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Problem : let $f:[0,\infty)\to\mathbb{R}$ be continuous on $[0,\infty)$, differentiable on $(0,\infty)$, Given that $f(0)=0$, and $\lim\limits_{x\to\infty}f(x)=0$. Prove that there is $c\in (0,\infty)$ such that $f'(c)=0$.

I need help on the above problem! Thanks. Also let me know if my proof provided below is true, if not please let me know the right answer.

Here is my first tentative: let $a>0$. Since $f$ is continuous and differentiable on $(0,\infty)$ and hence on $(0, a)$, then using the Mean Value Theorem we conclude that there is $b\in(0,a)$ such that: $f(a)-f(0)= f'(b)(a-0)$. as $b$ tends to $\infty$, $f(a)$ tends to $0$ which makes the left hand side of the equality equal to $0$, and this implies $f'(b)=0$.

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The problem is given an $a$, you get a $b$. So as $a$ tends toward infinity, so could $b$. – Bill Cook Nov 11 '11 at 1:59
I think you might need to use an epsilon argument where you let $a$ be large enough so that $f(a)$ is small. – Aleks Vlasev Nov 11 '11 at 2:24
There is a slight problem in your statement of how you are applying MVT: It is because $f$ is continuous on $[0,a]$ (not just on $(0,a)$) and differentiable on $(0,a)$ that there exists $b\in(0,a)$ such that $f(a)-f(0)=f'(b)(a-0)$. – Jonas Meyer Nov 11 '11 at 2:42
up vote 2 down vote accepted

First suppose $f$ is identically zero on $[0,\infty)$. This case is trivial. Now assume $f$ not constant. Without loss of generality, assume there is an $a>0$ s.t. $f(a)>0$. Now let $0<s<f(a)$. Then by IVT there is a $b$ in $[0,a]$ s.t. $f(b) = s$.

Now $f(x)$ tends to zero as $x$ tends to infinity so can find N s.t for all $x>N$, $f(x)<s$. Pick $c>N$ so $f(c)<s$. Apply IVT to get a $d$ in $[a,c]$ s.t $f(d)=s$. Now apply MVT to $b$ and $d$.

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Sorry for the notation. Haven't used Latex at all. – E.Lim Nov 11 '11 at 2:28
While the case is trivial, perhaps the asker would want a more thorough explanation of why it is trivial, as he is clearly working on this for practice. – analysisj Nov 11 '11 at 2:42
@analysisjb: For practice wouldn't it be best for the asker to think about that case him or herself? – Jonas Meyer Nov 11 '11 at 2:44
@JonasMeyer Absolutely. I do not think that he should prove that portion. It just may be more useful to offer an explanation as to why he considers it trivial, or provide a hint as to its triviality. I say this mainly because though E.Lim may consider it trivial, the original asker may not. – analysisj Nov 11 '11 at 2:50
@Zi2018Alpha: Yes, E.Lim's answer is good and correct. It is perhaps not fleshed out to the level of detail that an analysis instructor may expect of students' solutions in an introductory class (depending on the instructor), and in any case it is probably a good idea for you to work out every detail of why it is correct. If you have questions on details of the solution, you could post these questoins as comments here, and I am guessing that E.Lim or others who notice your comments would be happy to help further. – Jonas Meyer Nov 11 '11 at 3:26

You can use a proof by contradiction: if there is no such $c$, then the continuous function $f'$ is either always positive or always negative on $(0,\infty)$. Without loss of generality (since you can replace $f$ by $-f$), say $f'$ is always positive. Since $f(0)=0$, you will be able to disprove the hypothesis $\lim_{x\to\infty} f(x)=0$.

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It is not assumed in the statement that $f'$ is continuous. Derivatives do have the intermediate value property, but that is a theorem separate from the intermediate value theorem (for continuous functions), because not all derivatives are continuous. – Jonas Meyer Nov 11 '11 at 2:32
It depends on what "differentiable" means in the original poster's context: in some courses "differentiable" means "continuously differentiable". Point taken in general though. – Greg Martin Nov 11 '11 at 2:35

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