Question #f06e5

1 Answer
Sep 30, 2017

Use the fact that (f^{-1})'(27)=1/(f'(f^{-1}(27))), the Product Rule and Chain Rule, and the fact that f^{-1}(27)=3 since f(3)=27.

Explanation:

Note that the Product Rule and Quotient Rule imply that f'(x)=3x^{2}e^{-6(x-3)}-6x^{3}e^{-6(x-3)}.

Also note, by "educated guessing" (or looking at the graph), that f(3)=3^{3}e^{-6(3-3)}=27*1=27. Therefore, f^{-1}(27)=3. (These problems are typically set up in such a way that you will be "lucky" in your educated guessing on this part).

Since f'(3)=3*3^{2}*1-6*3^{3}*1=27-162=-135, it follows that (f^{-1})'(27)=1/(f'(f^{-1}(27)))=- 1/135.

The fact that (f^{-1})'(y)=1/(f'(f^{-1}(y)) follows from the Chain Rule by differentiating both sides of the equation f(f^{-1}(y))=y if we assume the differentiability of the functions in question.

We are also implicitly assuming that f is invertible here, which in may only be locally so. In fact, this particular function is not invertible overall, since it fails the horizontal line test. However, near x=3 it is invertible (the function has negative slope for x>1/2). See the graph below.

graph{x^3*e^(-6(x-3)) [-1, 7, -499973, 500027]}