2015 AP Calculus AB 5a | AP Calculus AB solved exams | AP Calculus AB | Khan Academy

The figure above shows the graph of f prime, the derivative of a twice differentiable function f on the interval. It's a closed interval from negative three to four. The graph of f prime has horizontal tangents at x equals negative one, x equals one, and at x equals three. So you have a horizontal tangent right over a horizontal tangent right over there. Let me draw that a little bit neater, right over there a horizontal tangent right over there and a horizontal tangent right over there.

All right, the areas of the regions bounded by the x-axis and the graph of f-prime on the intervals negative two to one, closed intervals from negative two to one, so this region right over here, and the region from one to four, so this region right over there, they tell us have the areas are 9 and 12 respectively. So that area is 9 and that area is 12.

So part a: find all x coordinates at which f has a relative maximum; give a reason for your answer. All x coordinates at which f has a relative maximum. So you might say, "Oh look, this looks like a relative maximum over here," but this is an f; this is the graph of f prime. So let's think about what needs to be true for f to have a relative maximum at a point.

So let's—we are probably familiar with what relative maxima look like; they look like a little lump like that. They could also actually look like that, but since this is a differentiable function over the interval, we're probably not dealing with a relative maximum that looks like that.

And so what do we know about a relative maximum point? So let's say that's our relative maximum. Well, as we approach our relative maximum from values below that x value, we see that we have a positive slope; our function needs to be increasing.

So over here, we see f is increasing going into the relative maximum point. f is increasing, which means that the derivative of f, the derivative of f must be greater than zero. And then after we pass that maximum point, we see that our function needs to be decreasing. This is another color; we see that our function is decreasing right over here.

So f decreasing, which means that f prime of x needs to be less than zero. So our relative maximum point should happen at an x value where our first derivative transitions from being greater than 0 to being less than 0.

So what x values? Let me say this: so we have f has relative—let me just write shorthand—relative maximum at x values where f prime transitions from positive to negative. Let me write this a little bit neater to negative. And where do we see f prime transitioning from positive to negative? Well, over here we see that only happening once.

We see right here f prime is positive, positive, positive, and then it goes negative, negative, negative. So we see f prime is positive over here, and then right when we hit x equals negative two, f prime becomes negative.

f prime becomes negative, so we know that the function itself—not f prime—f must be increasing here because f prime is positive, and then our function f is decreasing here because f prime is negative. And so this happens at x equals two, so let me write that down: this happens at x equals two, this happens at x equals two, and we're done.