finding the inverse function $f(x) = x^5+ x^3 + x$, then find what is $f^{-1}(3)$

Question

if $f(x) = x^5+ x^3 + x$, then $f^{-1}(3) = ?$

I TRIED to do it, and I got this answer : 3/91 . I don't know if it is correct or not?

I tried to do this work like this: $$ y = x^5+x^3+x\\ y = x(x^4+x^2+1)\\ y/(x^4+x^2+1) = x\\ y = x/(x^4+x^2+1)\\ f^{-1}(x)= x/(x^4+x^2+1)\\ f^{-1}(3)=3/91$$

Answer 1

Let $y=y(x)$ is an inverse function.

Thus, $y^5+y^3+y=3,$ which since $y^5+y^3+y$ increases, gives $y=1$ and $f^{-1}(3)=1.$

Answer 2

$f(x)-3 = x^5+x^3+x-3 = (x-1)(x^4+x^3+2x^2+2x+3) = 0$ $\Leftrightarrow x=1 \quad \lor \quad x^4+x^3+2x^2+2x+3 = 0$

Since $f'(x) = 5x^4 + 3x^2 + 1 \ge 1 > 0$ for all $x \in \mathbb R$, $f$ is strictly monotone increasing, thus $x=1$ must be the only (real) solution.