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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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    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 $f(a)-f(0)=f'(b)(a-0)$.2011-11-11

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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. 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). Pick $c>N$ so $f(c). 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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    @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.2011-11-11
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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 depends on what "differentiable" means in the original poster's context: in some courses "differentiable" means "continuously differentiable". Point taken in general though.2011-11-11