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Studying for a test? Prepare with these 3 lessons on AP Calculus AB solved exams.

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# 2015 AP Calculus AB 5a

Video transskription

- [Voiceover] The figure above
shows the graph of f prime, the derivative of a
twice-differentiable function f, on the interval, that'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 x equals three. So you have a horizontal
tangent right over, a horizontal tangent right over there. And let me draw that a little bit neater, right over there, a horizontal tangent right over there, and a horizontal tangent right over there. Alright. 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, have the areas are 9 and 12, respectively. So that area's 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 wait, wait this looks
like a relative maximum over here, but this isn't f. This is the graph of f prime." So let's think about when we used to we don't have a graph of f in front of us. So let's think about
what it needs to be true for f to have a relative
maximum at a point. We are probably familiar with what relative maximum will look like. It'd look like a little lump, like that. It could also actually look like that but since this is 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. As we approach our relative maximum from values, as we have x
values that are approaching the x value of our relative maximum point, as we approach it from
values below that x value, we see that we have a positive slope. Our function needs to be increasing. So over here. 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 past that maximum point, after we past that maximum point, we see that our function
needs to be decreasing. Do this in another color. We see that our funciton is decreasing right over here, so f decreasing, 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. It should happen at an x value where our first derivative transitions from being greater than zero to being less than zero. So what x value, so let me say this, so we have f has relative, let me
just write it shorthand, relative maximum at x values where f prime transitions, transitions, transitions from from positive, positive, to negative, to, let me write
this a little bit neater, to negative. to negative. And where do we see f prime transitioning from positive to negative? Well over here we only
see that happening once. We see right here f prime is
positive, positive, positive, and then it goes negative,
negative, negative. So we see, 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 at 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, happens at x equals two. And we're done.