$f(x)=3x^3-2x^2-5$, find $g'(-10)$ where $g$ is the inverse of $f$

Question

Two functions $f$ and $g$ are inverses. Given $f(x)=3x^3-2x^2-5$, find $g'(-10)$.

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I'm not exactly sure what to do here honestly. I set $f(x)=-10$ and found a viable $x$ to be $-1$, but I don't know what to do next.

I'm given the formula $g'(x)=\frac{1}{f'(g(x))}$ but I don't see any way to apply it.

Answer 1

From $f(-1)=-10$ we get $g(-10)=-1.$

Hence

$$g'(-10)=\frac{1}{f'(g(-10))}= \frac{1}{f'(-1))}.$$

Answer 2

I know you know how to find the derivative of the inverse function so, I'll not be solving the problem but will try to add why $g' = \frac 1{f'}$

Consider,

• $y = f(x)$ be a function and you found the slope of tangent at some point $x =x_p$
• Now, Interchange the X and Y axes
• The tangent to inverse function$g(x)$ should be perpendicular to the tangent of $f(x)$ however, because of rotation negative sign will not be there

Else:

• Analytical $$\frac {\delta x}{\delta y} = \frac 1{\frac {\delta y}{\delta x}}$$ Apply limit $$g' = \frac 1{f'}$$