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Let $\ f:]a,b[\rightarrow\mathbb{R}$ be a differentiable function on $]a,b[$ so that :

$$\lim_{x\rightarrow a}_{>}\ f(x) = \lim_{x\rightarrow b}_{<}\ f(x) = +\infty$$

Show that there exists $c\in ]a,b[$ so that $\ f'(c) = 0$.

What I did is that i separated this problem in three cases. Let $\delta>0$ and consider $a_1 = a+\delta$ and $b_1 = b-\delta$. Now we have the following cases: $f(a_1)=f(b_1)$, $\ f(a_1)<f(b_1)$ or $\ f(a_1)>f(b_1)$.

When $f(a_1)=f(b_1)$ we can directly apply Rolle's theorem on $]a_1,b_1[$ so $\exists c\in ]a_1,b_1[ $ such that $f'(c)=0$.

Now I'm not sure about the case where $\ f(a_1)<f(b_1)$ but here is what I did: Since we have that $\lim_{x\rightarrow a}_{+}\ f(x) = +\infty, \ \exists a_2\in]a,b[$ such that $f(a_2)>f(b_1)$. Now using the intermediate value theorem on $]a_2,b_1[$ we get that $\exists a_3$ such that $\ f(a_3)=f(b_1)$ and using Rolle's theorem we now have that $\exists c\in]a_3,b_1[$ such that $\ f'(c)=0$.

The last case is analogous to the previous one.

Could someone tell me if the proof is correct, especially for the case where $\ f(a_1)<f(b_1)$ ?

Thank you in advance.

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up vote 0 down vote accepted

Yes, this looks fine; though you need to be a bit more careful so that your interval notation makes sense: You should choose your $\delta$ small enough so that $a_1$ and $b_1$ are in $(a,b)$ with $a_1<b_1$. Also when selecting $a_2$, you should choose it from the interval $(a, b_1)$.

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Thank you for your comment. I try to be more careful but there's always something I forget. I hope I can improve with time and exercice. – akinn Dec 2 '12 at 15:01

Since there seems to be more cases in your proof than necessary, let me suggest the following approach. Choose any $c$ in $(a,b)$ and any real number $z$ such that $z\gt f(c)$.

Since $f(c)\lt z\lt\lim\limits_{x\to a^+}f(x)$, by the intermediate value theorem, there exists $a'$ in $(a,c)$ such that $f(a')=z$. Likewise, since $f(c)\lt z\lt\lim\limits_{x\to b^-}f(x)$, by the intermediate value theorem, there exists $b'$ in $(c,b)$ such that $f(b')=z$. Now, apply Rolle theorem to the restriction of $f$ to $[a',b']$.

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Your proof is indeed much simpler. Thank you for your answer. – akinn Dec 2 '12 at 14:54

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