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.

In order to find $f^{-1}(3)=x$ for some $x$. We can apply the function to both sides giving us $3=f(a)$ now we need to find $x^5+x^3+x=3\implies x^5+x^3+x-3=0$. There are many ways to solve this however you can go ahead and guess that one solution would be $x=1$. Now we can perform polynomial division on $x^5+x^3+x-3$ by dividing by $x-1$ as that is a root and we get $x^4+x^3+2x^2+3$. The rest of your roots are there however it seems they are not real.

There is almost an art to guessing however one way to get a good feel is to try and plot the graph for different values. I guessed 1 because there were 3 terms with no coefficients and they evaluate to 3.
