Is the inverse function of $f(x) = \frac{1}{f(x)}$? If not, why is the notation $f^{-1}(x)$ ? Also, explain $f^{-1}(x)' = \frac{1}{f'(x)}$? When I want to find the inverse of a function I replace the x with y and then isolate y.
However, I couldn't use that for the example below:
Let $f(x) = \frac{1}{2}x^3+3x-4$ and let $h(x)$ be the inverse function of $f(x)$
To find $h(x) $ I did:
$$x = \frac{1}{2}y^3+3y-4 \\ 
\Leftrightarrow x + 4 = \frac{y^3}{2}+3y \\
\Leftrightarrow x +4 = \frac{y^3+6y}{2}\\
\Leftrightarrow 2x + 8 = y^3 + 6y \\
\Leftrightarrow ???$$
I didn't know how to continue (still don't) so I tried a different approach. I know that 
$$h'(x) = \frac{1}{f'(x)}$$
And because the notation for the inverse function is $f^{-1}(x)$ I assumed that the inverse function could also be represented by $\frac{1}{f(x)}$, so I tried to do that with a simpler function like $g(x) = x^2$:
$$g^{-1}(x) = \frac{1}{x^2}$$
However, using the first method this I get $g^{-1}(x)= \sqrt{x}$ , which is the correct answer
(Wolfram Alpha confirms this )
Also, another thing that confuses me is that the formula for the derivative of the inverse function, considering $h(x) = f^{-1}(x)$, is
$$h(x)' = \frac{1}{f'(x)}$$
In my understanding, this means that the derivative of the inverse function is the inverse of the derivative (of the given function).
Logically, shouldn't the value of the inverse function be the inverse of the given function? In other words,
If
$$h(x)' = \frac{1}{f'(x)}$$,
shouldn't
$$h(x) = \frac{1}{f(x)}$$
?
I don't understand though why $h(x)' = \frac{1}{f'(x)}$
My questions:


*

*How do I continue $ \Leftrightarrow 2x + 8 = y^3 + 6y \Leftrightarrow ???$

*Is the inverse function of $f(x) = \frac{1}{f(x)}$? If not, why is the notation $f^{-1}(x)$ ? 

*What is the logic behind $h(x)' = \frac{1}{f'(x)}$ ? In other words, could you show proof of it?

 A: The problem relies on the fact that the symbol $f^n$, with $n\in\mathbb{N}$, can be given a priory two meanings:


*

*The function $$\overbrace{f\circ \cdots \circ f}^{n\text{ times}}$$
obtained by composition of $f:X\rightarrow X$, where $X$ is some set, with itself $n$ times.

*The function $$x\mapsto (f(x))^n$$ which sends $x$ to the $n$-th power of $f(x)$ and where $f:X\rightarrow \mathbb{R}$ is a real valued (or more generally valued in some ring) function.


Without a convention, $f^n$ and $f^n(x)$ seems a good notation for both of these functions. However, due to historical reasons, the first convention is the usual one. In this way, unless otherwise stated, $f^n$ usually means $f$ composed with itself $n$ times.
As a consequence of this convention, we write $f^{-1}$ for the inverse function of $f:X\rightarrow X$, i.e., the function such that $f\circ f^{-1}=\mathrm{id}$ and $f^{-1}\circ f=\mathrm{id}$, as it is the only possible value which allows the extension of the exponent rule $f^n\circ f^m=f^{n+m}$ to all integers in such a way that $f^0=\mathrm{id}$.
Of course, the above applies directly only to functions of the form $f:X\rightarrow X$. However, due to use, it has stick to all functions the above notation of the inverse function.
In conclusion, this is just a matter of convention motivated by an arbitrary choice at some moment in mathematics' history.
EDIT: Answering the other questions. About the computation of the inverse function $h$ of $f(x):=\frac{1}{2}x^3+3x-4$, you arrive to
$$y^3+6y=2x+8\text{.}$$
Here, applying the technique of Inverse of y=x3+xy=x3+x?, i.e. taking $u+v=y$, you obtain the system
$$\begin{cases}u^3+v^3&=2x+8\\u^3v^3&=-8\end{cases} $$
which gives, by transforming it nto a quadratic equation,
$$u^3=x+4+\sqrt{x^2+8x+24}\text{ and }v^3=x+4-\sqrt{x^2+8x+24}$$
or the other way around. And so
$$h(x)=\sqrt[3]{x+4+\sqrt{x^2+8x+24}}+\sqrt[3]{x+4-\sqrt{x^2+8x+24}}$$
should be the inverse as $x^2+8x+24\geq 0$ for all $x\geq 0$ and because $x^3+6x=t$ has a unique real solution for each $t\in\mathbb{R}$ as the derivative of $x^3+6x$ is always positive.
With respect the derivative, note that that the formula is
$$h'(x)=\frac{1}{f'(h(x))}$$
and not the one given by you, i.e. the derivative is evaluated at the inverse. This follows by derivating $f(h(x))=\mathrm{id}(x)=x$ using the chain rule.
A: I'm going to answer the second and third questions before the first.
As you have seen for yourself, the inverse function of $f(x)$ is very different than the reciprocal $\frac{1}{f(x)}$.  The meaning of the inverse function is that you are trying to find a function $g(x)$ such that $g(f(x)$ is the identity function (that always delivers the value $x$.  Let's assume that inverse were
$\frac{1}{f(x)}$; then we would have
$$
f\left( \frac{1}{f(x)} \right) = x 
$$
and by the chain rule,
$$
\frac{d}{dx} f\left( \frac{1}{f(x)} \right) = \frac{d}{dx}\,x = 1 \\
-\frac{df(x)}{dx} \frac{1}{[f(x)]^2} = 1 \\
\frac{df(x)}{dx} = -[f(x)]^2 
$$
The general solution of that differential equation is $f(x) = \frac1{x+C}$ where $C$ is an arbitrary constant. If we substitute that back into $f\left( \frac{1}{f(x)} \right) = x$ we get
$$
f\left( x+C \right) = x \\
\frac1{(x+C)+C} = x 
$$
and that is never going to be an identity, no matter what we take $C$ to be. 
So the inverse function is never the  reciprocal of the function.
The notation $f^{-1}(x)$ is just a convenience, kind of like $\sin^2\theta$ or for that matter $\frac{dy}{dy}$ where you can't "cancel the d's". I like the explanation given in the other answer, to explain why this particular notation "makes sense."
Now for the third question:
Say $f$ and $h$ are functions of $x$, and $h(f(x)) = x$. Now take the derivative of both sides, using the chain rule:
$$
\frac{d}{dx}\left(  h(f(x)) \right) = 1 \\
\frac{dh}{df} \frac{df}{dx} = 1 \\
\frac{dh}{df} = \left( \frac{df}{dx}\right)^{-1}
$$
The first question has a much harder answer.  To obtain the inverse function in closed form, you would need to apply the formula for the roots of a cubic equation.  ALthough you could do that, it is horribly messy.
