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:

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

Differentiate w.r.t $x$:

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

Now, we need:

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

Hence:

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

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

\(\displaystyle x^3-6x^2-3=3\)

\(\displaystyle x^3-6x^2-6=0\)

The only real root is:

\(\displaystyle x=2+\sqrt[3]{11-\sqrt{57}}+\sqrt[3]{11+\sqrt{57}}\)

Next, we find:

\(\displaystyle f'(x)=3x^2-12x\)

Hence:

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

And so we have:

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