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Let $f'$ exist for all $x$ on $[a,b]$, and suppose that $f'(a)=-1, f'(b)=1$. Prove that even if $f'$ is not continuous, there must exist a number $c, a<c<b$, with $f'(c)=0$

I'm not sure how to approach this problem. I thought about Rolle's Theorem, but can't really show it.

micad
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1 Answers1

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This is Darboux's theorem. To prove it consider the minimum of $f$ on $[a,b]$. Let it be achieved at $c$. We can't have $c=a$ since $f'(a)<0$. Neither can $c=b$. So $a<c<b$, and the usual argument shows $f'(c)=0$.

Angina Seng
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