Thread: Continuous function Mean value theorem

1. Continuous function Mean value theorem

Suppose that
f : [0, 1]--->R is continuously differentiable.

(a) Show that there is some number
M such that |f'(x)|< M for all x.

I understand mean value therorem but use it to solve this problem. Can you give me some hints please?

Thanks

2. A continuously differentiable function on a compact set means that it's continuous (in this case) on $\displaystyle [0,1]$ and with a continuous derivative on $\displaystyle [0,1].$ Hence, since the derivative is continuous on a compact set, it's bounded.

3. It seems to me like this would require the extreme value theorem (applied to $\displaystyle f'$), not the mean value theorem, but maybe I am just missing something.

4. From lecture notes: By the boundedness principle, a continuous function on a closed interval attains its maximum and minimum.

Therefore there exists a number M such that f(x)<=M for all x belong to [0,1]. How do i apply this to $\displaystyle |f'|$

I think i have to do something like.... how do i complete the answer? thanks
$\displaystyle 0<|f'|<max(m1,m2)$

5. Originally Posted by charikaar
From lecture notes: By the boundedness principle, a continuous function on a closed interval attains its maximum and minimum.

Therefore there exists a number M such that f(x)<=M for all x belong to [0,1]. How do i apply this to $\displaystyle |f'|$

I think i have to do something like.... how do i complete the answer? thanks
$\displaystyle 0<|f'|<max(m1,m2)$
You said "f : [0, 1]--->R is continuously differentiable." "Continuously differentiable" (as opposed to just "differentiable") means that f' is also continuous. Apply the boundedness principle to f', not f.