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Question

# Let f(x)=x^3-3 x^2-1, $x \geq 2$ Find the value of df^{-1} / dx at the point x=-1=f(3)

Solution

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Since $f(3) = -1$, $a = 3$ and $b = -1$. Since $f'(x) = 3x^2 - 6x$, $f'(3) = 3(3)^2 - 6(3) = 9$ and hence

$\left.\dfrac{df^{-1}}{dx}\right|_{x=-1} = \dfrac{1}{f'(f^{-1}(-1))} = \dfrac{1}{f'(3)} = \dfrac{1}{9}$

If $f^{-1}(x)$ is the inverse function of $f(x)$ and $f(a) = b$, then $df^{-1}/dx$ at $x = b$ is

$\left.\dfrac{df^{-1}}{dx}\right|_{x=b} = \dfrac{1}{f'(f^{-1}(b))} = \dfrac{1}{f'(a)}$

In other words, $(f^{-1})'(b) = \dfrac{1}{f'(a)}$ since $a = f^{-1}(b)$.

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