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

Let 𝑓 and 𝑔 be differentiable functions such that 𝑓⁻¹(π‘₯) = 𝑔(π‘₯) for all π‘₯. The given table shows selected values of 𝑓(π‘₯) and 𝑓′(π‘₯). What is the value of 𝑔′(3)?

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Video Transcript

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

Let’s begin by considering what we know about our functions 𝑓 and 𝑔. They’re differentiable functions. And we know that the inverse of 𝑓 is equal to this other function 𝑔. Geometrically, what does that mean? The inverse of a function can be interpreted as its reflection in the line 𝑦 equals π‘₯. So since 𝑔 and 𝑓 are inverses for one another, we can use the values in the table to find some values for 𝑔 of π‘₯.

We know that 𝑓 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 𝑦 equals π‘₯, we get four, three. And since our function 𝑔 is a reflection of the function 𝑓 in the line 𝑦 equals π‘₯, we now know that 𝑔 of four must be equal to three. We know that 𝑓 of five is equal to three. And this is represented by the point with Cartesian coordinates five, three. Reflecting this point in the line 𝑦 equals π‘₯ and we end up with the point with Cartesian coordinates three, five. So that tells us that 𝑔 of three is equal to five.

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

Well, 𝑓 of five corresponds to 𝑔 of three. So the slope of 𝑔 at 𝑓 of five is equal to one over 𝑓 prime of π‘₯, which we now know is one over 𝑓 prime of five. 𝑓 prime of five is negative four. So the slope of 𝑔 at 𝑓 of five, which we know is three, is equal to one over negative four. And that’s negative one-quarter.

So 𝑔 prime of three is negative one-quarter.

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