# Question f06e5

Sep 30, 2017

Use the fact that $\left({f}^{- 1}\right) ' \left(27\right) = \frac{1}{f ' \left({f}^{- 1} \left(27\right)\right)}$, the Product Rule and Chain Rule, and the fact that ${f}^{- 1} \left(27\right) = 3$ since $f \left(3\right) = 27$.

#### Explanation:

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

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

Since $f ' \left(3\right) = 3 \cdot {3}^{2} \cdot 1 - 6 \cdot {3}^{3} \cdot 1 = 27 - 162 = - 135$, it follows that $\left({f}^{- 1}\right) ' \left(27\right) = \frac{1}{f ' \left({f}^{- 1} \left(27\right)\right)} = - \frac{1}{135}$.

The fact that (f^{-1})'(y)=1/(f'(f^{-1}(y))# follows from the Chain Rule by differentiating both sides of the equation $f \left({f}^{- 1} \left(y\right)\right) = y$ 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 $x > \frac{1}{2}$). See the graph below.

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