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Current time:0:00Total duration:6:11

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.4 (EK)

, FUN‑4.A.5 (EK)

, FUN‑4.A.6 (EK)

Robert was asked to find where G of X which is equal to the cube root of x has inflection points this is his solution and then later we are asked is robert's work correct if not what's his mistake so pause this video and try to figure it out on your own all right now let's work through this together so our original G of X is equal to the cube root of x which is the same thing as X to the one-third so in step one it looks like Roberts trying to find the first and second derivative so the first trivet if we just do the power rule so we'll be one-third x to the decrement the exponent so this is looking good second derivative we take this multiply this times one third which would be negative two nights and then decrement negative 2/3 which would indeed be negative five-thirds so that looks right and then it looks like Roberts trying to rewrite it so we have the negative 2/9 still but then he recognized that this is the same thing as X to the 5/3 in the denominator and X to the 5/3 is the same thing as X as the cube root of x to the fifth so this is all looking good step one looks good and then step two it looks like he's trying to find the solution or he's trying to find x values where the second derivative is equal to zero and it is indeed true that the second that this has no solution that you can never make this second derivative equal to zero in order to be zero the numerator would have to be zero and well 2 is never going to be equal to zero so this is correct and then step three he says G doesn't have any inflection points now this is a little bit suspect it is in many cases our inflection point is a situation where our second derivative is equal to zero and even then we don't know it's an inflection point it would be a candidate inflection point we would have to confirm that our second derivative crosses signs or switches signs as we cross that x value but here we can't find a situation where a second derivative is equal to zero but we have to remind ourselves that other candidate inflection points are where our second derivative is undefined and so he can't make this statement without seeing where our second derivative could be undefined so for example he could say that G prime prime is undefined undefined when what well this is going to be undefined when X is equal to 0 X 0 to the fifth cube root of that that's going to be 0 but then you're dividing by 0 so G prime prime undefined when X is equal to 0 so therefore x equals so we could say candidate candidate candidate inflection point when x equals 0 and so then we would want to test it and we could set up a traditional table that you might have seen before where we have our interval or intervals we could have test values and our intervals and we have to be careful with those make sure that they are indicative and then we would say the sign of our second derivative of G prime prime and then we would have our concavity concavity of G and in order for x equals 0 to be an inflection point we would have to switch signs as our second derivative would have to switch signs as we cross x equals 0 and R which would mean our concavity of G switches signs as we go as we cross x equals 0 so let's do values less than 0 negative infinity to 0 and then values greater than 0 infinity 0 to infinity I could do test values let's say I'll use negative 1 and 1 and you have to be careful when you use these you have to make sure that we are close enough that nothing nothing unusual happens between these test values up until we get to that candidate inflection point and now what's the sign of our second derivative when X is equal to negative 1 when x equals negative 1 so let's see negative 1 to the fifth power is negative 1 cube root of negative 1 is negative 1 and so we're gonna have negative 2/9 divided by negative 1 it's gonna be positive two nights so our sign right over here is going to be positive and when and this is going to be in general when we're dealing with any negative value because if you take any negative value to the fifth power it's going to be negative and then you take that the cube root of that you're gonna have negative but then you have a negative value divided by that you're gonna get a positive value so you can feel good that this test value is indicative of actually this entire interval and if you're dealing with a positive value well that to the fifth power is going to be positive cube root of that still going to be positive but then you're gonna have negative 2/9 divided by that positive value so this is going to be negative so it is indeed the case that our concavity of G switches as we cross x equals zero or concave upwards when X is less than zero our second derivative is positive and we are concave downwards when X is greater than zero let me write that a little bit downwards down words when X is greater than 0 so we are switching concavity as we cross x equals zero and so this tells us that X so let's see we are switching signs switching let me say G prime prime switching signs as we cross x equals zero and our function is defined at x equals zero and function defined at x equals zero so we have an inflection point at x equals zero so inflection point at x is equal to zero and if you're familiar with the graph of the cube root you would indeed see an inflection point at that point so there we go he was wrong in step three there actually is an inflection point it's not when the second derivative is equal to zero it's actually where the second derivative is undefined

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