# Showing $f(0) = 0$ and $|f'(x)| \leq M$ implies $|f(x)| \leq M |x|$.

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? – DonAntonio Nov 28 '12 at 5:06

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)? – Maximiliano Nov 28 '12 at 6:39
is it |sin x| <= M |x|? – Maximiliano Nov 28 '12 at 6:42
@Maximiliano: When $f(x)=\sin(x)$, we have $f'(c)=\cos(c)$ which implies $|f'(c)|=|\cos(c)|\leq 1 = M.$ – Mhenni Benghorbal Nov 28 '12 at 8:36

$f(x) = \int_0^x f'(t) dt$ 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. – Katie Dobbs Nov 28 '12 at 13:32
Oh well, that's the way the cookie bounces. – marty cohen Dec 1 '12 at 6:07

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 it – Maximiliano Nov 28 '12 at 5:24
@Maximiliano Do you know the full statement of the mean value theorem? – Katie Dobbs Nov 28 '12 at 5:25
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 x – Maximiliano Nov 28 '12 at 6:24