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Suppose that $f$ is differentiable on a finite interval $(a,b)$ and $$ \lim_{x\to a^+}f(x)=\lim_{x\to b^-}f(x)=\infty. $$ Prove that, for any $r\in\mathbb{R}$ there exists $c\in (a,b)$ such that $f'(c)=r$.

MY THOUGHTS: This theorem seems painfully obvious on an intuitive level, but I am not sure how to begin formalizing this. The Mean Value Theorem seems likely to be used here. Let $r\in\mathbb{R}$ and then there is some $c\in(s,t)\subset(a,b)$ such that $$ f'(c)=\frac{f(t)-f(s)}{t-s}=r. $$ It isn't immediately clear to me how to choose the subinterval $(s,t)$. Perhaps there is a better way to proceed?

Laars Helenius
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    This works fine. Fix $s$ now, and let $t\to a$, $t\to b$. This shows that $f'$ takes arbitrarily large and small values. A derivative has the intermediate value property, so this gives the full claim. –  Mar 07 '15 at 05:38

1 Answers1

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You can use the fact that the derivative has the intermediate value property.

So, for the sake of simplicity, take $a=-1$ and $b = 1$, and consider $f(0)$

Then as f is unbounded near 1, for all M such that $\exists x \in ]\frac{1}{2},1[$ such that $f(x) > M$

You then have $\frac{f(x)-f(0)}{x} > f(x)-f(0) > M$

So there exist $c \in ]0,x[$ such that $f'(c) > M$

By the same reasonning, you get that there exist $d \in ]y,0[$ such that $f'(d) < -M$

And by the intermediate value theorem, $\forall z \in [-M,M], \exists e \in [d,c], f'(e) = z$

And as you can take any $M$, it's proved for all $\mathbb{R}$

Tryss
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