# Video: AP Calculus AB Exam 1 β’ Section I β’ Part A β’ Question 23

Let π and π be differentiable functions such that πβ»ΒΉ(π₯) = π(π₯) for all π₯. The given table shows selected values of π(π₯) and πβ²(π₯). What is the value of πβ²(3)?

02:34

### Video Transcript

Let π and π be differentiable functions such that the inverse of π of π₯ is equal to π of π₯ for all π₯. The given table shows selected values of π of π₯ and π prime of π₯. What is the value of π prime of three? And then, we have a table here which contains values for π₯, π of π₯, and π prime of π₯.

Letβs begin by considering what we know about our functions π and π. Theyβre differentiable functions. And we know that the inverse of π is equal to this other function π. Geometrically, what does that mean? The inverse of a function can be interpreted as its reflection in the line π¦ equals π₯. So since π and π are inverses for one another, we can use the values in the table to find some values for π of π₯.

We know that π of three is equal to four. So we can plot that as the point with the Cartesian coordinates three, four. If we reflect this point in the line π¦ equals π₯, we get four, three. And since our function π is a reflection of the function π in the line π¦ equals π₯, we now know that π of four must be equal to three. We know that π of five is equal to three. And this is represented by the point with Cartesian coordinates five, three. Reflecting this point in the line π¦ equals π₯ and we end up with the point with Cartesian coordinates three, five. So that tells us that π of three is equal to five.

Now, weβre trying to find the value of π prime of π₯. So next, we recall something called the inverse function theorem. And this says that if π prime of π₯ is not equal to zero, then the slope or the gradient of the inverse of π at π of π₯ is equal to one over π prime of π₯. Now, we know that the inverse of π is equal to π in our question. So we can replace the inverse of π with π in the theorem. And this time, we see that the slope of π, and thatβs remember π prime, at π of π₯ is equal to one over π prime of π₯.

Well, π of five corresponds to π of three. So the slope of π at π of five is equal to one over π prime of π₯, which we now know is one over π prime of five. π prime of five is negative four. So the slope of π at π of five, which we know is three, is equal to one over negative four. And thatβs negative one-quarter.

So π prime of three is negative one-quarter.