Justification using second derivative: maximum point | AP Calculus AB | Khan Academy

We're told that given that h prime of negative four is equal to zero, what is an appropriate calculus-based justification for the fact that h has a relative maximum at x is equal to negative four?

So, right over here we actually have the graph of our function h. This is the graph y is equal to h of x, and we don't have graphed the first derivative, but we do have graphed the second derivative right here in this orange color, h prime prime.

So they're telling us, given that h prime of negative four is equal to zero. So that's saying that given that the first derivative at x equals negative four is equal to zero, and you can see that the slope of the tangent line when x is equal to negative four does indeed equal zero.

So, given that, what is a calculus-based—let me underline that—a calculus-based justification for the fact that h has a relative maximum at x equals negative four?

So this first one says that the second derivative at x equals negative 4 is negative. Now, what does that tell us? If the second derivative is negative, that means that the first derivative is decreasing, which is another way of saying that we are dealing with a situation where, at least at x equals negative 4, we are concave downwards.

Which means that the general shape of our curve is going to look something like this around x equals negative 4. If the slope at x equals negative 4 is 0, well that tells us that yes, we indeed are dealing with a relative maximum point.

If the second derivative of that point was positive, then we would be concave upwards. And then if our derivative is 0 there, we'd say, okay, that's a relative minimum point. But this is indeed true: the second derivative is negative at x equals negative 4, which means we are concave downwards.

This means that we are an upside-down u, and that point where the derivative is 0 is indeed a relative maximum. So let me—sir, that is the answer, and we're done!

But let's just rule out the other ones. h increases before x equals negative four; that is indeed true. Before x equals negative four, we are increasing, and h decreases after it. That is true, and that is one rationale for thinking that, hey, we must have a maximum point assuming that our function is continuous at x equals negative four.

So this is true; it is a justification for a relative maximum, but it is not calculus-based, and so that's why we can rule this one out.

The second derivative has a relative minimum at x equals negative 4. Well, it does indeed seem to be true; there's a relative minimum there, but that's not a justification for why this is why h of negative 4, or why we have a relative maximum at x equals negative 4.

For example, you could have a relative minimum in your second derivative, but your second derivative could still be positive there. So what if the second derivative was like that? That would still be a relative minimum, but if it was positive at that point, then you would be concave upwards, which would mean that at x equals negative four, your original function wouldn't have a maximum point—it would have a minimum point.

And so just a relative minimum isn't enough in order to know that you are dealing with a relative maximum. You would have to know that the second derivative is negative there. Now, this fourth choice: h prime prime is concave up.

It does indeed look like the second derivative is concave up, but that by itself does not justify that the original function is concave up. For example, well, I could use this example right here. This is a potential second derivative that is concave upwards, but it is positive the entire time.

If your second derivative is positive the entire time, that means that your first derivative is increasing the entire time, which means that your original function is going to be concave upwards the entire time. And so if you're concave upwards the entire time, then you would not have a relative maximum at x equals negative four.

So we would rule that one out as well.