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