2
$\begingroup$

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

  • 0
    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6544/discussion-between-martini-and-christopher-washington)2012-11-29

2 Answers 2

2

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)".

0

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

  • 0
    The second para of OP states that $f$ is differentiable. Is something different implied by continuously differentiable?2014-03-12