242.7x.01 d/dx of inverse equation

[karush]

$\tiny{242.7x.01}$
$\textsf{Find the value of $df^{-1}/dx$ at $f(a)$, $x\ge4$, $a=3$}$
\begin{align*}\displaystyle
f(x)&=x^3-6x^2-3 \\
\frac{df^{-1}}{dx}\biggr\rvert_{3}
&=\frac{1}{{\frac{d}{dx}}\biggr\rvert_{3}} \\
&=\frac{1}{3x^2-12x}
=\frac{1}{3(3)^2 - 12(3)}\\
f^{-1}(3)&=\color{red}{-\frac{1}{9}}
\end{align*}
$\textit{if ok,,, comments on notation ??}$

 

[MarkFL]

I would begin by writing:

$$f^{-1}\left(f(x)\right)=x$$

Differentiate w.r.t $x$:

$$[f^{-1}]'\left(f(x)\right)f'(x)=1$$

Now, we need:

$$f(x)=a\implies x=f^{-1}(a)$$

Hence:

$$[f^{-1}]'(a)=\frac{1}{f'\left(f^{-1}(a)\right)}$$

To find $f^{-1}(a)$ where $a=3$, we need to solve the following for $x$:

$$x^3-6x^2-3=3$$

$$x^3-6x^2-6=0$$

The only real root is:

$$x=2+\sqrt[3]{11-\sqrt{57}}+\sqrt[3]{11+\sqrt{57}}$$

Next, we find:

$$f'(x)=3x^2-12x$$

Hence:

$$f'\left(2+\sqrt[3]{11-\sqrt{57}}+\sqrt[3]{11+\sqrt{57}}\right)=3\left(\left(\sqrt[3]{11-\sqrt{57}}+\sqrt[3]{11+\sqrt{57}}\right)^2-4\right)$$

And so we have:

$$[f^{-1}]'(3)=\frac{1}{3\left(\left(\sqrt[3]{11-\sqrt{57}}+\sqrt[3]{11+\sqrt{57}}\right)^2-4\right)}\approx0.025080128988$$