When
f(x)=x^7+2x^3+x-1
I have found
f^-1(3)=1
But I need to prove this.
Can anyone help?
By definition of $\displaystyle f^{-1}$ , $\displaystyle f^{-1}(\{y\})=\{x\in\mathbb{R}:f(x)=y\}$ . As $\displaystyle f(1)=3$ we can conclude that $\displaystyle 1\in f^{-1}(\{3\})$ . If we prove that $\displaystyle f$ is strictly increasing then, no $\displaystyle x\neq 1$ satisfies $\displaystyle f(x)=3$ so, $\displaystyle f^{-1}(\{3\})=\{1\}$ . That was your initial question. Now, if you want a closed formula $\displaystyle f^{-1}(x)$ for all $\displaystyle x\in\mathbb{R}$ , better forget it.
If finding $\displaystyle f^{-1}(*)$ is a 'categoric imperative' , then a method based on advanced tecniques of complex analysis will be supplied without demonstration...
If the function the inverse of which has to be found is...
$\displaystyle y= x^{7} + 2\ x^{3} + x -1$ (1)
... then first with the variable substitution $\displaystyle z=1+y$ we obtain...
$\displaystyle z=f(x)= x^{7} + 2\ x^{3} + x$ (2)
Now the inverse $\displaystyle x=f^{-1} (z)$ of a function for which is $\displaystyle f(0)=0$ can be written in series expansion as...
$\displaystyle x= \sum_{n=1}^{\infty} a_{n}\ z^{n}$ (3)
... where the $\displaystyle a_{n}$ are given by...
$\displaystyle a_{n}= \frac{1}{n!}\ \lim_{x \rightarrow 0} \frac{d^{n-1}}{d x^{n-1}} \{\frac{x}{f(x)}\}^{n}$ (4)
In our case of course is...
$\displaystyle \frac{x}{f(x)} = \frac{1}{x^{6}+ 2\ x^{2} +1}$ (5)
... and the x as function of y will be...
$\displaystyle x= \sum_{n=1}^{\infty} a_{n}\ (1+y)^{n}$ (6)
Details of computation of the $\displaystyle a_{n}$ are left to the [young I suppose...] starter of the thread...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$