Can you inverse a funcion by rotating it? In school i sometimes run on some excercises where you need to calculate something that has an inverse function in it but you cannot find the inverse and you need to work your way around it. I know that the inverse function is symmetrical to the original one through the $y=x$, so i figured that if you rotate the function by 90 degrees positive turn (counter-clockwise) and then flip it through the $y$-axis you get the inverse. If you have a 1-1 function like $f(x)=e^x - e^{-x}  + c$, is there a way to find the function that gives the same plot but rotated? I am not talking about simply rotating the graph, but about finding the funcion that gives the rotated plot.
 A: 
if you rotate the [graph of a] function by $90$ degrees positive turn (counter-clockwise) and then flip it through the $y$-axis you get the [graph of the] inverse

That is correct.  I wouldn't express it as "the same plot but rotated" because there is also that "flip", or reflection, that you did after rotating it.
A: Well...the easiest way to find the function $f$ rotated 90 degrees is simply to first find $f^{-1}$, and then to set $g(x) = f^{-1}(-x)$.
But that kind of defeats the purpose, since you were trying to rotate as an intermediate step in finding the inverse.
To rotate it 90 degrees directly, you could convert it into polar form as a function $r(\theta)$. Then $r'(\theta) = r(\theta- 90)$ rotates it 90 degrees clockwise. Then convert $r'(\theta)$ back into polar form.
But I think it will be easier to just do the standard thing, which is to write it as $y = f(x)$, set $x = f(y)$, and then solve for $y$.
A: if you have a parametric representation of a function
$$x = f_1(t)$$ $$y = f_2(t)$$ where $t \in \mathbb{R}$
then the inversed function would have parametrisation reversed / rotated by 180 ° along the $y=x$ diagonal:
$$x = f_2(t)$$ $$y = f_1(t)$$ where $t \in \mathbb{R}$
you may transform each point of the function by matrix $R=\begin{bmatrix} 0&1\\ 1&0\end{bmatrix}$. 
A: In the simplest words, the plot of inverse of a given function is actually a mirror image of the plot of the given function about line y=x. 
This is because, x-axis and y-axis are also the mirror images of each other about line y=x. 
A: Given a function $f$, it leaves a trace $(x,f(x))^\top$ in the plane. Rotating this trace gives the new graph
$$(\underbrace{x\cos\phi+f(x)\sin\phi}_{g(x)},\;\underbrace{-x\sin\phi+f(x)\cos\phi}_{h(x)})^\top.$$
If you are able to invert $g(x)$, then $\bar f(x):=h(g^{-1}(x))$ will be your rotated function by an angle of $\phi$. Note that $g$ is only invertible if $f'$ nowhere equals $\pm\cot \phi$. Especially functions with unbounded derivative cannot be rotated by any angle.
