# Rolle theorem on infinite interval

We have : f(x) is continues in $[1,\infty]$ and differentiable in $(1,\infty)$

$\lim_{x \to \infty}f(x) = f(1)$

we have to $\textbf{prove}$ that : there is $b\in(1,\infty)$ such that $f'(b) = 0$

I'm sure we have to use Rolle's theorem so I tried using Mean Value theorem and using the limit definition at $\infty$

Any ideas how can I use them ?

Update :

after seeking the answers that I've got :

I'm having trouble finding x1$\neq$x2 such that f(x1)=f(x2)

*I need a formal solution

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how much large intervals and can you explain more – Was Fr Feb 6 '14 at 14:20
Sorry. I had an idea that didn't work. – 5xum Feb 6 '14 at 14:25

Let $g(x)=f\bigl(\frac 1x\bigr)$ for $x\in (0,1]$ and $g(0)=f(1)$. Then $g$ is continuous in $[0,1]$ and derivable in $(0,1)$. By Rolle's Theorem there exists $c\in (0,1)$ such that $g'(c)=0$, hence $$0=g'(c)=-f'\Bigl(\frac 1c\Bigr)\frac 1{c^2}$$ thus if $b=\frac 1c$, then $f'(b)=0$ and $b\in (1,+\infty)$.

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Outline:

You may, and should, assume that $f$ is non-constant.

Pick $a$ so that $f(a)\ne f(1)$ (if no such $a$ exists $f$ is a constant function).

Choose a value $c$ that is in between $f(1)$ and $f(a)$. Use the Intermediate Value Theorem to show there is a $x_1\in(1,a)$ with $f(x_1)=c$.

Use the Intermediate Value Theorem and your limit condition to show there is an $x_2\in(a,\infty)$ with $f(x_2)=c$.

Now you're set to apply Rolle's Theorem.

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I came to this Idea and I'm sure it should lead to the answer but still I'm having trouble using the limit condition to show that there is x2 such that f(x2)=c can you please show me ? – Was Fr Feb 6 '14 at 14:39
@WasFr Since $\lim_{x\rightarrow\infty} f(x)=1$, you can find a $y>a$ such that $f(y)$ is in between $c$ and $f(1)$. Apply the IVT to $f$ on $[a,y]$ to find $x_2$. – David Mitra Feb 6 '14 at 14:57
Let's be more specific : What Epsilon shall I choose ? to make it done. – Was Fr Feb 6 '14 at 15:20
I solved it by choosing a minimal M for a given epsilon – Was Fr Feb 6 '14 at 16:29

An alternative to the existing answers:

Because of your hypotheses, the function defined by: $$g(t) = \begin{cases} f(\tan(t)) & t \in [\pi/4, \pi/2) \\ f(1) = \lim_{x \to \infty} f(x) & t = \pi/2 \end{cases}$$ is continuous on $[\pi/4, \pi/2]$, differentiable on $(\pi/4, \pi/2)$, and $g(\pi/4) = f(1) = g(\pi/2)$. Apply Rolle's theorem to get $t \in [\pi/4, \pi/2]$ such that $g'(t) = 0$.

Compute $g'(t) = \tan'(t) f'(\tan(t)) = (1 + \tan(t)^2) f'(\tan(t)) = 0$. Therefore for $x = \tan(t) \in (1, \infty)$, $$f'(x) = 0$$

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can you explain why : tan'(t)= $(1 + \tan(t)^2)$ ? – Was Fr Feb 6 '14 at 15:09
That's the derivative of the tangent function? – Najib Idrissi Feb 6 '14 at 15:13
absolutely I missed that. yes nice answer thank you. – Was Fr Feb 6 '14 at 15:21

Think about where $f(1)$ is. Pick a spot, any spot. Then think about all the points afterward. It's continuous and differentiable literally everywhere but possibly $f(1)$, and so we need to think about how it could not be flat somewhere.

First, we know it would be flat everywhere if $f(x)=f(1)$ a constant.

Next, we draw a line between $x = 1$ and $\infty$. At infinity, we have $f(\infty)=f(1)=K$ some finite constant. Then we think about how we draw a line from 1 to $\infty$ . If it's flat, ie $f(x)=f(1)=K$, then it is obviously true that $f'(x)=0$ everywhere between $1$ and $\infty$.

Now consider the case where it is non-constant.

Then we know there is a max (or min, by symmetry) $m$ such that $|f(m)|=|M|>k = f(1)$ on the number line. Now since $f$ is differentiable at $m \in (1,\infty)$, and $M$ is a max/min, it must be smooth and both sides smoothly approach it's flatness. This $m$ is your $b$ you are seeking to prove exists.

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I agree with you . but I needed a more formal approach. – Was Fr Feb 6 '14 at 14:41

If $f$ is constant, the statement is trivial. If it is not, then we have some $x_0$ such that $f(x_0)\neq f(1)$.

• If $f(x_0) > f(1)$: Define the midpoint between $f(1)$ and $f(x_0)$ as $y=\frac{f(x_0)+f(1)}{2}$. Because the function has values $f(1)$ and $f(x_0)$ on the interval $[1,x_0]$, there exists such $a$ that $f(a) = y$. Because the limit of $f(x)$ as $x\rightarrow\infty$ is $f(1)$, there exists such an $M$ that $f(x) < y$ for all $x\geq M$. Because the function $f$ has values $f(x_0)$ and $f(M)<y$ on the interval $[x_0, M]$, there exists such a point $b$ that $f(b)=y$. Now use Rolle's theorem on $[a,b]$.

• If $f(x_0) < f(1)$: Define the function $g(x) = -f(x)$ and use the previous argument on $g$.

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why we can deduct that there is an x such that f(x)>f(1) maybe f(1) is an upper least bound ? – Was Fr Feb 6 '14 at 14:29
That's why I wrote without loss of generality. Either you have $x$ for which $f(x)>f(1)$ or you have $x$ for which $f(x)<f(1)$, in which case you can look at the function $g(x) = 2 - f(x)$ and use the same argument on $f$ as you did on $f$. – 5xum Feb 6 '14 at 14:31
you mean you want to use Fermat's theorem for derivative on maximum or minimum ? – Was Fr Feb 6 '14 at 14:34
Yes. And the fact that the limit of $f$ is equal to $f(1)$ guarantees that either a maximum or a minimum exists. – 5xum Feb 6 '14 at 14:35
but, the challenge is to show the solution in a formal way. I agree with the idea that you are presenting but I'm in doubt of writing it formally. see the answer down I think it should lead to a formal solution and read my comment maybe you can help me. Thank you – Was Fr Feb 6 '14 at 14:45