Inverse of $f^{-1}(x)=x^5+2x^3+3x+1$ question Let $f$ be a one-to-one function whose inverse function is $f^{-1}(x)=x^5+2x^3+3x+1$.
Compute the value of $x_0$ such that $f(x_0)=1$.
I am confused as to what this question is asking me, particularly since I don't understand the subscript under the $x$ variable. 
 A: The point of the question is to see whether you understand how $f$ and $f^{-1}$ are related to each other: $$f\big(f^{-1}(x)\big)=x\;,\tag{1}$$ and $$f^{-1}\big(f(x)\big)=x\tag{2}$$ for any $x$. You want to find an $x_0$ such that $f(x_0)=1$. If $f(x_0)=1$, then $$f^{-1}\big(f(x_0)\big)=f^{-1}(1)\;.$$ Now use $(2)$ the formula for $f^{-1}$ that you've been given, and you'll have your $x_0$.
A: $f^{-1}(x)=x^5+2x^3+3x+1$.
$f(f^{-1}(x))=f(x^5+2x^3+3x+1)$.
$f(f^{-1}(x))=x$  as function property

$x=f(x^5+2x^3+3x+1)$.  
$f(x^5+2x^3+3x+1)=x$
$f(x_0)=1$
You can see that $x=1$ so   $x_0=x^5+2x^3+3x+1$
$x_0=1^5+2.1^3+3.1+1=7$
A: One way to find the inverse of an equation is to reverse the variables $x$ and $y$ (this corresponds to reflection over the line $y=x$) and then solve the resulting equation for $y$. In this case, you don't need to actually find the inverse, just compute a specific value. If we have
$$
y=x^5+2x^3+3x+1
$$
then we can write the inverse as
$$
x=y^5+2y^3+3y+1
$$
Since this second equation is the inverse of the inverse, it is the original equation $f(x)$. We want to find the $x_0$ such that $f(x_0)=1$. By the identification of $y$ with the function, all we have to do to find $x_0$ is plug in $y=1$.
EDIT: Brian's answer is exactly what I've done, only written with better notation. Use that since it makes that relationships much clearer than hiding it behind another variable $y$.
