# $f$ is derivative in $(0,2)$ that appiles $\lim_{x \to 0+}f(x)=\lim_{x \to 2-}f(x)= +\infty$ Proof there is $c \in (0,2)$ such $f'(c)=0$.

I have this problem :

$f$ is derivative in $(0,2)$ that appiles $$\lim_{x \to 0+}f(x)=\lim_{x \to 2-}f(x)= +\infty$$

Proof there is $c \in (0,2)$ such $f'(c)=0$.

I'll write the proof and I'll be very glad to receive feedback, because when it's comes to using the definition it pretty hard to me.

The intuitive explanation behind the proof :

$\lim_{x \to 0+}f(x)=\lim_{x \to 2-}f(x)= +\infty$, since both of the limits goes to infinity, therefore exist such $a,b \in (0,2) \implies f(a)=f(b)$, then using Rolle's theorem and we receive that there is $c \in (a,b)\subseteq (0,2) \implies c \in (0,2)$ such that $f'(c)=0$.

Proof :

Lets choose $n \in \mathbb{R}$ and demand that $n>0$ and also $f(1)<n$ exist such $n \in \mathbb{R}$ because $\mathbb{R}$ is unbound.

• $\lim_{x \to 0+}f(x)= \infty$ For $M>n>0$ exists $\delta_1>0$ so for all $x$ that implies $0<x<\delta_1 \implies f(x)>M>n$
• $\lim_{x \to 2-}f(x)= \infty$ For $M>n>0$ exist $\delta_2>0$ so for all $x$ that implies $2-\delta_2<x<2 \implies f(x)>M>n$

We found that $f(1)<n$ and for $f(2-\frac{\delta_2}{2})>M>n$ and as well $f(\frac{\delta_1}{2})>M>n$

Since $f$ is derivative in $(0,2)$ is also continuous in $(0,2)$ using the intermediate value theorem we can find such $a$ in the interval $[\frac{\delta_1}{2},1]$ and also we can find such $b$ in the interval $[1,2-\frac{\delta_2}{2}]$

Such that $f(a)=f(b)$

(I used the intermediate value theorem in $[\frac{\delta_1}{2},1]$ and $[1,2-\frac{\delta_2}{2}]$)

Since $[a,b] \subset (0,2)$ we can conclude that $f$ is continuous in $[a,b]$ and derivative in $[a,b] \subset (a,b)$ and since $f(a)=f(b)$, using Rolle's theorem there exist such $c \in (a,b) \implies c \in (0,2)$ such that $f'(c)=0$

Any help will be appreciated.

• Since you only know infinite as $x \to 0^+,$ it's a bad idea to use specifically the point $1$. Just use that near $0$ there is $\delta$ so $f$ is arbitrarily large in $(0,\delta)$ and do like you have started. – coffeemath Jan 4 '15 at 18:18
• @coffeemath Can you explain why use specifically the point 1 is a bad idea?$\\$ As for your method I agree its easiler that method, saying that near 0 there are arbitrarily large values and near 2 there are arbitrarily large values and since that we can find $f(a)=f(b)$ and continue the way I did, but it hard for me to translate it into well constructed mathematical form, so I used my method. – JaVaPG Jan 4 '15 at 18:30
• JaVaPG -- I didn't really follow your argument, some steps seemed not immediately valid to me (but maybe could be made to work). Anyway see my "answer" below for a fill-in about what I had in mind. – coffeemath Jan 5 '15 at 0:16

It could be useful to consider $f(1)=k.$ Since $f(x) \to +\infty$ as $x$ goes to $0$ or to $2$ from the appropriate sides, there are small disjoint intervals $(0,\delta_1)$ and $(2-\delta_2,2)$ in which there exist points $a,b$ respectively for which $f(a)>k$ and $f(b)>k.$ If it happens that $f(a)=f(b)$ then Rolles Theorem applies as you mention. Otherwise one of them is greater than the other. If it is $f(a)>f(b),$ then you have the arrangement $k=f(1)<f(b)<f(a)$ so using the intermediate value theorem there is some $\xi$ in the open interval $(a,1)$ for which $f(\xi)=f(b),$ and now Rolles can be applied using the two points $\xi,b.$ Similar argument for the other case $f(a)<f(b).$