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Current time:0:00Total duration:6:13

AP.CALC:

FUN‑4 (EU)

, FUN‑4.A (LO)

, FUN‑4.A.7 (EK)

so what I want to do in this video is familiarize ourselves with the second derivative test and before I even get into the nitty-gritty of it I really just want to get an intuitive feel for what the second derivative test is telling us so let me just draw some axes here so let's say that's my y-axis let's say this is my x-axis and let's say I have a function that has a a relative maximum value at x equals C so let's say we have a situation that looks something like that and x equals C is right over so that's the Point C F of C so I can draw some straighter dotted line so that is X being equal to C and we visually see that we have a local a local maximum point there and we already hit we can use our calculus tools to think about what's going on there well one thing that we know we know that the slope of the tangent line at least the way I've drawn it right over here is equal to 0 so we could say F prime of C is equal to 0 and the other thing we can see is that we are concave downward in the neighborhood around x equals C so notice our slope is constantly decreasing and since our slope is notice it's it's positive it's less positive even less positive it goes to 0 then it becomes negative more negative and even more negative so we know that F prime prime we know that F prime prime of C is less than 0 and so you know I haven't done a deep mathematical proof here but if I have a critical point where F prime where our critical point at x equals C so f prime of C is equal to 0 and we also see that the second derivative there is less than 0 well intuitively this makes sense that we are at a maximum value and we could go the other way if we are at a local a local minimum point at x equals C or a relative minimum point so our first derivative should still be equal to 0 because our slope of a tangent line right over there is still zero so f prime of C is equal to zero but in this second situation we are concave upwards the slope is constantly is constantly increasing we have an upward-opening bowl and so here we have a relative minimum value or we could say our second derivative is greater than zero visually we see it's a relative minimum value and we can tell just looking at our derivatives at least the way I've drawn it first derivative is equal to zero and we are concave upwards second derivative is greater than 0 and so this intuition that we hopefully just built up is what the second derivative test tells us so it says hey look if we're dealing with some function f let's say it's a twice differentiable function so that means that over some interval so that means that you could take find its its first and second derivatives are defined and so let's say there's some point x equals C where its first derivative is equal to zero so the slope of the tangent line is equal to 0 and the derivative exists in a neighborhood around C and most of the functions we deal with if it's differentiable at C it tends to be differentiable in the neighborhood around C and then we also assume that second derivative exists is twice differentiable well then we might be dealing with a maximum point we might be dealing with a minimum point or we might not know what we're dealing with and it might it might be neither a minimum or a maximum point but using the second derivative test if we take the second derivative and if we see that the second derivative is indeed less than zero then we have a relative maximum point then so this is a situation that is that we started with right up there if our second derivative is greater than 0 then we are in this situation right here we're concave upwards with a slope is zero that's the bottom of the bowl we have a relative minimum point and if our second derivative 0 it's inconclusive there we don't know what is actually going on at that point we can't make any strong statement so with that on the way let's just do a quick example just to see if if this has gelled let's say that I have some twice differentiable function H and let's say that I tell you that H of 8 is equal to 5 I tell you that H prime of 8 is equal to 0 and I tell you that the second derivative at X equals 8 is equal to negative 4 so given this can you tell me whether whether the point 8 comma 5 so the point 8 comma 5 is that a relative is that a relative minimum relative minimum maximum point or not enough info not enough info or inconclusive and like always pause the video and see if you can figure it out well we're assuming it's twice differentiable I think it's safe to assume that it's and will for the sake of this art problem we're going to assume that the derivative exists in a neighborhood around x equals 8 so this example C is 8 so the 0.85 is definitely on the curve the derivative is equal to 0 so we're dealing potentially with one of these scenarios and our second derivative is less than zero second derivative is less than 0 so this through us so the fact that the second derivative so H prime prime of 8 is less than zero tells us that we fall into this situation right over here so just with the information they've given us we can say that at the point 8 comma 5 we have a relative maximum value or that this is a relative maximum point for this if somehow that if they tell us the second derivative was 0 then we would say it's inconclusive and they told us and that's all they told us and if they told us the second derivative is greater than 0 then we would be dealing with a relative minimum value at x equals 8