How to Find the Derivative of the Inverse Function f(x) for a Given Polynomial?

[karush]

Let $f(x)={x}^{3}-3{x}^{2}-1, x\ge2$
$\text{find} \ {df}^{-1}/dx$
$ \text{at the point} \, \, x=-1=f(3)$
Not really sure what this is asking for

 

[HallsofIvy]

$$f^{-1}$$ is the "inverse" of function f. If the function, f, changes "a" to "b" ($$f(a)= b$$), then, if f has an inverse, $$f^{-1}$$ changes "b" to a- it "reverses" the function.
Was that what you didn't understand? It seems peculiar that you would be taking a Calculus course without having seen that before.

Here, $$f(x)= x^3- 3x^2- 1$$, for $$x\ge 2$$. Notice that $$f(3)= 3^3- 3(3^2)- 1= 27- 27- 1= -1$$. That is, f change 3 to -1 so the inverse function changes -1 to 3: $$f^{-1}(-1)= 3$$. This problem asks you to find the derivative of $$f^{-1}(x)$$ at $$x= -1$$.

Your Calculus text should have, probably in the same section where you found this problem, a discussion of "inverse" functions as well as a statement (and probably a proof) of the theorem that $$\frac{df^{-1}}{dx}(b)= \frac{1}{\frac{df}{dx}(a)}$$ where f(a)= b.