Derivative of an Inverse, Can't find Inverse

Question

Taken from a single variable calc book.

Find $g'(a)$, where g is the inverse function of the given function

$f(x)=x^5-x^3+2x, a=2$

I intend to use the formula $g'(a)=\frac{1}{f'(g(a))}$, and I know that $f$ has an inverse. Obviously, $f'(x)=5x^4-3x^2+2$, but I can't seem to find $g$. Here's what I tried

$$y=x^5-x^2+2x$$ $$y+x^2=x^5+2x$$ $$(y+x^2)^{\frac{1}{5}}=(x^5+2x)^{\frac{1}{5}}$$

After this, I couldn't decide on an optimal way to proceed. I know that the binomial theorem extends to non-integer values, but I feel that this would be overkill for the problem...is it? Is there a simple way to find this inverse, or do I have to work my way through some nasty algebra? Thanks

Edit: it was just made clear to me that I overlooked the obvious and forgot that $g(2) \implies f(x) = 2$; however, I am still curious as to how to find the inverse of the given function.

Answer 1

You just need to know the inverse for that particular value. Let's find $x=g(2)\iff f(x)=2$:

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

since $f'(x)>0~~\forall~ x$ and thus it is invertible and attains each real number as a value only once. We have that $g(2)=1$ and finally we can calculate the derivative to be

$$g'(2)=\frac{1}{f'(g(2))}=\frac{1}{4}$$

Answer 2

To find $g(2) = f^{-1}(2)$, we solve the following: \begin{align} 0=x^5-x^3+2x-2 &= x^3(x+1)(x-1)+2(x-1)\\ &=(x^4+x^3+2)(x-1). \end{align} Clearly, $x^4+x^3>0$ for all $x\notin[-1,0]$, and hence, $x^4+x^3+2>0$ in that interval. Also, for $x\in[-1,0]$, $x^4,x^3>-1$, and thus, $x^4+x^3+2>0$.

Therefore, $g(2)=1$.