Section four point three calculus one how derivatives affect the shape of a graph so first fact if f prime of X is greater than zero on an interval then that means that f of X our original function is increasing on that interval and here we have a graph of some arbitrary function and notice wherever the derivative is positive meaning graphically the slopes of our tangent lines would be positive right all of the tangent lines here as well as over here they're ill astray today highlighted line anywhere that the slope of the tangent line is positive we can tell that our function is increasing from left to right similarly if f prime of X is less than zero on an interval then that means our function f of X is decreasing on that interval and you can see here Illustrated below that these tangent line strong to the curve would have negative slopes meaning f prime of X at these tangent lines would be negative and on this interval that's highlighted here the function is decreasing so this is a fact that we're going to use now in order to identify our local extrema or local mins and maxes for our functions so here is the first derivative test if F prime of X changes from positive to negative at C then f of X has a local maximum at C and notice this is illustrated below F prime of X changes from positive to negative so if you have a positive derivative that means the function is increasing right the slope of the tangent would be positive and then it switches to negative well exactly where that switch happened is where we have a local Max and similarly if F prime of X changes from negative to positive at C then f of X has a local minimum at C and here the local minimum is illustrated below so the derivative went from being negative slopes of tangent lines were negative to positive so the function went from decreasing to increasing and where it's switched from decreasing to increasing we have a local min now notice both of these the local Max and the local min are critical numbers since the derivative F prime of X is equal to 0 in this case or it could also not exist all right save this first derivative test and we're going to apply it for some of the examples later in this section now moving on to concavity if the graph of f of X lies above all its tangent lines on an interval I then f of X is concave up on the interval and what this means when we're speaking in terms of derivatives is that the second derivative for all the X values on that interval would be positive well what is something that concave up look like if you notice graph down here below this portion of our function is concave up it resembles a parabola opening upwards if you want to think of it that way and notice here the graph of the function lies above all of its tangent lines so if we draw tangent lines to this function right everywhere I draw a tangent line those lines are below the graph so that's how we base our definition for concave up that the function lies above all of its tangent lines similarly if the graph of f of X lies below all of its tangent lines on an interval then f of X is concave down on that interval and notice here this portion of our function in blue is concave down see how the function lies below all of the purple lines all of the tangent lines and anymore that you would draw and what this means when we're looking at its derivative is that the second derivative F double prime of X is less than zero now where the graph switches from being concave up to concave down or vice versa is called an inflection point okay so a point P on f of X is called an inflection point in f of X if f of X is continuous at p and f of X changes in concavity okay so there has to be a change in concavity in order for a point to be considered an inflection point and just a little something to help you remember which way it looks like concave up like a cup and concave down like a frown alright last little tidbit is the second derivative test the second derivative test is applying the second derivative to identify where we have local mins and maxes on our graph so s of X has a local minimum at C if F prime of C equals zero and the second derivative is positive now remember if the second derivative is positive that means the graph is concave up so it would look something like this and also we know that F prime of C is equal to zero so f prime of C would equal zero here say that's C here's C F prime of C equals zero and clearly this is a local min and we can also use the second derivative test to identify local maxes so f of X has a local max at C if F prime of C equals zero so we have a horizontal tangent and F double prime of C is less than zero so that means it's concave down like a frown and here would be C where F prime of C is zero there C and clearly this is a local Max typically for this section the first derivative test is usually more useful to identify local mins and maxes but when we get to later section on optimization the second derivative test can be quite nice if F double prime of C is zero the second derivative test gives you no information so you have to actually go I can use the first derivative test to determine whether you have a local min or max or nothing at all all right so let's look at an example find the intervals of increasing decreasing find the local min and maxes find the intervals of concavity and find the inflection points before we proceed it's really important to identify the domain of your function because you can't have the min or a Max or an inflection point or anything like that at values that are outside of the domain so if we notice here f of X is a rational function but there are no restrictions because the denominator is never going to be 0 so the domain for f of X is all real numbers that'll make things a little bit easier ok so to start off we're gonna find the intervals of increase and decrease in order to do that let's first find f prime of X so f prime of X is equal to I'm gonna need to use the quotient rule so I have low D high x squared plus 3 times derivative of the numerator 2x minus high D low x squared times derivative of the denominator which is 2x over the denominator squared x squared plus 3 quantity squared alright now from here let's distribute and clean up the numerator simplifying and combining like terms so this is gonna give us 2x cubed plus 6x minus 2x cubed over x squared plus 3 quantity squared and then I'm just gonna be left with 6x over x squared plus 3 squared so here's my first derivative now I want to identify all the critical values so let's find the critical values they come from two places remember where F prime of X is equal to zero and where F prime of X does not exist always list out and explore both cases so if F prime of X were to equal zero remember all you need to do is focus on when the numerator is 0 because the de nom being zero would just make the expression undefined so that would mean 6x would equal zero so x equals zero is a critical value F prime of X does not exist would mean that x squared plus three squared is zero so x squared plus three would be zero but that's not possible that would mean x squared has to be negative three so there's none from this case no real numbers don't worry we're only dealing with real numbers in our calculus career for right now okay so what we want to do is test now and see if x equals zero this critical value is perhaps a local min or max and the way you want to do that is by making a number line so you make a number line and when we're applying the first derivative test this is a number line testing the sine of f prime of x okay you're going to list your critical value or your critical number on the number line and then we're gonna substitute values on either side on the two intervals into the derivative into whatever we labeled our number line in this case f prime of x to see whether it's positive or negative so if I plug in something smaller than zero say x equals negative one into the derivative so if I plug in x equals negative one into the derivative remember here's my derivative well the denominator is always positive and if I plug in negative one into the numerator six times negative one that's gonna be negative six so the derivative is gonna be negative on the interval from negative infinity to zero well what does that mean for f of X that means f of X is decreasing very good now let's plug in something on the other side of zero anything from 0 to infinity would do how about x equals 1 so going back to the derivative here if I plug in 1 into the numerator that's gonna be positive 6 the denominator is always positive so this is going to give me a positive number which means my function is increasing so this describes the behavior of f of X well look at the graphs look at what's happening if my function f of X goes from decreasing to increasing what does that mean happened at zero well we have a local min so we have a local min f of zero and let's go back to our original function and see what would F of 0 be so I'm gonna substitute in 0 for X here so I'm just gonna have 0 over 3 which is 0 do I have a local max no in this case I do not so state it that way it shows that you considered the case not that you were careless all right and then we're gonna list the intervals of increase and decrease so the function is increasing on the interval from 0 to infinity you never include the critical values or the mins and maxes when you're listing intervals of increase and decrease those are turning points right technically the function is not increasing or decreasing there and then the function is decreasing from negative infinity to 0 all right so that's the answer for Part A and B increasing decreasing and mins and maxes okay so that's parts and E now Part C and D asks for the intervals of concavity and the inflection points in order to find that I need my second derivative so I'll just write out again that F prime of X was equal to 6x over x squared plus 3 squared so now let's go ahead and take another derivative so f double prime of X is gonna be the denominator x squared plus 3 squared times derivative of the numerator which is 6 minus the numerator so 6 x times the derivative of the denominator so I'm gonna need to use the chain rule I'm gonna times two times x squared plus three which is now to the first and then times the derivative of x squared plus three which is 2x over in the denominator I'm gonna have x squared plus three squared squared so x squared plus three to the fourth all right now here's where a lot of students struggle to simplify their second derivative before you start getting all gung-ho and multiplying everything out you want to factor out the GCF the greatest common factor to save yourself lots of heartache so what do these two terms in the numerator have in common well I can take out an x squared plus 3 from both right and I can also take out a 6 that's it but that will do me a lot of help so I'm going to take out a six and an x squared plus three and then let's see what's left over from each of the terms so from this first term here if I take out a six and an x squared plus three and left with just x squared plus three to the first so I don't need any parentheses on it - and then now I've taken out from the second term the six and the x squared plus three so what's left over well I have two X and another 2x so that's the quantity two x squared or 4x squared over the denominator which is still x squared plus three to the fourth all right now let's start simplifying and notice I don't have to do a bunch of foiling and multiplying and combining like terms I'm already in a pretty good place so first thing I noticed is x squared plus three and the numerator cancels with one of the x squared plus berries in the denominator so now it's only going to be cubed and then I still have here six times the quantity 3 minus 3x squared over x squared plus 3 cubed and remember we're going to find the zeros of the second derivative in the numerator and the denominator you want it as factored and simple as possible when you're in order to know how far to simplify your derivative so that you can find those zeros easily meaning I want to take out this 3 right and now I have 6 times 3 so 18 times 1 minus x squared over x squared plus 3 cubed okay lovely so here's my second derivative and then now this is going to help me identify the intervals of concavity and potential inflection points okay so for potential inflection points we find them similar to the process that we use when we find critical values so they come from where f double prime of x equals zero or where f double prime of x does not exist so if F double prime of x equals zero that means the numerator equals zero 18 is never going to be zero so it means 1 minus x squared would be zero which means X would equal plus or minus 1 if F double Prime doesn't exist that would mean the denominator is zero but remember the denominator is x squared plus 3 cubed so x squared plus 3 would equal zero and we already said that this doesn't have any real solutions so my only possible or potential inflection points are at positive and negative one all right we're gonna proceed similar to how we conducted the first derivative test so I'm gonna make a number line list my potential inflection points which are negative 1 and positive 1 and then I'm gonna label the number line F double prime of X so that's where I'm gonna be substituting in the values when I'm testing my intervals into the second derivative always make sure you label your number line so you don't get confused where you're substituting things all right so let's go ahead pick a value in between negative infinity and negative 1 how about negative 2 so I know again the denominator is always gonna be positive because when I square a number it's positive if I add three it's still positive and then if I cube it things aren't changing so really I only need to pay attention to the numerator if I plug in something like negative two and square it it's going to be larger than one so the numerator is going to be negative so if I substitute in x equals negative two the numerator is negative what does this mean well it means f of X is concave down since the second derivative is negative so this is what f of X is doing all right between negative 1 and 1 what's a nice number to plug in I vote for zero again I'm substituting this into the second derivative so 18 times 1 minus zero that's gonna be positive which means f of X is concave up lastly the interval from 1 to infinity we can try something like positive 2 and 1 minus 4 that's going to be negative so f of X is concave down on that interval alright so we can list our intervals of concavity so the function is concave down from negative infinity to negative 1 Union 1 to infinity they have to be in order from left to right and then the function is concave up from negative 1 to 1 and what does this mean that I have two inflection points right there was a change in concavity so I have inflection points of negative 1 and a positive 1 and we want to also give what the function value is at those points case you forgot I'll remind you f of X is x squared over x squared plus 3 so if I plug in 1 or substitute 1 into the function I'll have 1 over 1 plus 3 so that's 1/4 and the same thing for negative 1 so these are both 1/4 all right very nice that concludes that example oh that's not quite the shape I was going hold on yeah let's do one more with perhaps a more complicated derivative and some domain restrictions just to spice things up and show you what's out there okay so we'll go through the same process we're gonna find the intervals of increase decrease local mins and maxes intervals of concavity and inflection points so next example bonus in this case our function f of X is x times the square root of x plus 1 now be careful here I have a domain restriction remember for radical functions the argument cannot be negative which means X plus 1 has to be greater than or equal to 0 so X has to be greater than or equal to negative 1 so my domain is the interval from negative 1 to infinity always always write that out first because it will affect your answers all right great now let's go ahead and find the derivative so f prime of X is gonna equal now I need to use the product rule here so I'm gonna leave X alone and then take the derivative of the square root of x plus 1 so that's gonna be 1/2 X plus 1 to the negative 1/2 plus derivative of X is 1 times X plus 1 to the 1/2 leave it alone ok now I'm gonna rewrite everything using radicals so I'm gonna have x over 2 rad X plus 1 plus and then this is rad X plus 1 now I want to get a common denominator so the common denominator is 2 rad X plus 1 so I need to multiply this second term by 2 rad X plus 1 over 2 rad X plus 1 and then now I'm gonna be left with x over 2 rad x plus 1 plus no notice if I have rad X plus 1 times R add X plus 1 that's just gonna be X plus 1 so I have 2 times X plus 1 over 2 rad X plus 1 and then I can go ahead distribute and combine like terms in the numerator so I have 2 X plus 2 plus another X which is gonna leave me with 3 X plus 2 over 2 rad X plus 1 so there's our first derivative and now we're ready to roll okay let's find our critical values so remember critical values come from two places first place to consider is 1 F prime of X is equal to 0 so that would mean the numerator 3x plus 2 is equal to 0 solving for x I get x equals negative 2/3 and that's what's in my domain remember my domain is from negative 1 to infinity so this we can consider now what about where F prime of X does not exist well F prime of X does not exist if the denominator is negative 1 but that's an endpoint that's not within there's no open interval around negative 1 it's an endpoint on my domain so since this is an endpoint it cannot be a local min or max okay if you need a review on that you can check the video on section for point one that discusses that in more detail alright so actually I only have one critical value to consider so that's not a critical value at negative 1 only negative 2/3 is a critical value now let's make our number line be careful no arrow on the left-hand side because my domain is only from negative one to positive infinity so there's negative 1 this is f prime of X and I'm gonna list my one-and-only critical value right here at negative 2/3 okay so I'm gonna substitute in values into the derivative notice again like last time this is actually pretty convenient the denominator this is always gonna be positive right I have a radical and then I multiply by 2 that's gonna be positive so say I plug in something between negative 1 and negative 2/3 like negative point-eight or something then the numerator is going to be negative which means my function is decreasing and then if I plug in something larger than negative 2/3 like 0 or 1 or anything else then the derivative is gonna be positive which means the function is increasing so we can list our intervals of increase and decrease the function is increasing from negative 2/3 to positive infinity and it's decreasing from be careful negative ones not including negative 1 to negative 2/3 these are always open intervals do I have any local extrema well it looks like here I have a local min right look at what the function is doing its decreasing and then increasing so we have a local min at f of negative 2/3 I'm gonna substitute that into my function remember f of X is X so negative 2/3 times R at X plus 1 which is gonna be rad 1/3 so this is approximately negative point 4 do I have any local maxes in this case no but list it out so that your teacher knows you've thought about it ok good so we took care of increasing decreasing local min or Max and we've used the first derivative to completion now it's time to look the second derivative all right here we go just as a reminder F prime of X I'll rewrite it here was three X plus two over two rad X plus 1 so f double prime of X using the quotient rule I'm gonna have two rad X plus one times the derivative of the numerator which is three minus the numerator 3x plus two times the derivative of the denominator so I'm gonna have two times 1/2 times X plus one to the negative 1/2 all over the denominator squared so that's gonna be 4 times X plus one all right very nice how to begin simplifying well as soon as I notice I have a negative exponent technically I have a complex fraction right because that negative exponent I would rewrite that quantity in the denominator with a positive exponent so what you want to do is just multiply the numerator and denominator by X plus 1 to the 1/2 so I can clear up that complex fraction okay and then now some really nice stuff is gonna happen have no fear so when I distribute X plus 1 to the 1/2 to the first term notice I have another rad X plus 1 and that's pretty cool because rad X plus 1 times another rad X plus 1 is just gonna give me X plus 1 and then what else is going on I have 2 times 3 which is 6 so this is 6 times the quantity X plus 1 don't forget your parentheses so far so good - and then let's see what else is going on in the numerator we actually have a bit of a cancel party so 2 and 1/2 cancel and then now X plus 1 to the negative 1/2 is going to cancel with X plus 1 to the positive 1/2 so all I'm left with it's 3 X plus 2 just don't forget your parentheses so the numerator is looking a lot better already denominator I'm just gonna rewrite it as four times the quantity X plus 1 to the three-halves now and then if i distribute this is 6x plus 6 minus 3x minus 2 so I'm just going to be left with a positive 3x plus 4 over 4 times X plus 1 to the 3 house and there's our second derivative all cleaned up and snazzy there we go all right now let's find our possible inflection points so similar to critical critical values they come from where f double prime of x equals 0 or where f double prime of X does not exist ok so f double prime of X is equal to 0 that would mean the numerator is equal to 0 which means X would equal negative 4/3 but negative 4/3 is smaller than negative 1 and remember my domain is only from negative 1 to infinity so this is not an element of our domain so it's not a possible inflection point it wouldn't even make it on the number line we're not considering it similar to f double prime of X not existing notice that would mean 4 times X plus 1 to the three-halves equals 0 but that would mean x equals negative 1 and that's outside that's the end point so that's not a contender so we're not gonna consider that one either oh no what do we do just proceed as usual it's actually gonna be even easier than last time ok so no possible inflection points that means the concavity is consistent throughout the whole graph so we just got to determine what kind of creme cavity we're talking about so make your number line but just be careful the domains only from negative 1 to infinity so don't put arrows on both sides negative 1 this is my number line for f double Prime and look at my second derivative you actually don't even need to plug anything in the denominator is always gonna be positive right if I raise something to the three-halves power and my domains only from negative one to infinity then that's going to be positive now the numerator keep in mind you're only plugging in X's from negative one to infinity so the smallest the numerators ever going to be is one so it's always positive that means my second derivative is positive so my graph is concave up always but don't write always for your answer right the graph is concave up only where it actually exists meaning on its domain so it's gonna be concave up from negative one but not including negative ones because the second derivative doesn't exist there all the way to infinity it's not concave down anywhere just list it so we know you thought about it and there's no inflection points okay if you're curious I recommend trying to sketch this by in hand putting all the information together we're gonna actually do it later on in the chapter or you can just type it in to some online graphing tool just to get an idea of what the graph looks like I think you might find it interesting so that concludes section 4.3 don't forget to comment like and subscribe