- #1

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Use the Mean Value Theorom (for the derivative of a function) to prove that f(x)>=1-x for x>=0

also

Mean Value Theorom states:

[f(b)-f(a)]/ [b-a]= f'(c) where c is an element of [a,b]

- Thread starter sean/mac
- Start date

- #1

- 8

- 0

Use the Mean Value Theorom (for the derivative of a function) to prove that f(x)>=1-x for x>=0

also

Mean Value Theorom states:

[f(b)-f(a)]/ [b-a]= f'(c) where c is an element of [a,b]

- #2

daniel_i_l

Gold Member

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what do you get for f(x) when you use the MVT with b=x, a=0? What are the limits on c?

- #3

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gives

[1/(1+x) - 1]/x=-1/(1+c)^2

which when i do the algebraic manipulation gives

1+x=(1+c)^2

i don't know how to make a relation between 1-x and 1/(1+x)

- #4

HallsofIvy

Science Advisor

Homework Helper

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What is the largest possible value for 1/(1+c)

gives

[1/(1+x) - 1]/x=-1/(1+c)^2

which when i do the algebraic manipulation gives

1+x=(1+c)^2

i don't know how to make a relation between 1-x and 1/(1+x)

- #5

- 350

- 1

You already went too far, you want to keep f (don't substitute) on the lhs to get that inequality, so keep it like this:[f(b) - f(a)]/[b-a]=f'(c) for b=x and a=0

gives

[1/(1+x) - 1]/x=-1/(1+c)^2

[tex](f(x) -1)/x = -1/(1+c^2)[/tex]

and then follow HallsofIvy's hint.

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