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Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f(0)=0$ and $|f'(x)|\leq M$. Prove that $|f(x)|\leq M|x|$. Apply this to the function $f(x)=\sin x$.

I'm unsure of how to prove this problem. This problem is from the Mean Value Theorem section chapter. I will ask question if in doubt of the proof provided.

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    Of course, you're assuming $\,f\,$ is differentiable, right?2012-11-28

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A related problem.

Mean Value Theorem: If a function $f$ is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that $$ f'(c) = \frac{f(b) - f(a)}{b-a}. $$

Apply this to your problem gives

$$ f'(c) = \frac{f(x) - f(0)}{x-0},\quad c\in(0,x) $$

$$ \implies \left| \frac{f(x)}{x}\right|=|f'(c)| \leq M \implies |f(x)|\leq M |x|.$$

Now, apply this to $\sin(x)$ and figure out what $M$.

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    would we still use (0,x)?2012-11-28
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    is it |sin x| <= M |x|?2012-11-28
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    @Maximiliano: When $f(x)=\sin(x)$, we have $f'(c)=\cos(c)$ which implies $ |f'(c)|=|\cos(c)|\leq 1 = M. $2012-11-28
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$f(x) = \int_0^x f'(t) dt$ so $|f(x)| = |\int_0^x f'(t) dt| \le \int_0^x |f'(t)| dt \le \int_0^x M dt = M x $.

Note: this seems too easy, so I might be misapplying something or assuming something that is not necessarily true. Or I might actually be right - that sometimes happens.

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    Unfortunately, not this time :(. You assumed the derivative is integrable. See [this](http://en.wikipedia.org/wiki/Volterra's_function).2012-11-28
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    Oh well, that's the way the cookie bounces.2012-12-01
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Suppose you had an $x$ such that $|f(x)| > M|x|.$ See what the mean value theorem gives you applied to the interval $(0,x).$

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    Im unsure how to do it2012-11-28
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    @Maximiliano Do you know the full statement of the mean value theorem?2012-11-28
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    Yes its the what mhenni wrote up above and it what I have in my textbook. But its that last part which Im stuck how to apply it to the problem f(x)=sin x2012-11-28