- #1

kidia

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

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- Thread starter kidia
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- #1

kidia

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

- #2

lurflurf

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wellkidia said:

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

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

kidia

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please lurflurf clarify more and give me the values

- #4

lurflurf

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Okkidia said:please lurflurf clarify more and give me the values

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

kidia

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Thanx a lot lurflurf

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