Now this video is for those of you who are taking the calculus one final exam, that is if you're in college, or if you're in high school, you can also use this if you'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's go ahead and begin. Number one, evaluate the limits shown below.
Now, it's always good to see if you could use direct substitution. However, if we plug in 3, we're going to get 9 squared minus 9, which will give us 0 in the denominator. So we can'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 3 times x minus 3. The square root of x squared is x, and the square root of 9 is 3. And so that's how you can get those numbers. Now what about factoring this trinomial?
What two numbers multiply to negative 15, but add to positive 2? So this is going to be positive 5 and negative 3. 5 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's going to be 3 plus 5 divided by 3 plus 3. 3 plus 5 is 8. 3 plus 3 is 6. And then if you divide both numbers by 2, you could reduce that to 4 over 3. So A is the correct answer. Now let's move on to number two. Evaluate the expression shown below.
So for each of these problems, pause the video and work on it. And then when you'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's going to be that constant times the variable raised to the constant minus 1. So just to review, the derivative of x to the 4th power is going to be, in this case n is 4, 4x to the 4 minus 1, which is 4x cubed.
So now let's try this problem. So what we're going to do is differentiate each one individually. The derivative of x to the 6th power is going to be 6x to the 6th minus 1, which is 5. Now, for 3 over x, we need to rewrite it. If we move the variable to the top, it becomes 3x to the minus 1. And so now we can use the power rule. So the derivative of x to the minus 1 is going to be negative 1x.
to the negative 1 minus 1, which is negative 2. Now, the square root of x, we also need to rewrite it. And we can rewrite it as x to the 1 half. So using the power rule, it's going to be 1 half x, and then it's going to be 1 half minus 1. So far we have 6x to the 5th power, minus 3x to the minus 2, and then 1 half minus 1. Negative 1 is the same as negative 2 over 2, so 1 minus 2 is negative 1. So 1 half minus 1 is negative 1 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's going to be x to the positive 1 half. Now x to the 1 half is the same as the square root of x. So instead of writing x to the 1 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 3. So step 1, set these two parts equal to each other. So we're going to have 2cx minus 6 is equal to x squared plus cx.
Now the next thing we need to do is replace x with 3. So it's going to be 2c 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 6c minus 6. 3 squared is 9, so we have 9 plus 3c on the right side. Now, let's subtract both sides by 3c, and at the same time, let's add 6 to both sides.
So we could cancel this and that. So here we have 6c minus 3c, which is 3c. And then 9 plus 6, that's 15. Dividing both sides by 3. I don't know what just happened there, but let's fix that. We get 15 over 3, which is 5. And so that's the answer. c is equal to 5. So that'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'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'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's say if I wanted to differentiate e to the x squared. It's going to be e to the x squared times the derivative of x squared, which is 2x.
Or, let's say if I wanted to differentiate e to the 7x. It's going to be e to the 7x times the derivative of 7x, which is 7. Now let'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'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's going to be x on the bottom, so we get 1 over x. Now let's say 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's going to be 2x over x squared plus 7. Now let's talk about the product rule. The derivative of, let's say we have two things multiplied to each other, f times g, it's going to be the derivative of the first part, so that's f prime, times the second part, which is g, plus the first part times the derivative of the second. So let's say that f is e to the 4x.
So what's f prime going to be? Using the equation that we had before, it's e to the u times u prime. So it'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's ln 2x plus 5. Now g prime, using the formula u prime over u, it'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'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 4e to the 4x and then times g so that's ln 2x plus 5 and then plus F, which is e to the 4x. times g prime, which is that, 2 over 2x plus 5. Now, we can basically combine those two. So, I'm going to rewrite this as 2e to the 4x over 2x plus 5. So, we can leave our final answer like this. And so, that'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'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'm also going to place a link to my calculus playlist. Let's say if you need help with a specific topic, and you don'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's start with this one.
So how can we evaluate the following integral? What is the antiderivative of 4x to the 5th plus x to the 4th minus 3x squared divided by x squared? What we need to do is simplify the expression first.
So let's begin by dividing each term in the numerator of the fraction by x squared. So x to the 5th divided by x squared is going to be x cubed times 4. Because 5 minus 2 is 3. Now x to the 4th divided by x squared is x squared. 4 minus 2 is 2. And then negative 3x squared divided by x squared. The x squared stuff will cancel. And so it's going to be negative 3. 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 1. The antiderivative of x to the n is going to be, so instead of subtracting 1, we're going to add 1 to the constant, and then we're going to divide by that result. And don't forget to add the constant of integration. So this is the power rule when dealing with integration.
Add 1 and then divide by that result. So what is the antiderivative of x cubed? The constant in the front, just rewrite it.
So what we're going to do to find the antiderivative of x cubed is add 1 to the exponent. 3 plus 1 is 4. And then divide by that number. Here we're going to do the same. 2 plus 1 is 3, and then divide by 3. Now the antiderivative of negative 3 is simply negative 3x. Negative 3 is the same as just negative 3 times x to the 0, because anything raised to the 0 power is 1. And so you can think of it this way.
If we add 1 to 0, and then divide by the result, we're going to get negative 3x. So that's the antiderivative 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 4, and so we're going to have x to the 4th plus, we can write this as 1 3rd. x cubed and then minus 3x plus c.
So this is the final answer for the problem. So that's how we can evaluate this particular integral. Number 6. Find the equation of the tangent line to the curve x cubed plus 4xy squared plus y cubed equals 107 at the point 2,3 using implicit differentiation.
So now let's talk about this. Let'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'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's going to equal dy dx at the point 2,3. So we need to use implicit differentiation to find dy dx.
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's go ahead and begin.
Let'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, 4xy squared, we need to use the product rule. So let'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's going to be y squared. And then plus. Now we'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's 2y times dy dx. So remember, when dealing with implicit differentiation, every time you differentiate a y variable with respect to x, you need to add dy dx to it. The derivative of y cubed is 3y squared times dy dx. And the derivative of a constant, 107, is 3y squared times dy dx.
is 0. Now at this point we do not need to isolate dy dx. In fact to make it easier I recommend replacing x and y with their respective values. So let's replace x with 2 and let's replace y with 3. And then let's calculate the value of dy dx.
2 squared is 4 times 3, that's 12. 3 squared is 9 times 4 is 36. 4 times 2 is 8. 2 times 3 is 6. 3 squared is 9 times 3, that's 27. So right now, we can add these two. 12 plus 36 is 48. And if we move that to the other side of the equation, it's going to be negative 48 on this side. Now, 8 times 6 is 48. And then we have plus 27 dy dx. 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 dy dx, and that's equal to negative 48. So now we can divide both sides by 75. So dy dx 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're going to use the point-slope formula.
y minus y1 is equal to m times x minus x1. 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's divide the top and the bottom by 3. So this is negative 16 over 25. So let's replace m with that. Now x1 is 2. 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'll be done 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't have that anywhere in any of the answer choices.
So let's multiply both sides by 25. So on the right side, these two will cancel. On the left side, we're going to have 25y minus 75. And then it's going to be negative 16x. And then negative 16 times negative 2, that's positive 32. Now let's rearrange what we have.
I'm going to move the negative 16x from the right side to the left side. So on the left side, it's going to be positive 16x, and then we're going to have plus 25y. Now the negative 75, I'm going to move it to this side.
On the right side, it'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's really no way in which we can work out this limit. We have no numbers to plug in, there's nothing to factor, we can'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 0 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's what you're looking for. So what is f prime of x?
Because that'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'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's work on number 8. Evaluate the integral shown below. So how can we do this? We can't use the power rule for this example, but we could use something known as u-substitution.
What we're going to do is make u equal to 3x squared plus 5. And so du 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're going to replace 3x squared plus 5 with the u variable. And at the same time, we're going to replace the dx with what we have here. So that's du over 6x. So notice that all of the x variables have been cancelled. 2 over 6, we can reduce that to 1 over 3. And I'm going to move the constant in the front.
The square root of u is the same as u to the 1 half. So now we can find the antiderivative of u to the 1 half. So we need to add 1 to it. 1 half plus 1 is 3 over 2. And then instead of dividing by 3 over 2, we can multiply by the reciprocal.
2 over 3. And don't forget to put plus C. So multiplying these two fractions, it's going to give us 2 over 9, and then we'll have u to the 3 halves plus c. Now our next step is to replace u with what we have here, 3x squared plus 5. So the final answer is 2 over 9, 3x squared plus 5, raised to the 3 over 2, plus c. So that's how you could use u substitution. to find the antiderivative of certain functions.
Number nine, water is flowing into a cylinder with a diameter of six feet and a height of 10 feet. If the height of the water in the cylinder is increasing at three 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 going to the search box and type in related rates problem organic chemistry tutor. If you type in that phrase, you'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 rates problems, what you need to do is you need to write a formula. So what we have is a cylinder.
So let's draw a picture. So we know the diameter of the cylinder, so that'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's 3 feet per minute. And it's positive because it's increasing. Now the next thing we need to do is write the formula for this situation.
So we'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... So... which letters are constants and which ones are variables. So V, the volume, would you say it's a variable or a constant? In this case, V is a variable because the volume of the water in the cylinder is changing.
It's not constant. Pi is a constant. It's always 3.14. Now what about the radius of the water in the cylinder?
Is that constant? Imagine this 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 3 feet, the radius of the water in the cylinder will always be 3 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're dealing with related rates, you're not differentiating with respect to X as you would use an implicit differentiation. You're differentiating with respect to time. And the only thing you need to worry about are the variables and 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's the same as differentiating 5x.
You wouldn't use the product rule for that. Or 8x cubed. You don'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's say if x was a 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's one variable expression, but using x.
And ln x is also using x, but that'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 thus be pi r squared, and the derivative of h is 1, and then times dh dt.
So this is a constant like 5, 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't change.
r is going to be 3 feet. And dh dt, that's changing at 3 feet per minute. So 3 squared times 3, that'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's like cubic feet, cubic meters, cubic inches.
So in this case, it's cubic feet since h is in feet. Now the time dt is in minutes. So this is going to be the same. That's going to be in minutes as well.
Thus, it's 27 pi cubic feet per minute. Number 10. Identify all intervals where the function f is increasing given f is equal to x cubed plus 3 over 2 x squared minus 36x minus 9. 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 0 and solving for x.
So let's do that. The derivative of x cubed is 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 3x squared and then The 2's will cancel, so this is going to be plus 3x minus 36. So let's set this equal to 0. Now, we can take out the GCF, the greatest common factor. So if we take out a 3, we're going to have x squared plus x minus 12. Now, let's factor the trinomial. What two numbers multiply to negative 12 but add to 1?
So this is going to be positive 4 and negative 3. 4 plus negative 3 is positive 1, and 4 times negative 3 is negative 12. Now what we need to do at this point is set each factor equal to 0. If you set x plus 4 equal to 0, the critical point that you'll get will be negative 4. And if you set x minus 3 equal to 0, you'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's pick a number greater than 3. Let's say 5. If we plug in 5 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 4 and 3, like 0? Will it be positive or negative?
0 minus 3 is negative. 0 plus 4 is positive. A positive number times a negative number will give us a negative result.
Now let's pick something less than negative 4, like negative 5. Negative 5 plus 4 is negative. Negative 5 minus 3 is a negative number. A negative times a negative number will give us a positive number. All the way to the left, we'll have negative infinity, and to the right, positive infinity. Now, let'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 4. and it's also increasing between 3 and infinity.
It's decreasing between negative 4 and 3. So the answer, we could say the function is increasing between negative 4, I mean between negative infinity to negative 4, union, and then between 3 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's decreasing. So this looks like a local maximum. Here it's decreasing, and then increasing.
So this looks like a local minimum. So that'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 16x minus x squared plus 5. 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's only one location where it'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 0. And so this is our critical point. Let'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 2. 16 divided by 2 is 8. Negative 2x divided by 2 is just negative x. And if we set 8 minus x equal to 0, if we add x to both sides, x is 8. So the critical point is 8. So automatically, that tells us that answer choice B is the right answer. But let's finish the problem.
So if you were to plot this on a number line, and let's say if you picked a number that's greater than 8, like 9. 8 minus 9 is negative. If we plug in something less than 8, like 7, 8 minus 7 is positive. So whenever the signs alternate from positive to negative, going left to right, you have a maximum.
It's increasing, and then it's decreasing. So we have a local max at 8. 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 8. 10 times 8 is 80. 6 times 8 is 48. So 80 plus 48, that's 128. 8 squared is 64. 128 minus 64 is 64. 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 It's located at the x-coordinate, or at x equals 8. So if 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'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's the basic idea behind calculating the average value. So let's say we have some generic function f of x.
So basically what we're doing is we're taking the area under the curve and we're going to divide it by basically the width of the interval. which is an x value, and that's going to give us an average y value. So our goal is to calculate this average y value of the function. So here'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 1, b is 5. So this is going to be 1 over 5 minus 1 times the definite integral from 1 to 5 and then the function is x cubed plus 8x minus 4 dx.
5 minus 1 is 4 so this becomes 1 over 4. Now we need to find the antiderivative of that expression. The antiderivative of x cubed is x to the fourth divided by four. And for 8x, it's going to be 8x squared divided by two, which is 4x squared. And the antiderivative of negative four is going to be negative 4x. Now we need to evaluate this from one to five.
Now before we do so, it might be advantageous to distribute the 1 fourth to everything inside the brackets. So it's going to be 1 over 16 times x to the fourth, and then these fours will cancel. So that's going to be plus x squared, and these fours will cancel as well, so minus x, evaluated from 1 to 5. So first, let's plug in 5. So it's going to be 1 over 16 times 5 to the 4th power, and then plus 5 squared minus 5, and then we can plug in 1. So it's going to be 1 over 16 plus 1 minus 1. Now, 5 to the 4th power, that's 625. and then that'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 624 over 16, and 25 minus 5 is 20. Now 624 divided by 16, that's 39. So it's 39 plus 20, which is 59. So this is the average y value of the function over the interval 1 to 5. Therefore, c is the right answer. Number 13, evaluate the expression shown below.
So what is the derivative of 2x cubed minus 7x squared raised to the eighth power? For this problem, we need to use something called a chain rule. Perhaps you've seen it expressed this way. The derivative of a composite function let'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're going to multiply by the derivative of the inside of the function.
We're going to follow that same process to differentiate this expression. So think of the outside portion of the function as x to the 8. That's like f of x. And think of the inside function g of x.
as being 2x cubed minus 7x squared. So let's differentiate the outside portion. The derivative of x to the 8 would be 8x to the 7th power. So what we're going to do is we're going to move the 8 to the front.
Now instead of writing x, we're going to keep everything that we have on the inside exactly the same. And then we're going to subtract this by 1. So that's going to be 7. Next, we're going to take the derivative of the inside function to get our g prime. So the derivative of 2x cubed is 6x 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's going to give us 16x.
And then we have 2x cubed minus 7x squared raised to the 7th power. And now that we'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'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. you need more help on this topic. So let's move on to the next problem.
Number 14, evaluate the limit expression shown below. Now, let'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's a free response problem and you can't use a calculator, you just have to know the techniques involved. But let's say if you do have access to a calculator. What you could do is plug in a number that's close to 4. Let'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.060975. Now do the same thing for a number that's closer to 4. So if we try 4.01, if you type this in, you should get negative.06234 notice that it's converging to a value it's getting close to negative.06 something and you can go even further you can try 4.001 I'm going to try that So you should get negative 0.06248 with some other numbers. So we can see that it's around negative 0.06.
The answer can't be A. 1 over 4 is 0.25. It can't be B.
Negative 2 thirds is negative 0.6 repeated. Negative 8 and 12 are out. Now, E is the answer.
This is negative 0.0625, if you type in negative 1 over 16 in your calculator. And so that's one simple method that you could use to evaluate any limit if you don't know what to do, if you have access to a calculator. But now, let'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's distribute the 4x. So 4x times 1 over x, the x variables will cancel. And so we're just going to get 4 on top.
Now, 4x times 1 fourth, the 4's will cancel. And so we're just going to get negative x on top. And on the bottom, don't distribute it.
Just rewrite it as 4x times x minus 4. Now, what'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 1. from 4 minus x.
If you take out a negative 1 from negative x, negative x changes to positive x. And if you take out a negative 1 from positive 4, it becomes negative 4. So notice that we can cancel x minus 4. So now we have the limit as x approaches 4 for negative 1 over 4x. So now we can use direct substitution.
So replace an x with 4. we're going to get negative 1 over 16. 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's write down some notes.
Let'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 0, 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 is said to be concave up.
When the second derivative is negative, the function f is said to be concave down. And when it's equal to 0, f 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't change sine, it's not an inflection point.
So for this particular problem, we are concerned with this information. So let's go ahead and find the second derivative and set it equal to zero. Let'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's negative 12x. And the derivative of 5x is 5. And the derivative of the constant 1 is 0. Now, let's calculate the second derivative.
The derivative of 3x squared is going to be 6x. And the derivative of negative 12x is just negative 12. Let's set the second derivative equal to 0, and let's factor out a 6. So now, if we set x minus 2 equal to 0, we can see that we have a point of interest at x equals 2. So we're going to make a number line, and we'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'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's when the second derivative is negative.