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Current time:0:00Total duration:3:47

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.10 (EK)

, FUN‑4.A.11 (EK)

the twice differentiable function G and its second derivative G prime prime our graph and you can see it right over here I'm actually working off of the article on Khan Academy called justifying using second derivatives so we see our function G and we see not its first derivative but its second derivative here in this brown color so then the article goes on to say or the problem goes on to say for students were asked to give an appropriate calculus based justification for the fact that G has an inflection point at x equals negative 2 so let's just feel good that it at least intuitively feels right so x equals negative 2 remember what an inflection point is it's where we're going from concave downwards to concave upwards or concave upwards to concave downwards or another way to think about it it's a situation where our slope goes from decreasing to increasing or from increasing to decreasing and when we look at it over here it looks like look how it looks like our slope is decreasing it's positive but it's decreasing goes to 0 then it keeps decreasing it becomes it's negative now keep decreasing until we get to about x equals negative 2 and then it seems that it's increasing it's getting less and less and less and less negative it looks like it's a zero right over here then it just keeps increasing gets more and more and more positive so it does indeed look like at x equals negative 2 we go from being concave downwards to concave upwards now a calculus based justification is we could look at it at the second derivative and see where the second derivative crosses the x axis because where the second derivative is negative that means our slope is decreasing we are concave downwards and where the second derivative is positive it means our first derivative is increasing our slope of our original function is increasing and we're concave upwards so notice we do indeed the second derivative does indeed cross the x-axis at x equals negative 2 it's not enough for it to just be zero or touch the x-axis it needs to cross the x axis in order for us to have an inflection point there so given that let's look at the students justifications and see what we can if we kind of put and play that put the teach had in our mind and say what a teacher would say for the different justifications so the first student says the second derivative of G changes signs at x equals negative 2 well that's exactly what we were just talking about if the second derivative changes signs goes in this case it goes from negative to positive that means our first derivative went from decreasing to increasing which is indeed good for saying this is a calculus based justification so at least for now I'm gonna put kudos you are correct there it crosses the x-axis so this is ambiguous what what is crossing the x-axis if a student wrote this I'd say what are they talking about the function are they talking about the first derivative the second derivative and so I would say please use more precise language this cannot be accepted as a correct justification all right let's do the other ones the second derivative of G is increasing at x equals negative 2 well no that doesn't justify why you have an inflection point there for example the second derivative is increasing at x equals negative 2 point 5 the second derivative is even increasing at x equals negative 1 but you don't have inflection points at those places so I would say this doesn't justify why G has an inflection point and then the last student responds the graph of G changes concavity at x equals negative 2 that is true but that isn't a calculus based justification we'd want to use our second derivative here

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