Mean Value Theorem: Proving Inequalities with f'(x)

[kidia]

Please help me on this.

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`]

 

[lurflurf]

Please help me on this.

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`]

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)

 

[kidia]

please lurflurf clarify more and give me the values

 

[lurflurf]

please lurflurf clarify more and give me the values

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')

 

[kidia]

Thanx a lot lurflurf