# 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 $(f^{-1})'(b)=1/f'(a)$

• 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

## 2 Answers

It is a well known fact that if a function is differentiable, then it is continuous. All you need is this

"A continuous function ƒ is one-to-one (and hence invertible) if and only if it is either strictly increasing or decreasing (with no local maxima or minima)".

Inverse function theorem requires continuous differentiability, which you don't assume.

• Welcome to stackExchange henry, this is just a comment and it doesn't answer the question. – Semsem Mar 12 '14 at 7:21
• 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