# Mean Value Theorem

1. Aug 4, 2005

### kidia

If f(x) is continuous on [a,b],apply the Mean Value Theorem to prove the inequalities min[f] $$\leq$$ f(b)-f(a)$$/_$$ b-a $$\leq$$ [Max f]

2. Aug 4, 2005

### lurflurf

well
f'(x)=(f(b)-f(a))/(b-a)
for some x in (a,b)
so what are the biggest and smallest possible values of
(f(b)-f(a))/(b-a)

3. Aug 4, 2005

### kidia

please lurflurf clarify more and give me the values

4. Aug 4, 2005

### lurflurf

Ok
remember a<x<b
x* is some number in the interval so
f'(x*)=(f(b)-f(a))/(b-a)
M>=f'(x) for all x
M>=(f(b)-f(a))/(b-a)
max(f')>=f'(x) for all x
so
max(f')>=f(x*)=(f(b)-f(a))/(b-a)
the min case is analogous

The idea is f'(x) may be greater than, less than, or equal to (f(b)-f(a))/(b-a).
min(f')<=f'(x)<=max(f')
so since for some x*
f'(x*)=(f(b)-f(a))/(b-a)
and for all x
min(f')<=f'(x)<=max(f')
then
min[f]<=f(b)-f(a) b-a <=[Max f`]
since
min(f')<=f'(x)<=max(f')
is true for all x
including
min(f')<=f'(x*)<=max(f')

5. Aug 5, 2005

### kidia

Thanx a lot lurflurf