If $f(x) = x^5 + x^3 + x$ find $f^{-1}(3)$ and $f(f^{-1}(2))$ How do I go about solving a problem like this? Do I find the inverse formula first? If so I am stuck at $\frac{x}{y} = y^3(y^2+1)$. Help is welcomed. 
 A: Note that for the second one, $f$ and $f^{-1}$ essentially cancel. So $f(f^{-1}(2))=2$. 
For the former, letting $b=f^{-1}(3)$, we have $f(b)=f(f^{-1}(3))=3$. So we need to find $b$ such that $f(b)=3$. Then
$$b^5+b^3+b=3$$
which I think has at least one pretty clear solution.
A: OK, I admit I didn't see this at first, but here's a hint:
If you're looking for $f^{-1}(3)$ then you want to solve this equation for $x$:
$$3 = x^5 + x^3 + x.$$
Can you see a solution for $x$ by inspection?
A: Given the phrasing of the question, it's unlikely that you are expected to find the inverse; you have been asked a question to which the answer is very guessable (in particular, $f^{-1}(3)=1$), and for the second, we know that the composition of a function with its inverse is the identity. The question seems to be checking your understanding of the notion of an inverse function, so don't overthink it!
However, you do need to check or show that the inverse for the former is unique!
A: The continuos function  $f:\mathbb{R} \to \mathbb{R}$ defined by rule $f(x)=x^5+x^3+x$ is stricly incrasing, then the equation $f(x)=3$ has only a single solution. But $1^5+1^3+1=3$, then $x=1$ is single solution the equation $f(x)=3$ therefore $f^{-1}(3)=1$. Clearly $f^{-1}(f(2))=2$ because $f$ is bijection.
