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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?2012-11-29
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    I edited the question to include the Inverse Function Theorem.2012-11-29
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    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?2012-11-29
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    It implies a sign on $f'$ due to the Mean Value Theorem right?2012-11-29
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    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.2012-11-29
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    Why can you apply the MVT? $f'$ need not be continuous! That's the point I also wanted to make above.2012-11-29
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    Well, suppose $x_1,x_2\in I$ and $f'(x_1)>0$ and $f'(x_2)<0$, then, since $f'(x)\neq 0$ for all $x\in I$, $f$ would have to be discontinuous at some point in $I$, which contradicts the assumptions of the exercise. Therefore $f'(x)$ is either positive of negative for all $x\in I$.2012-11-29
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    let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/6544/discussion-between-martini-and-christopher-washington)2012-11-29

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