# Application of the Inverse Function Theorem

I was given an exercise in my real analysis class that reads as follows.

Suppose that $I$ is a nondegenerate interval, that $f:I\rightarrow\mathbb{R}$ is differentiable, and that $f'(x)\neq 0$ for all $x\in I$. Prove that $f^{-1}$ exists and is differentiable on $f(I)$.

I don't see how the assumptions in this exercise are any different than the assumptions of the Inverse Function Theorem. That is, since $f$ is differentiable on $I$ and $f'(x)\neq 0$ for all $x\in I$, $f$ is continuous on $I$, $f$ is strictly monotone on $I$ and therefore is injective. It seems to me that all that this exercise is just an extension of the Inverse Function Theorem to entire intervals as opposed to a single point in an interval. So would it be sufficient to say that since the Inverse Function Theorem holds for an arbitrary $x\in I$ then it holds for all $x\in I$ and therefore $f^{-1}$ exists and is differentiable on $f(I)$? I'm just not sure how to prove this rigorously.

Inverse Function Theorem: Let $I$ be an open interval and $f:I\rightarrow \mathbb{R}$ be 1-1 and continuous. If $b=f(a)$ for some $a\in I$ and if $f'(a)$ exists and is nonzero, then $f^{-1}$ is differentiable at $b$ and $(f^{-1})'(b)=1/f'(a)$

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Could you state your version of the Inverse Function Theorem? – martini Nov 29 '12 at 9:05
I edited the question to include the Inverse Function Theorem. – kaiserphellos Nov 29 '12 at 9:08
You can apply this at the end as you say. But I think you should argue why, as you say, $f$ is strictly monotone. Note that $f'$ needn't be continuous on $I$, so $f'(x)\ne 0$ for all $x \in I$, does not directly imply a sign on $f'$. It does, but why? – martini Nov 29 '12 at 9:12
It implies a sign on $f'$ due to the Mean Value Theorem right? – kaiserphellos Nov 29 '12 at 9:17
That combined with the fact that $I$ is a nondegenerate interval. That is, let $I=(a,b)$ where $a\neq b$, then we can apply the Mean Value Theorem using the fact that $f'(x)\neq 0$ for any $x\in I$. That then implies that $f(x_2)-f(x_1)$ is nonzero for $x_1, x_2\in I$ and $a\leq x_1 < x_2 \leq b$ and therefore, $f$ is strictly monotone. – kaiserphellos Nov 29 '12 at 9:23

The second para of OP states that $f$ is differentiable. Is something different implied by continuously differentiable? – wendy.krieger Mar 12 '14 at 7:26