopics in those chapters so let&#39;s go ahead and begin number one evaluate the limit shown below now it&#39;s always good to see if you could use direct substitution however if we plug in 3 we&#39;re going to get 9 squared minus 9 which will give us 0 in the denominator so we can&#39;t use direct substitution in this case what we need to do first is we need to factor the expression so to factor a difference of perfect squares x squared minus 9 we can write that as x plus three times x minus three the square root of x squared is x and the square root of nine is three and so that&#39;s how you can get those numbers now what about factoring this trinomial what two numbers multiply to negative 15 but add to positive two so this is going to be positive five and negative three five plus negative 3 adds up to 2 5 times negative 3 multiplies to negative 15. so at this point we could cancel the x minus 3 factor and so what we have left over is the limit as x approaches 3 of x plus 5 over x plus 3. so now at this point we could use direct substitution so it&#39;s going to be 3 plus 5 divided by three plus three three plus five is eight three plus three is six and then if you divide both numbers by two you can reduce that to four over three so a is the correct answer now let&#39;s move on to number two evaluate the expression shown below so for each of these problems pause the video and uh work on it and then when you&#39;re ready just hit the play button and see if you have the right answer now for this problem you need to be familiar with the power rule the derivative of a variable raised to a constant such as x to the n it&#39;s going to be that constant times the variable raised to the constant minus one so just to review the derivative of x to the fourth power is going to be in this case n is four four x to the four minus one which is four 4x cubed so now let&#39;s try this problem so what we&#39;re going to do is differentiate each one individually the derivative of x to the sixth power is going to be six x to the sixth minus one which is five now for three over x we need to rewrite it if we move the variable to the top it becomes three x to the minus one and so now we can use the power rule so the derivative of x to the minus one is going to be negative one x to the negative one minus one which is negative two now the square root of x we also need to rewrite it and we can rewrite it as x to the one half so using the power rule it&#39;s going to be one half x and then it&#39;s gonna be one half minus one so far we have six x to the fifth power minus three x to the minus two and then one half minus one negative one is the same as negative two over two so one minus two is negative one so one half minus one is negative a half now we need to get rid of the negative exponents and so what we can do is bring the variable back to the bottom and the negative exponent is going to change to a positive exponent so this is going to be negative 3 over x squared and this is going to be plus 1. the 2 is already on the bottom now we need to move the x variable to the bottom and so that&#39;s going to be x to the positive one-half now x to the one-half is the same as the square root of x so instead of writing x and one-half we can write the square root of x here thus this is the final answer number three find the value of c that makes f of x continuous to make the piecewise function continuous the two parts of the piecewise function must be equal to each other when x is equal to three so step one set these two parts equal to each other so we&#39;re going to have two cx minus six is equal to x squared plus c x now the next thing we need to do is replace x with three so it&#39;s going to be two c times 3 minus 6 is equal to 3 squared plus c times 3. at this point we only have one variable in the equation which is c so we can solve for it 2 times 3 is 6. so on the left side we have six c minus six three squared is nine so we have nine plus three c on the right side now let&#39;s subtract both sides by three c and at the same time let&#39;s add six to both sides so we could cancel this and that so here we have 6c minus 3c which is 3c and then a 9 plus 6 that&#39;s 15. dividing both sides by three i don&#39;t know what just happened there but let&#39;s fix that we get 15 over three which is five and so that&#39;s the answer c is equal to five so that&#39;s the value of c that makes f of x continuous so e is the correct answer choice number four find the derivative of the expression shown below now there&#39;s three things that you need to know in order to do this problem you need to know how to differentiate exponential functions logarithmic functions and you need to know how to use the product rule so let&#39;s talk about how to find the derivative of exponential functions the derivative of e to the u where u is a function of x is going to be the same thing e to the u times u prime so let&#39;s say if i wanted to differentiate e to the x squared it&#39;s going to be e to the x squared times the derivative of x squared which is 2x or let&#39;s say if i want to differentiate e to the 7x it&#39;s going to be e to the 7x times the derivative of 7x which is 7. now let&#39;s talk about how to differentiate logarithmic functions specifically natural log functions the derivative of ln u with respect to x is going to be u prime divided by u so let&#39;s say if i want to differentiate ln x so u is x the derivative of x is 1 and since u is x that&#39;s going to be x on the bottom so we get 1 over x now let&#39;s see if i have the derivative of ln x squared plus 7. the derivative of x squared plus 7 is 2x on the bottom whatever is inside the natural log expression we just need to rewrite that on the bottom so it&#39;s going to be 2x over x squared plus 7. now let&#39;s talk about the product rule the derivative of let&#39;s say we have two things multiplied to each other f times g it&#39;s going to be the derivative of the first part so that&#39;s f prime times the second part which is g plus the first part times the derivative of the second so let&#39;s say that f is e to the 4x so what&#39;s f prime going to be using the equation that we had before it&#39;s e to the u times u prime so it&#39;s going to be e to the 4x times the derivative of 4x which is 4. now g is going to be the second part of that expression so that&#39;s ln 2x plus 5. now g prime using the formula u prime over u it&#39;s going to be the derivative of 2x plus 5 which is just 2 divided by whatever is inside the natural log expression so that&#39;s 2x plus 5. so now all we need to do is follow this formula so the derivative of this expression is going to be f prime which is 4 e to the 4x and then times g so that&#39;s ln 2x plus 5 and then plus f which is e to the 4x times g prime which is that two over two x plus five now we can basically combine those two so i&#39;m gonna rewrite this as two e to the four x over two x plus five so we can leave our final answer like this and so that&#39;s it for this problem number five evaluate the following integral now i do want to mention a few things before we start this problem one is to check the links in the description section below i&#39;m going to post a few links to some other videos like on limits derivatives and integration for those of you who need more example problems i&#39;m also going to place a link to my calculus playlist let&#39;s say if you need help with a specific topic and you don&#39;t see it in this exam so feel free to take a look at that when you get a chance now these problems may seem simple to you but they will get harder as we progress in this video but let&#39;s start with this one so how can we evaluate the following integral what is the anti-derivative of 4x to the fifth plus x to the fourth minus 3x cubed i mean minus 3x squared divided by x squared what we need to do is simplify the expression first so let&#39;s begin by dividing each term in the numerator of the fraction by x squared so x to the fifth divided by x squared is going to be x cubed times four because five minus two is three now x to the fourth divided by x squared is x squared four minus two is two and then negative three x squared divided by x squared the x squared stuff will cancel and so it&#39;s going to be negative three now we have to use the power rule version for integration so just to review we said that the derivative of a variable raised to a constant is going to be the constant times the variable raised to the constant minus one the antiderivative of x to the n is going to be so instead of subtracting one we&#39;re going to add one to the constant and then we&#39;re going to divide by that result and don&#39;t forget to add the constant of integration so this is the power rule when dealing with uh integration add one and then divide by that result so what is the anti-derivative of x cubed the constant in the front just rewrite it so what we&#39;re going to do to find the anti-derivative of x cubed is add 1 to the exponent 3 plus 1 is 4 and then divide by that number here we&#39;re going to do the same 2 plus 1 is 3 and then divide by three now the antiderivative of negative three is simply negative three x negative three is the same as just negative three times x to the zero because anything raised to the zero power is one and so you could think of it this way if we add one to zero and then divide by the result we&#39;re going to get negative three x so that&#39;s the anti-derivative of negative 3. now the next thing we need to do is add the constant of integration finally we need to simplify our answer so we could cancel the four and so we&#39;re gonna have x to the fourth plus we can write this as one third x cubed and then minus 3x plus c so this is the final answer for the problem so that&#39;s how we can evaluate this particular integral number six find the equation of the tangent line to the curve x cubed plus 4 x y squared plus y cubed equals 107 at the point 2 comma 3 using implicit differentiation so now let&#39;s talk about this let&#39;s say if we have some generic curve it may not be this exact curve the tangent line is a line that touches the curve at exactly one point a secant line touches the curve at two points now in order to find the equation of the tangent line we&#39;ll need three things we need the x and y coordinates of this point of intersection which we have already and we need the slope of the tangent line which we do not have now the slope of the tangent line is equal to the first derivative of the function at some x value but in this particular problem it&#39;s going to equal d y d x at the point two comma three so we need to use implicit differentiation to find d y dydx plug in this point that will give us the slope of the tangent line and then we could use the point slope formula to write the equation of the tangent line so let&#39;s go ahead and begin let&#39;s start by differentiating this expression with respect to x the derivative of x cubed with respect to x is 3x squared now for this expression 4x y squared we need to use the product rule so let&#39;s say f is 4x and g is y squared now the basic idea behind the product rule is that you differentiate the first part of the function the derivative of 4x is 4 and then you leave the second part alone so that&#39;s going to be y squared and then plus now we&#39;re going to leave the first part alone and then differentiate the second part the derivative of y squared with respect to x be careful with this it&#39;s 2y times dydx so remember when dealing with implicit differentiation every time you differentiate a y variable with respect to x you need to add dydx to it derivative of y cubed is three y squared times d y d x and the derivative of a constant 107 is zero now at this point we do not need to isolate dydx in fact to make it easier i recommend replacing x and y with their respective values so let&#39;s replace x with two and let&#39;s replace y with three and then let&#39;s calculate the value of d y d x two squared is four times three that&#39;s twelve three squared is nine times four is thirty six four times two is eight two times three is six three squared is nine times three that&#39;s twenty seven so right now we can add these two twelve plus thirty six is forty 48 and if we move that to the other side of the equation it&#39;s going to be negative 48 on this side now 8 times 6 is 48. and then we have plus 27 dydx so at this point we can get rid of this stuff since we have like terms we can add these two numbers 48 plus 27 40 plus 20 is 60 8 plus 7 is 15 60 plus 15 is 75. so we have 75 dydx and that&#39;s equal to negative 48. so now we can divide both sides by 75. so dydx is equal to negative 48 over 75 when x is 2 and when y is 3. so this is equal to the slope of the tangent line at that point so now we have enough information to write the equation of the tangent line so we&#39;re going to use the point slope formula y minus y one is equal to m times x minus x one y1 is 3 x1 is 2. so this is going to be y minus 3 is equal to m m is negative 48 over 75 but we can reduce that let&#39;s divide the top and the bottom by three so this is negative 16 over 25 so let&#39;s replace m with that now x1 is two so this is the equation of the tangent line in point slope form so if this was a free response problem you could leave your answer like this and you&#39;ll be done with with it but this is a multiple choice problem and looking at the answers all of them are in standard form now sometimes you may need to solve for y sometimes the answers may be in slope intercept form y equals mx plus b but we need to get it in this form first so how can we do that one thing i recommend doing is getting rid of the fraction because we don&#39;t have that anywhere in any of the answer choices so let&#39;s multiply both sides by 25 so on the right side these two will cancel on the left side we&#39;re going to have 25y minus 75 and then it&#39;s going to be negative 16x and then negative 16 times negative 2 that&#39;s positive 32. now let&#39;s rearrange what we have i&#39;m going to move the negative 16x from the right side to the left side so on the left side it&#39;s going to be positive 16x and then we&#39;re going to have plus 25y now the negative 75 i&#39;m going to move it to this side on the right side it&#39;s going to be positive 75. so 32 plus positive 75 is positive 107. so this is the answer 16x plus 25y equals 107. that is the equation of the tangent line in standard form and so this matches answer choice d number seven which of the following answer choices is equivalent to the expression shown below so go ahead and think about it work out this problem now there&#39;s really no way in which we can work out this limit we have no numbers to plug in there&#39;s nothing to factor we can&#39;t rationalize the denominator or get any common denominators or none of that the only way to find the answer to this problem is to understand what it really means you need to know that f prime of x is equal to the limit as h approaches zero of f of x plus h minus f of x divided by h so this entire thing is equal to f prime of x and that&#39;s what you&#39;re looking for so what is f prime of x because that&#39;s going to give us the answer to this entire expression now first we need to know what f of x is notice that f of x lines up with sine x so if f of x is sine x what&#39;s f prime of x the derivative of sine x is cosine x and so this is the answer this entire thing is equal to cosine x so a is the right answer now let&#39;s work on number eight evaluate the integral shown below so how can we do this we can&#39;t use the power rule for this example but we could use something known as u-substitution what we&#39;re going to do is make u equal to three x squared plus five and so d u is going to be the derivative of that which is 6x times dx the derivative of a constant is 0. now what i recommend doing is dividing both sides by 6x to solve for dx so dx is going to be du divided by 6x in the next step we&#39;re going to replace 3x squared plus 5 with the u variable and at the same time we&#39;re going to replace the dx with what we have here so that&#39;s du over 6x so notice that all of the x variables have been canceled two over six we can reduce that to one over three and i&#39;m going to move the constant in the front the square root of u is the same as u to the one half so now we can find the antiderivative of u to the one-half so we need to add one to it one-half plus one is three over two and then instead of dividing by three over two we can multiply by the reciprocal two over three and don&#39;t forget to put plus c so multiplying these two fractions it&#39;s going to give us two over nine and then we&#39;ll have u to the three halves plus c now our next step is to replace u with what we have here three x squared plus five so the final answer is two over nine three x squared plus five raised to the three over two plus c so that&#39;s how you could use u-substitution to find the anti-derivative of certain functions number 9 water is flowing into a cylinder with a diameter of 6 feet and a height of 10 feet if the height of the water in the cylinder is increasing at 3 feet per minute at what rate is the volume of the water in the cylinder changing now for those of you who want harder related rates problems you can check out the videos that i have on youtube so what i recommend doing is go into the search box and type in related rates problem organic chemistry tutor if you type in that phrase you&#39;re going to see all of the related rates problems that i have so for those of you who wish to improve your skills in this area feel free to take a look at that now whenever solving a related excuse me a related rates problems uh what you need to do is you need to write a formula so what we have is a cylinder so let&#39;s draw a picture so we know the diameter of the cylinder so that&#39;s 6 feet which means the radius of the cylinder has to be half of that so the radius is 3 feet the height of the cylinder is 10 feet now we have some water in the cylinder and more water is entering the cylinder now we know the height of the water in the cylinder is increasing at a rate of 3 feet per minute so what variable does that represent so this is the rate at which the height is changing with respect to time so it&#39;s three feet per minute and it&#39;s positive because it&#39;s increasing now the next thing we need to do is write the formula for this situation so we&#39;re dealing with volume and cylinder so what is the volume of a cylinder the volume of a cylinder is equal to the area of the base times the height of the cylinder and the base is a circle the area of a circle is pi r squared so the volume is pi r squared times height now we need to know which letters are constants and which ones are variables so v the volume would you say it&#39;s a variable or constant in this case v is a variable because the volume of the water in the cylinder is changing it&#39;s not constant pi is a constant it&#39;s always 3.14 now what about the radius of the water in the cylinder is that constant imagine the cylinder being filled up with water will the radius of the water change now we know that liquids they take up the shape of their container so because the radius of the cylinder is always three feet the radius of the water in the excuse me in the cylinder will always be three feet so the radius is a constant so pi r squared that whole expression is a constant now the height of the water is changing because as you add water the water level will rise so h is a variable now this is important because when you&#39;re dealing with related rates you&#39;re not differentiated with respect to x as you would using implicit differentiation you&#39;re differentiating with respect to time and the only thing you need to worry about are the variables not the constants because if r was changing when we differentiate it we would have to use the product rule but because this is a constant we do not have to use the product rule it&#39;s the same as differentiating 5x you wouldn&#39;t use the product rule for that or 8x cubed you don&#39;t use the product rule for that but if you have like x cubed times y squared where both of these are variables then you need to use the product rule but let&#39;s say if x was constant and y was a variable then you do not have to use the product rule you only need to use the product rule if you have two different variables being multiplied or if you have two variable expressions of the same type so e to the 4x that&#39;s one variable expression but using x and ln x is also using x but that&#39;s also a variable expression so you would still use the product rule in that case but if you have a constant times a variable you do not need to use the product rule so keep that in mind so the derivative of v with respect to time is dv dt now what is the derivative of pi r squared times h with respect to time if we were to differentiate 5h it would simply be 5 times dh dt the derivative of pi r squared times h will just be pi r squared and the derivative of h is one and then times dht so this is the constant like five so we just need to rewrite it now we have everything that we need to get the answer our goal is to calculate dv dt r doesn&#39;t change r is going to be 3 feet and dhdt that&#39;s changing at 3 feet per minute so 3 squared times 3 that&#39;s 27 thus the answer is going to be 27 pi now what are the units of dv dt because you also need to get that right the unit for h height would be like feet meters things like that now for volume it&#39;s like cubic feet cubic meters cubic inches so in this case it&#39;s cubic feet since h is in feet now the time dt is in minutes so this is going to be the same that&#39;s going to be in minutes as well thus it&#39;s 27 pi cubic feet per minute number 10 identify all intervals where the function f of x is increasing given f of x is equal to x cubed plus three over two x squared minus 36 x minus nine so in order to find the intervals where the function is increasing we need to identify the critical points and one way we can do that is by finding the first derivative and setting it equal to zero and solving for x so let&#39;s do that the derivative of x cubed is three 3x squared the derivative of x squared is 2x the derivative of x is 1 and the derivative of a constant is 0. so we have three x squared and then the twos will cancel so this is going to be plus three x minus thirty six so let&#39;s set this equal to zero now we can take out the gcf the greatest common factor so if we take out a three we&#39;re going to have x squared plus x minus 12. now let&#39;s factor the trinomial what two numbers multiply to negative 12 but add to one so this is going to be positive four and negative three four plus negative three is positive one and four times negative three is negative twelve now what we need to do at this point is set each factor equal to zero if you set x plus four equal to zero the critical point that you&#39;ll get will be negative 4 and if you set x minus 3 equal to 0 you&#39;ll get the critical point positive 3. so now what we need to do is make a number line by putting these critical points on it next we need to check the signs so let&#39;s pick a number greater than three let&#39;s say five if we plug in five into this expression into the first derivative will we get a positive number or a negative number now 5 minus 3 is a positive number 5 plus 4 is a positive number and this is positive so the result will be positive now what if we pick a number between negative four and three like zero will it be positive or negative zero minus three is negative zero plus four is positive a positive number times a negative number will give us a negative result now let&#39;s pick something less than negative four like negative five negative five plus four is negative negative five minus three is a negative number a negative times a negative number will give us a positive number all the way to the left we&#39;ll have negative infinity and to the right positive infinity now let&#39;s get rid of this one thing i do want to mention is that whenever the first derivative is equal to a positive number the function f of x is increasing whenever the first derivative is equal to a negative number so you have a negative slope then the function is decreasing so these are the signs for the first derivative so the function is increasing between negative infinity and negative four and it&#39;s also increasing between three and infinity it&#39;s decreasing between negative four and three so the answer we could say the function is increasing between negative four i mean between negative infinity to negative four union and then between three and infinity so this is the answer to the problem now something i do want to mention is that you could use this to determine the location of any relative maximum or relative minimum some textbooks might use the term local maximum or local minimum so here the function is increasing and then it&#39;s decreasing so this looks like a local maximum here it&#39;s decreasing and then increasing so this looks like a local minimum so that&#39;s how you can identify the relative extrema on a function using the same process number 11 identify the location and maximum value of the function f of x is equal to 16 x minus x squared plus five now the leading term is negative x squared it has the highest degree and the general shape of negative x squared is basically a downward parabola so for this particular graph there&#39;s only one location where it&#39;s a maximum the location of the maximum is the x value and the actual maximum value is the y coordinate so basically we need to find both the x y coordinate of our answer so how can we do it how can we use calculus to locate the maximum value of the function we need to find the critical point because at this region where we have a horizontal tangent line the slope is zero and so this is our critical point let&#39;s go ahead and find it so f prime of x is going to be the derivative of 16x is just 16. the derivative of x squared is 2x and the derivative of 5 is 0. so if we set this equal to 0 all we need to do is factor out a two 16 divided by two is eight negative two x divided by two is just negative x and if we set eight minus x equal to zero if we add x to both sides x is eight so the critical point is eight so automatically that tells us that answer choice b is the right answer but let&#39;s finish the problem so if you were to plot this on a number line and let&#39;s say if you picked a number that&#39;s greater than eight like nine eight minus nine is negative if we plug in something less than eight like seven eight minus seven is positive so whenever the signs alternate from positive to negative going left to right you have a maximum it&#39;s increasing and then it&#39;s decreasing so we have a local max at eight now all we need to do to find the y coordinate is plug in the x value so f of 8 is going to be 16 times 8 minus 8 squared plus 5. now 16 times eight ten times eight is eighty six times eight is forty eight so eighty plus forty eight that&#39;s one twenty eight eight squared is sixty four one twenty eight minus sixty four is sixty four and 64 plus 5 is 69. now in case this problem was a free response problem and not a multiple choice you need to know that this is the maximum value of the function the y-coordinate now the location of the maximum value is located at the x-coordinate or at x equals eight so they ask you for the location of the critical point or the maximum value this is your answer if they ask you for the maximum value itself this is your answer so the function has a maximum value of 69. so b is the right answer for this problem now let&#39;s move on to 12. calculate the average value of the function f of x is equal to x cubed plus 8x minus 4 over the interval 1 to 5. so here&#39;s the basic idea behind calculating the average value so let&#39;s say we have some generic function f of x so basically what we&#39;re doing is we&#39;re taking the area under the curve and we&#39;re going to divide it by basically the width of the interval which is an x value and that&#39;s going to give us an average y value so our goal is to calculate this average y value of the function so here&#39;s the formula that we need the average function value or the average y value is going to be 1 divided by the width of the interval times the area under the curve in that interval which is the definite integral from a to b of f of x to dx so this right here represents the area under the curve and this portion represents the width of the interval which as we said is an x value and that will give us the average function value so what is a and what is b a as we can see is one b is five so this is going to be one over five minus one times the definite integral from one to five and then the function is x cubed plus eight x minus four dx five minus one is four so this becomes one over four now we need to find the anti-derivative of that expression the anti-derivative of x cubed is x to the fourth divided by four and for eight x is going to be eight x squared divided by two which is four x squared and the antiderivative of negative four is uh going to be negative 4x and now we need to evaluate this from 1 to 5. now before we do so it might be advantageous to distribute the 1 4 to everything inside the brackets so it&#39;s going to be 1 over 16 times x to the fourth and then these fours will cancel so that&#39;s going to be plus x squared and these fours will cancel as well so minus x evaluated from one to five so first let&#39;s plug in five so it&#39;s going to be one over 16 times 5 to the fourth power and then plus 5 squared minus 5 and then we could plug in 1. so it&#39;s going to be 1 over 16 plus 1 minus 1. now five to the fourth power that&#39;s 625 and then that&#39;s divided by 16. 5 squared is 25 and then we have negative 1 over 16 and 1 minus 1 is 0. now 625 over 16 minus 1 over 16 that becomes 6 24 over 16. and 25 minus 5 is 20. now 624 divided by 16 that&#39;s 39. so it&#39;s 39 plus 20 which is 59. so this is the average y value of the function over the interval one to five therefore c is the right answer number thirteen evaluate the expression shown below so what is the derivative of two x cubed minus seven x squared raised to the eighth power for this problem we need to use something called the chain rule perhaps you&#39;ve seen it expressed this way the derivative of a composite function let&#39;s say f of g of x is going to be f prime of g of x times g prime of x so notice that we differentiate the outside portion of the function keeping the inside the same and then we&#39;re going to multiply by the derivative of the inside of the function we&#39;re going to follow that same process to differentiate this expression so think of the outside portion of the function as x to the eighth that&#39;s like f of x and think of the inside function g of x as being two x cubed minus seven x squared so let&#39;s differentiate the outside portion the derivative of x to the eight would be eight x to the seventh power so we&#39;re gonna do is we&#39;re gonna move the eight to the front now instead of writing x we&#39;re going to keep everything that we have on the inside exactly the same and then we&#39;re going to subtract this by one so that&#39;s going to be seven next we&#39;re going to take the derivative of the inside function to get our g prime so the derivative of two x cubed is six x squared and the derivative of 7x squared is 14x and since this is a free response question we can leave our answer like this now for those of you who may want to simplify the answer we could take out the gcf in this expression so we could take out a 2x so multiplying 8 by 2x that&#39;s going to give us 16x and then we have 2x cubed minus 7x squared raised to the seventh power and now that we&#39;ve taken out 2x we need to divide everything there by 2x 6x squared divided by 2x is 3x negative 14x divided by 2x is negative 7. so you could leave your answer like this if you want to so that&#39;s how you could use the chain rule now for those of you who want more examples on the chain rule particularly harder examples i already have a video on that on youtube just type in chain rule organic chemistry tutor and you can get more examples on this if you need more help on this topic so let&#39;s move on to the next problem number 14 evaluate the limit expression shown below now let&#39;s say if you get a limit problem like this and you have no idea what to do it turns out that you can evaluate any limit if you have access to a calculator if that is allowed on your test if it&#39;s a free response problem and you can&#39;t use the calculator you just have to know the techniques involved but let&#39;s say if you do have access to a calculator what you could do is plug in a number that&#39;s close to four let&#39;s say 4.1 so type this in your calculator and make sure to use parentheses so you should get this f of 4.1 is negative point zero six zero nine seven five now do the same thing for a number that&#39;s closer to four so if we try four point zero one if you type this in you should get negative point zero six two three four notice that it&#39;s converging to a value it&#39;s getting close to negative point zero six something and you can go even further you can try 4.001 i&#39;m going to try that so you should get negative point zero six two 4 8 with some other numbers so we could see that it&#39;s around negative point zero six the answer can&#39;t be a one over four is point two five it can&#39;t be b negative two thirds is negative point six repeating negative eight and twelve are out now e is the answer this is negative point zero six two five if you type a negative 1 over 16 in your calculator and so that&#39;s one simple method that you could use to evaluate any limit if you don&#39;t know what to do if you have access to a calculator but now let&#39;s solve this the appropriate way what should we do if we have fractions within fractions if you have a complex fraction eliminate the smaller fractions by multiplying the top and the bottom by the common denominator of these two small fractions which is going to be 4x now let&#39;s distribute the 4x so 4x times 1 over x the x variables will cancel and so we&#39;re just going to get 4 on top now 4x times 1 4 the fours will cancel and so we&#39;re just going to get negative x on top and on the bottom don&#39;t distribute it just rewrite it as 4x times x minus 4. now what&#39;s our next step here what do you think we need to do at this point what i recommend doing is factoring out a negative one from four minus x if you take out a negative one from negative x negative x changes to positive x and if you take out a negative one from positive four it becomes negative four so notice that we can cancel x minus four so now we have the limit as x approaches four for negative one over four x so now we can use direct substitution so replacing x with four we&#39;re going to get negative one over sixteen so this is the final answer answer choice e number 15 identify all intervals where the function f of x is concave downward now before we attempt this problem let&#39;s write down some notes let&#39;s summarize what we need to know when the first derivative is positive you need to know that the function f of x is increasing and when the first derivative is negative the function f of x is decreasing now when f prime of x is equal to zero you have a horizontal tangent line and you also have a critical point what about the second derivative when the second derivative is positive the function f of x is said to be concave up when the second derivative is negative the function f of x is said to be concave down and when it&#39;s equal to zero f of x has an inflection point if the concavity changes that is if the second derivative changes from negative to positive or positive to negative if it doesn&#39;t change sign it&#39;s not an inflection point so for this particular problem we are concerned with this information so let&#39;s go ahead and find the second derivative and set it equal to zero let&#39;s start with the first derivative f prime of x the derivative of x cubed is 3x squared the derivative of x squared is 2x times negative 6 that&#39;s negative 12x and the derivative of 5x is 5. and the derivative of the constant one is a zero now let&#39;s calculate the second derivative the derivative of three x squared is going to be six x and the derivative of negative twelve x is just negative twelve let&#39;s set the second derivative equal to zero and let&#39;s factor out a six so now if we set x minus two equal to zero we could see that we have a point of interest at x equals two so we&#39;re going to make a number line and we&#39;re going to test the signs so if we pick a number greater than 2 like 3 3 minus 2 is positive if we pick a number less than 2 like 1 1 minus 2 is negative so the second derivative will be negative if you plug in 1 into this expression always put your infinity signs so notice that 2 is an inflection point because the concavity the second derivative changes sign on the left side the second derivative is negative which means that the function is concave down on the right side because the second derivative is positive it&#39;s concave up there so for this particular problem the interval where the function is concave down is from negative infinity to 2. so that&#39;s when the second derivative is negative you

now this video is for those of you who are taking the calculus one final exam that is if you&amp;#39;re in college or if you&amp;#39;re in high school you can also use this if you&amp;#39;re taking the ap calculus exam you could benefit from this video as well so this video is going to cover continuity limits derivatives integration and all the subtopics in those chapters so let&amp;#39;s go ahead and begin number one evaluate the limit shown below now it&amp;#39;s always good to see if you could use direct substitution however if we plug in 3 we&amp;#39;re going to get 9 squared minus 9 which will give us 0 in the denominator so we can&amp;#39;t use direct substitution in this case what we need to do first is we need to factor the expression so to factor a difference of perfect squares x squared minus 9 we can write that as x plus three times x minus three the square root of x squared is x and the square root of nine is three and so that&amp;#39;s how you can get those numbers now what about factoring this trinomial what two numbers multiply to negative 15 but add to positive two so this is going to be positive five and negative three five plus negative 3 adds up to 2 5 times negative 3 multiplies to negative 15. so at this point we could cancel the x minus 3 factor and so what we have left over is the limit as x approaches 3 of x plus 5 over x plus 3. so now at this point we could use direct substitution so it&amp;#39;s going to be 3 plus 5 divided by three plus three three plus five is eight three plus three is six and then if you divide both numbers by two you can reduce that to four over three so a is the correct answer now let&amp;#39;s move on to number two evaluate the expression shown below so for each of these problems pause the video and uh work on it and then when you&amp;#39;re ready just hit the play button and see if you have the right answer now for this problem you need to be familiar with the power rule the derivative of a variable raised to a constant such as x to the n it&amp;#39;s going to be that constant times the variable raised to the constant minus one so just to review the derivative of x to the fourth power is going to be in this case n is four four x to the four minus one which is four 4x cubed so now let&amp;#39;s try this problem so what we&amp;#39;re going to do is differentiate each one individually the derivative of x to the sixth power is going to be six x to the sixth minus one which is five now for three over x we need to rewrite it if we move the variable to the top it becomes three x to the minus one and so now we can use the power rule so the derivative of x to the minus one is going to be negative one x to the negative one minus one which is negative two now the square root of x we also need to rewrite it and we can rewrite it as x to the one half so using the power rule it&amp;#39;s going to be one half x and then it&amp;#39;s gonna be one half minus one so far we have six x to the fifth power minus three x to the minus two and then one half minus one negative one is the same as negative two over two so one minus two is negative one so one half minus one is negative a half now we need to get rid of the negative exponents and so what we can do is bring the variable back to the bottom and the negative exponent is going to change to a positive exponent so this is going to be negative 3 over x squared and this is going to be plus 1. the 2 is already on the bottom now we need to move the x variable to the bottom and so that&amp;#39;s going to be x to the positive one-half now x to the one-half is the same as the square root of x so instead of writing x and one-half we can write the square root of x here thus this is the final answer number three find the value of c that makes f of x continuous to make the piecewise function continuous the two parts of the piecewise function must be equal to each other when x is equal to three so step one set these two parts equal to each other so we&amp;#39;re going to have two cx minus six is equal to x squared plus c x now the next thing we need to do is replace x with three so it&amp;#39;s going to be two c times 3 minus 6 is equal to 3 squared plus c times 3. at this point we only have one variable in the equation which is c so we can solve for it 2 times 3 is 6. so on the left side we have six c minus six three squared is nine so we have nine plus three c on the right side now let&amp;#39;s subtract both sides by three c and at the same time let&amp;#39;s add six to both sides so we could cancel this and that so here we have 6c minus 3c which is 3c and then a 9 plus 6 that&amp;#39;s 15. dividing both sides by three i don&amp;#39;t know what just happened there but let&amp;#39;s fix that we get 15 over three which is five and so that&amp;#39;s the answer c is equal to five so that&amp;#39;s the value of c that makes f of x continuous so e is the correct answer choice number four find the derivative of the expression shown below now there&amp;#39;s three things that you need to know in order to do this problem you need to know how to differentiate exponential functions logarithmic functions and you need to know how to use the product rule so let&amp;#39;s talk about how to find the derivative of exponential functions the derivative of e to the u where u is a function of x is going to be the same thing e to the u times u prime so let&amp;#39;s say if i wanted to differentiate e to the x squared it&amp;#39;s going to be e to the x squared times the derivative of x squared which is 2x or let&amp;#39;s say if i want to differentiate e to the 7x it&amp;#39;s going to be e to the 7x times the derivative of 7x which is 7. now let&amp;#39;s talk about how to differentiate logarithmic functions specifically natural log functions the derivative of ln u with respect to x is going to be u prime divided by u so let&amp;#39;s say if i want to differentiate ln x so u is x the derivative of x is 1 and since u is x that&amp;#39;s going to be x on the bottom so we get 1 over x now let&amp;#39;s see if i have the derivative of ln x squared plus 7. the derivative of x squared plus 7 is 2x on the bottom whatever is inside the natural log expression we just need to rewrite that on the bottom so it&amp;#39;s going to be 2x over x squared plus 7. now let&amp;#39;s talk about the product rule the derivative of let&amp;#39;s say we have two things multiplied to each other f times g it&amp;#39;s going to be the derivative of the first part so that&amp;#39;s f prime times the second part which is g plus the first part times the derivative of the second so let&amp;#39;s say that f is e to the 4x so what&amp;#39;s f prime going to be using the equation that we had before it&amp;#39;s e to the u times u prime so it&amp;#39;s going to be e to the 4x times the derivative of 4x which is 4. now g is going to be the second part of that expression so that&amp;#39;s ln 2x plus 5. now g prime using the formula u prime over u it&amp;#39;s going to be the derivative of 2x plus 5 which is just 2 divided by whatever is inside the natural log expression so that&amp;#39;s 2x plus 5. so now all we need to do is follow this formula so the derivative of this expression is going to be f prime which is 4 e to the 4x and then times g so that&amp;#39;s ln 2x plus 5 and then plus f which is e to the 4x times g prime which is that two over two x plus five now we can basically combine those two so i&amp;#39;m gonna rewrite this as two e to the four x over two x plus five so we can leave our final answer like this and so that&amp;#39;s it for this problem number five evaluate the following integral now i do want to mention a few things before we start this problem one is to check the links in the description section below i&amp;#39;m going to post a few links to some other videos like on limits derivatives and integration for those of you who need more example problems i&amp;#39;m also going to place a link to my calculus playlist let&amp;#39;s say if you need help with a specific topic and you don&amp;#39;t see it in this exam so feel free to take a look at that when you get a chance now these problems may seem simple to you but they will get harder as we progress in this video but let&amp;#39;s start with this one so how can we evaluate the following integral what is the anti-derivative of 4x to the fifth plus x to the fourth minus 3x cubed i mean minus 3x squared divided by x squared what we need to do is simplify the expression first so let&amp;#39;s begin by dividing each term in the numerator of the fraction by x squared so x to the fifth divided by x squared is going to be x cubed times four because five minus two is three now x to the fourth divided by x squared is x squared four minus two is two and then negative three x squared divided by x squared the x squared stuff will cancel and so it&amp;#39;s going to be negative three now we have to use the power rule version for integration so just to review we said that the derivative of a variable raised to a constant is going to be the constant times the variable raised to the constant minus one the antiderivative of x to the n is going to be so instead of subtracting one we&amp;#39;re going to add one to the constant and then we&amp;#39;re going to divide by that result and don&amp;#39;t forget to add the constant of integration so this is the power rule when dealing with uh integration add one and then divide by that result so what is the anti-derivative of x cubed the constant in the front just rewrite it so what we&amp;#39;re going to do to find the anti-derivative of x cubed is add 1 to the exponent 3 plus 1 is 4 and then divide by that number here we&amp;#39;re going to do the same 2 plus 1 is 3 and then divide by three now the antiderivative of negative three is simply negative three x negative three is the same as just negative three times x to the zero because anything raised to the zero power is one and so you could think of it this way if we add one to zero and then divide by the result we&amp;#39;re going to get negative three x so that&amp;#39;s the anti-derivative of negative 3. now the next thing we need to do is add the constant of integration finally we need to simplify our answer so we could cancel the four and so we&amp;#39;re gonna have x to the fourth plus we can write this as one third x cubed and then minus 3x plus c so this is the final answer for the problem so that&amp;#39;s how we can evaluate this particular integral number six find the equation of the tangent line to the curve x cubed plus 4 x y squared plus y cubed equals 107 at the point 2 comma 3 using implicit differentiation so now let&amp;#39;s talk about this let&amp;#39;s say if we have some generic curve it may not be this exact curve the tangent line is a line that touches the curve at exactly one point a secant line touches the curve at two points now in order to find the equation of the tangent line we&amp;#39;ll need three things we need the x and y coordinates of this point of intersection which we have already and we need the slope of the tangent line which we do not have now the slope of the tangent line is equal to the first derivative of the function at some x value but in this particular problem it&amp;#39;s going to equal d y d x at the point two comma three so we need to use implicit differentiation to find d y dydx plug in this point that will give us the slope of the tangent line and then we could use the point slope formula to write the equation of the tangent line so let&amp;#39;s go ahead and begin let&amp;#39;s start by differentiating this expression with respect to x the derivative of x cubed with respect to x is 3x squared now for this expression 4x y squared we need to use the product rule so let&amp;#39;s say f is 4x and g is y squared now the basic idea behind the product rule is that you differentiate the first part of the function the derivative of 4x is 4 and then you leave the second part alone so that&amp;#39;s going to be y squared and then plus now we&amp;#39;re going to leave the first part alone and then differentiate the second part the derivative of y squared with respect to x be careful with this it&amp;#39;s 2y times dydx so remember when dealing with implicit differentiation every time you differentiate a y variable with respect to x you need to add dydx to it derivative of y cubed is three y squared times d y d x and the derivative of a constant 107 is zero now at this point we do not need to isolate dydx in fact to make it easier i recommend replacing x and y with their respective values so let&amp;#39;s replace x with two and let&amp;#39;s replace y with three and then let&amp;#39;s calculate the value of d y d x two squared is four times three that&amp;#39;s twelve three squared is nine times four is thirty six four times two is eight two times three is six three squared is nine times three that&amp;#39;s twenty seven so right now we can add these two twelve plus thirty six is forty 48 and if we move that to the other side of the equation it&amp;#39;s going to be negative 48 on this side now 8 times 6 is 48. and then we have plus 27 dydx so at this point we can get rid of this stuff since we have like terms we can add these two numbers 48 plus 27 40 plus 20 is 60 8 plus 7 is 15 60 plus 15 is 75. so we have 75 dydx and that&amp;#39;s equal to negative 48. so now we can divide both sides by 75. so dydx is equal to negative 48 over 75 when x is 2 and when y is 3. so this is equal to the slope of the tangent line at that point so now we have enough information to write the equation of the tangent line so we&amp;#39;re going to use the point slope formula y minus y one is equal to m times x minus x one y1 is 3 x1 is 2. so this is going to be y minus 3 is equal to m m is negative 48 over 75 but we can reduce that let&amp;#39;s divide the top and the bottom by three so this is negative 16 over 25 so let&amp;#39;s replace m with that now x1 is two so this is the equation of the tangent line in point slope form so if this was a free response problem you could leave your answer like this and you&amp;#39;ll be done with with it but this is a multiple choice problem and looking at the answers all of them are in standard form now sometimes you may need to solve for y sometimes the answers may be in slope intercept form y equals mx plus b but we need to get it in this form first so how can we do that one thing i recommend doing is getting rid of the fraction because we don&amp;#39;t have that anywhere in any of the answer choices so let&amp;#39;s multiply both sides by 25 so on the right side these two will cancel on the left side we&amp;#39;re going to have 25y minus 75 and then it&amp;#39;s going to be negative 16x and then negative 16 times negative 2 that&amp;#39;s positive 32. now let&amp;#39;s rearrange what we have i&amp;#39;m going to move the negative 16x from the right side to the left side so on the left side it&amp;#39;s going to be positive 16x and then we&amp;#39;re going to have plus 25y now the negative 75 i&amp;#39;m going to move it to this side on the right side it&amp;#39;s going to be positive 75. so 32 plus positive 75 is positive 107. so this is the answer 16x plus 25y equals 107. that is the equation of the tangent line in standard form and so this matches answer choice d number seven which of the following answer choices is equivalent to the expression shown below so go ahead and think about it work out this problem now there&amp;#39;s really no way in which we can work out this limit we have no numbers to plug in there&amp;#39;s nothing to factor we can&amp;#39;t rationalize the denominator or get any common denominators or none of that the only way to find the answer to this problem is to understand what it really means you need to know that f prime of x is equal to the limit as h approaches zero of f of x plus h minus f of x divided by h so this entire thing is equal to f prime of x and that&amp;#39;s what you&amp;#39;re looking for so what is f prime of x because that&amp;#39;s going to give us the answer to this entire expression now first we need to know what f of x is notice that f of x lines up with sine x so if f of x is sine x what&amp;#39;s f prime of x the derivative of sine x is cosine x and so this is the answer this entire thing is equal to cosine x so a is the right answer now let&amp;#39;s work on number eight evaluate the integral shown below so how can we do this we can&amp;#39;t use the power rule for this example but we could use something known as u-substitution what we&amp;#39;re going to do is make u equal to three x squared plus five and so d u is going to be the derivative of that which is 6x times dx the derivative of a constant is 0. now what i recommend doing is dividing both sides by 6x to solve for dx so dx is going to be du divided by 6x in the next step we&amp;#39;re going to replace 3x squared plus 5 with the u variable and at the same time we&amp;#39;re going to replace the dx with what we have here so that&amp;#39;s du over 6x so notice that all of the x variables have been canceled two over six we can reduce that to one over three and i&amp;#39;m going to move the constant in the front the square root of u is the same as u to the one half so now we can find the antiderivative of u to the one-half so we need to add one to it one-half plus one is three over two and then instead of dividing by three over two we can multiply by the reciprocal two over three and don&amp;#39;t forget to put plus c so multiplying these two fractions it&amp;#39;s going to give us two over nine and then we&amp;#39;ll have u to the three halves plus c now our next step is to replace u with what we have here three x squared plus five so the final answer is two over nine three x squared plus five raised to the three over two plus c so that&amp;#39;s how you could use u-substitution to find the anti-derivative of certain functions number 9 water is flowing into a cylinder with a diameter of 6 feet and a height of 10 feet if the height of the water in the cylinder is increasing at 3 feet per minute at what rate is the volume of the water in the cylinder changing now for those of you who want harder related rates problems you can check out the videos that i have on youtube so what i recommend doing is go into the search box and type in related rates problem organic chemistry tutor if you type in that phrase you&amp;#39;re going to see all of the related rates problems that i have so for those of you who wish to improve your skills in this area feel free to take a look at that now whenever solving a related excuse me a related rates problems uh what you need to do is you need to write a formula so what we have is a cylinder so let&amp;#39;s draw a picture so we know the diameter of the cylinder so that&amp;#39;s 6 feet which means the radius of the cylinder has to be half of that so the radius is 3 feet the height of the cylinder is 10 feet now we have some water in the cylinder and more water is entering the cylinder now we know the height of the water in the cylinder is increasing at a rate of 3 feet per minute so what variable does that represent so this is the rate at which the height is changing with respect to time so it&amp;#39;s three feet per minute and it&amp;#39;s positive because it&amp;#39;s increasing now the next thing we need to do is write the formula for this situation so we&amp;#39;re dealing with volume and cylinder so what is the volume of a cylinder the volume of a cylinder is equal to the area of the base times the height of the cylinder and the base is a circle the area of a circle is pi r squared so the volume is pi r squared times height now we need to know which letters are constants and which ones are variables so v the volume would you say it&amp;#39;s a variable or constant in this case v is a variable because the volume of the water in the cylinder is changing it&amp;#39;s not constant pi is a constant it&amp;#39;s always 3.14 now what about the radius of the water in the cylinder is that constant imagine the cylinder being filled up with water will the radius of the water change now we know that liquids they take up the shape of their container so because the radius of the cylinder is always three feet the radius of the water in the excuse me in the cylinder will always be three feet so the radius is a constant so pi r squared that whole expression is a constant now the height of the water is changing because as you add water the water level will rise so h is a variable now this is important because when you&amp;#39;re dealing with related rates you&amp;#39;re not differentiated with respect to x as you would using implicit differentiation you&amp;#39;re differentiating with respect to time and the only thing you need to worry about are the variables not the constants because if r was changing when we differentiate it we would have to use the product rule but because this is a constant we do not have to use the product rule it&amp;#39;s the same as differentiating 5x you wouldn&amp;#39;t use the product rule for that or 8x cubed you don&amp;#39;t use the product rule for that but if you have like x cubed times y squared where both of these are variables then you need to use the product rule but let&amp;#39;s say if x was constant and y was a variable then you do not have to use the product rule you only need to use the product rule if you have two different variables being multiplied or if you have two variable expressions of the same type so e to the 4x that&amp;#39;s one variable expression but using x and ln x is also using x but that&amp;#39;s also a variable expression so you would still use the product rule in that case but if you have a constant times a variable you do not need to use the product rule so keep that in mind so the derivative of v with respect to time is dv dt now what is the derivative of pi r squared times h with respect to time if we were to differentiate 5h it would simply be 5 times dh dt the derivative of pi r squared times h will just be pi r squared and the derivative of h is one and then times dht so this is the constant like five so we just need to rewrite it now we have everything that we need to get the answer our goal is to calculate dv dt r doesn&amp;#39;t change r is going to be 3 feet and dhdt that&amp;#39;s changing at 3 feet per minute so 3 squared times 3 that&amp;#39;s 27 thus the answer is going to be 27 pi now what are the units of dv dt because you also need to get that right the unit for h height would be like feet meters things like that now for volume it&amp;#39;s like cubic feet cubic meters cubic inches so in this case it&amp;#39;s cubic feet since h is in feet now the time dt is in minutes so this is going to be the same that&amp;#39;s going to be in minutes as well thus it&amp;#39;s 27 pi cubic feet per minute number 10 identify all intervals where the function f of x is increasing given f of x is equal to x cubed plus three over two x squared minus 36 x minus nine so in order to find the intervals where the function is increasing we need to identify the critical points and one way we can do that is by finding the first derivative and setting it equal to zero and solving for x so let&amp;#39;s do that the derivative of x cubed is three 3x squared the derivative of x squared is 2x the derivative of x is 1 and the derivative of a constant is 0. so we have three x squared and then the twos will cancel so this is going to be plus three x minus thirty six so let&amp;#39;s set this equal to zero now we can take out the gcf the greatest common factor so if we take out a three we&amp;#39;re going to have x squared plus x minus 12. now let&amp;#39;s factor the trinomial what two numbers multiply to negative 12 but add to one so this is going to be positive four and negative three four plus negative three is positive one and four times negative three is negative twelve now what we need to do at this point is set each factor equal to zero if you set x plus four equal to zero the critical point that you&amp;#39;ll get will be negative 4 and if you set x minus 3 equal to 0 you&amp;#39;ll get the critical point positive 3. so now what we need to do is make a number line by putting these critical points on it next we need to check the signs so let&amp;#39;s pick a number greater than three let&amp;#39;s say five if we plug in five into this expression into the first derivative will we get a positive number or a negative number now 5 minus 3 is a positive number 5 plus 4 is a positive number and this is positive so the result will be positive now what if we pick a number between negative four and three like zero will it be positive or negative zero minus three is negative zero plus four is positive a positive number times a negative number will give us a negative result now let&amp;#39;s pick something less than negative four like negative five negative five plus four is negative negative five minus three is a negative number a negative times a negative number will give us a positive number all the way to the left we&amp;#39;ll have negative infinity and to the right positive infinity now let&amp;#39;s get rid of this one thing i do want to mention is that whenever the first derivative is equal to a positive number the function f of x is increasing whenever the first derivative is equal to a negative number so you have a negative slope then the function is decreasing so these are the signs for the first derivative so the function is increasing between negative infinity and negative four and it&amp;#39;s also increasing between three and infinity it&amp;#39;s decreasing between negative four and three so the answer we could say the function is increasing between negative four i mean between negative infinity to negative four union and then between three and infinity so this is the answer to the problem now something i do want to mention is that you could use this to determine the location of any relative maximum or relative minimum some textbooks might use the term local maximum or local minimum so here the function is increasing and then it&amp;#39;s decreasing so this looks like a local maximum here it&amp;#39;s decreasing and then increasing so this looks like a local minimum so that&amp;#39;s how you can identify the relative extrema on a function using the same process number 11 identify the location and maximum value of the function f of x is equal to 16 x minus x squared plus five now the leading term is negative x squared it has the highest degree and the general shape of negative x squared is basically a downward parabola so for this particular graph there&amp;#39;s only one location where it&amp;#39;s a maximum the location of the maximum is the x value and the actual maximum value is the y coordinate so basically we need to find both the x y coordinate of our answer so how can we do it how can we use calculus to locate the maximum value of the function we need to find the critical point because at this region where we have a horizontal tangent line the slope is zero and so this is our critical point let&amp;#39;s go ahead and find it so f prime of x is going to be the derivative of 16x is just 16. the derivative of x squared is 2x and the derivative of 5 is 0. so if we set this equal to 0 all we need to do is factor out a two 16 divided by two is eight negative two x divided by two is just negative x and if we set eight minus x equal to zero if we add x to both sides x is eight so the critical point is eight so automatically that tells us that answer choice b is the right answer but let&amp;#39;s finish the problem so if you were to plot this on a number line and let&amp;#39;s say if you picked a number that&amp;#39;s greater than eight like nine eight minus nine is negative if we plug in something less than eight like seven eight minus seven is positive so whenever the signs alternate from positive to negative going left to right you have a maximum it&amp;#39;s increasing and then it&amp;#39;s decreasing so we have a local max at eight now all we need to do to find the y coordinate is plug in the x value so f of 8 is going to be 16 times 8 minus 8 squared plus 5. now 16 times eight ten times eight is eighty six times eight is forty eight so eighty plus forty eight that&amp;#39;s one twenty eight eight squared is sixty four one twenty eight minus sixty four is sixty four and 64 plus 5 is 69. now in case this problem was a free response problem and not a multiple choice you need to know that this is the maximum value of the function the y-coordinate now the location of the maximum value is located at the x-coordinate or at x equals eight so they ask you for the location of the critical point or the maximum value this is your answer if they ask you for the maximum value itself this is your answer so the function has a maximum value of 69. so b is the right answer for this problem now let&amp;#39;s move on to 12. calculate the average value of the function f of x is equal to x cubed plus 8x minus 4 over the interval 1 to 5. so here&amp;#39;s the basic idea behind calculating the average value so let&amp;#39;s say we have some generic function f of x so basically what we&amp;#39;re doing is we&amp;#39;re taking the area under the curve and we&amp;#39;re going to divide it by basically the width of the interval which is an x value and that&amp;#39;s going to give us an average y value so our goal is to calculate this average y value of the function so here&amp;#39;s the formula that we need the average function value or the average y value is going to be 1 divided by the width of the interval times the area under the curve in that interval which is the definite integral from a to b of f of x to dx so this right here represents the area under the curve and this portion represents the width of the interval which as we said is an x value and that will give us the average function value so what is a and what is b a as we can see is one b is five so this is going to be one over five minus one times the definite integral from one to five and then the function is x cubed plus eight x minus four dx five minus one is four so this becomes one over four now we need to find the anti-derivative of that expression the anti-derivative of x cubed is x to the fourth divided by four and for eight x is going to be eight x squared divided by two which is four x squared and the antiderivative of negative four is uh going to be negative 4x and now we need to evaluate this from 1 to 5. now before we do so it might be advantageous to distribute the 1 4 to everything inside the brackets so it&amp;#39;s going to be 1 over 16 times x to the fourth and then these fours will cancel so that&amp;#39;s going to be plus x squared and these fours will cancel as well so minus x evaluated from one to five so first let&amp;#39;s plug in five so it&amp;#39;s going to be one over 16 times 5 to the fourth power and then plus 5 squared minus 5 and then we could plug in 1. so it&amp;#39;s going to be 1 over 16 plus 1 minus 1. now five to the fourth power that&amp;#39;s 625 and then that&amp;#39;s divided by 16. 5 squared is 25 and then we have negative 1 over 16 and 1 minus 1 is 0. now 625 over 16 minus 1 over 16 that becomes 6 24 over 16. and 25 minus 5 is 20. now 624 divided by 16 that&amp;#39;s 39. so it&amp;#39;s 39 plus 20 which is 59. so this is the average y value of the function over the interval one to five therefore c is the right answer number thirteen evaluate the expression shown below so what is the derivative of two x cubed minus seven x squared raised to the eighth power for this problem we need to use something called the chain rule perhaps you&amp;#39;ve seen it expressed this way the derivative of a composite function let&amp;#39;s say f of g of x is going to be f prime of g of x times g prime of x so notice that we differentiate the outside portion of the function keeping the inside the same and then we&amp;#39;re going to multiply by the derivative of the inside of the function we&amp;#39;re going to follow that same process to differentiate this expression so think of the outside portion of the function as x to the eighth that&amp;#39;s like f of x and think of the inside function g of x as being two x cubed minus seven x squared so let&amp;#39;s differentiate the outside portion the derivative of x to the eight would be eight x to the seventh power so we&amp;#39;re gonna do is we&amp;#39;re gonna move the eight to the front now instead of writing x we&amp;#39;re going to keep everything that we have on the inside exactly the same and then we&amp;#39;re going to subtract this by one so that&amp;#39;s going to be seven next we&amp;#39;re going to take the derivative of the inside function to get our g prime so the derivative of two x cubed is six x squared and the derivative of 7x squared is 14x and since this is a free response question we can leave our answer like this now for those of you who may want to simplify the answer we could take out the gcf in this expression so we could take out a 2x so multiplying 8 by 2x that&amp;#39;s going to give us 16x and then we have 2x cubed minus 7x squared raised to the seventh power and now that we&amp;#39;ve taken out 2x we need to divide everything there by 2x 6x squared divided by 2x is 3x negative 14x divided by 2x is negative 7. so you could leave your answer like this if you want to so that&amp;#39;s how you could use the chain rule now for those of you who want more examples on the chain rule particularly harder examples i already have a video on that on youtube just type in chain rule organic chemistry tutor and you can get more examples on this if you need more help on this topic so let&amp;#39;s move on to the next problem number 14 evaluate the limit expression shown below now let&amp;#39;s say if you get a limit problem like this and you have no idea what to do it turns out that you can evaluate any limit if you have access to a calculator if that is allowed on your test if it&amp;#39;s a free response problem and you can&amp;#39;t use the calculator you just have to know the techniques involved but let&amp;#39;s say if you do have access to a calculator what you could do is plug in a number that&amp;#39;s close to four let&amp;#39;s say 4.1 so type this in your calculator and make sure to use parentheses so you should get this f of 4.1 is negative point zero six zero nine seven five now do the same thing for a number that&amp;#39;s closer to four so if we try four point zero one if you type this in you should get negative point zero six two three four notice that it&amp;#39;s converging to a value it&amp;#39;s getting close to negative point zero six something and you can go even further you can try 4.001 i&amp;#39;m going to try that so you should get negative point zero six two 4 8 with some other numbers so we could see that it&amp;#39;s around negative point zero six the answer can&amp;#39;t be a one over four is point two five it can&amp;#39;t be b negative two thirds is negative point six repeating negative eight and twelve are out now e is the answer this is negative point zero six two five if you type a negative 1 over 16 in your calculator and so that&amp;#39;s one simple method that you could use to evaluate any limit if you don&amp;#39;t know what to do if you have access to a calculator but now let&amp;#39;s solve this the appropriate way what should we do if we have fractions within fractions if you have a complex fraction eliminate the smaller fractions by multiplying the top and the bottom by the common denominator of these two small fractions which is going to be 4x now let&amp;#39;s distribute the 4x so 4x times 1 over x the x variables will cancel and so we&amp;#39;re just going to get 4 on top now 4x times 1 4 the fours will cancel and so we&amp;#39;re just going to get negative x on top and on the bottom don&amp;#39;t distribute it just rewrite it as 4x times x minus 4. now what&amp;#39;s our next step here what do you think we need to do at this point what i recommend doing is factoring out a negative one from four minus x if you take out a negative one from negative x negative x changes to positive x and if you take out a negative one from positive four it becomes negative four so notice that we can cancel x minus four so now we have the limit as x approaches four for negative one over four x so now we can use direct substitution so replacing x with four we&amp;#39;re going to get negative one over sixteen so this is the final answer answer choice e number 15 identify all intervals where the function f of x is concave downward now before we attempt this problem let&amp;#39;s write down some notes let&amp;#39;s summarize what we need to know when the first derivative is positive you need to know that the function f of x is increasing and when the first derivative is negative the function f of x is decreasing now when f prime of x is equal to zero you have a horizontal tangent line and you also have a critical point what about the second derivative when the second derivative is positive the function f of x is said to be concave up when the second derivative is negative the function f of x is said to be concave down and when it&amp;#39;s equal to zero f of x has an inflection point if the concavity changes that is if the second derivative changes from negative to positive or positive to negative if it doesn&amp;#39;t change sign it&amp;#39;s not an inflection point so for this particular problem we are concerned with this information so let&amp;#39;s go ahead and find the second derivative and set it equal to zero let&amp;#39;s start with the first derivative f prime of x the derivative of x cubed is 3x squared the derivative of x squared is 2x times negative 6 that&amp;#39;s negative 12x and the derivative of 5x is 5. and the derivative of the constant one is a zero now let&amp;#39;s calculate the second derivative the derivative of three x squared is going to be six x and the derivative of negative twelve x is just negative twelve let&amp;#39;s set the second derivative equal to zero and let&amp;#39;s factor out a six so now if we set x minus two equal to zero we could see that we have a point of interest at x equals two so we&amp;#39;re going to make a number line and we&amp;#39;re going to test the signs so if we pick a number greater than 2 like 3 3 minus 2 is positive if we pick a number less than 2 like 1 1 minus 2 is negative so the second derivative will be negative if you plug in 1 into this expression always put your infinity signs so notice that 2 is an inflection point because the concavity the second derivative changes sign on the left side the second derivative is negative which means that the function is concave down on the right side because the second derivative is positive it&amp;#39;s concave up there so for this particular problem the interval where the function is concave down is from negative infinity to 2. so that&amp;#39;s when the second derivative is negative you

Transcript for:Calculus 1 Final Exam Review

Transcript for:
Calculus 1 Final Exam Review