# proof of inequality with absolute-value and function

Prove that if $$|f'(x)| \leq M$$, for all $$x \in [a,b]$$, then $$f(a)-M(b-a)\leq f(b) \leq f(a)+M(b-a)$$

What I tried so far:

Proof. Let $$x \in [a,b]$$

Assume $$|f'(x)|\leq M$$

Show $$f(a)-(Mb-Ma)\leq f(b) \leq f(a)+(Mb-Ma)$$

By assumption that $$f'$$ defined on $$[a,b]$$, we have following:

$$f$$ is cts on [a,b]

Also differentiable on (a,b)

By MVT have $$\exists c \in (a,b) s.t.$$

$$f'(c)=\frac{f(b)-f(a)}{b-a}$$

by assumption also have $$-M\leq f'(x) \leq M$$

That $$c \in (a,b), have -M \leq f'(c) \leq M$$

Have $$-M \leq \frac{f(b)-f(a)}{b-a} \leq M$$

Therefore $$f(a)-M(b-a)\leq f(b) \leq f(a)+M(b-a)$$

• You are not using the hypothesis at all. How do you get $|f(x)| \leq M$? – Kavi Rama Murthy Jun 9 at 0:31
• @Kavi Rama Murthy Thanks, i just fixed the typo and rewrote my proof. – Manx Jun 9 at 3:11

This is false. (Perhaps you have copied $$|f'(x)| \leq M$$ as $$|f(x)| \leq M$$).
Counterexample: if $$f(x)=\sqrt x$$ on $$[0,1]$$ then the hypothesis holds with $$M=1$$ but the right hand inequality in the conclusion fails for $$a$$ and $$b$$ close to $$0$$.
Answer to the edited question: just apply MVT: $$f(b)-f(a)=(b-a)f'(x)$$ for some $$x$$ and $$-M \leq f'(x) \leq M$$. Can you complete the proof using this?
• You are asked to prove some inequalities involving $f(a)$ and $f(b)$. You are not asked to prove anything that involves an arbitrary point $x \in [a,b]$. My $x$ is same as your $c$. – Kavi Rama Murthy Jun 9 at 4:40