Derivative of an Inverse Function $f(x)=x^5+2x^3+1$

Question

Is my solution correct?

$f:{\mathbb{R}}\rightarrow{\mathbb{R}},f(x)=x^5+2x^3+1$. Show that $f$ has an inverse. Find $(f^{-1})'(49)$.

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$f'(x)=5x^4+6x^2$

$f'(x)>0$, this means that $f$ has an inverse.

We need $(f^{-1})'(49)=\frac{1}{f'(f^{-1}(49)}$.

Find $f^{-1}(49)$

$49=x^5+2x^3+1$

$x=2$

$\Rightarrow f(2)=49$

$\Rightarrow 2=f^{-1}(49)$

$(f^{-1})'(49)=\frac{1}{f'(2)}=\frac{1}{104}$

Answer 1

Yes, your solution is correct.

Answer 2

$f'(x)>0$ is not quite correct, since it is not true when $x=0$. Even so, since this only happens at one point (in fact it would be okay as long as this is a set of measure $0$), you can still argue that your function is strictly increasing, hence one-to-one (which means that your function has an inverse).

Essentially what you are doing is that for $x<y$: $$f(y)-f(x)=\int_{x}^y f'(t) dt>0$$

Since $f'(t)>0$ for almost every $t$.