How do I solve $x^5 +x^3+x = y$ for $x$? I understand how to solve quadratics, but I do not know how to approach this question. Could anyone show me a step by step solution expression $x$ in terms of $y$?
The explicit question out of the book is to find $f^{-1}(3)$ for $f(x) = x^5 +x^3+x$
So far I have reduced $x^5 +x^3+x = y$ to $y/x - 3/4 = (x^2 + 1/2)^2$ or $y = x((x^2+1/2)^2 + 3/4)$ but Im still just as lost.

Solving it for a general $y$ will be difficult, but solving for $y=0$ is more doable.
If $x^5+x^3+x=0$, then because the left hand side is divisible by $x$, either $x=0$ or we can divide by $x$ to get $(x^2)^2+x^2+1=0$.  This is a quadratic in $x^2$, so by the quadratic formula, $x^2=\frac{-1\pm \sqrt{-3}}{2}.$
Thus, the five solutions are $x=0$ and $x=\pm\sqrt{\frac{-1\pm \sqrt{-3}}{2}}$.

If we want to show that the function is invertible, then because polynomial functions are continuous, it is enough to show that the function is either always increasing or always decreasing.  Because polynomial functions are continuously differentiable and polynomials cannot be locally constant without being completely constant, this is equivalent to showing that the first derivative is always non-negative or always non-positive.
The derivative is $5(x^2)^2+3(x^2)+1$.  Because $x^2\geq 0$ for all $x$, the derivative is always positive (in fact, always at least $1$), and therefore the function must be invertible.
