How do you Use the Mean Value Theorem to prove this

**maximus101** · February 17th 2011, 04:26 AM

Use the Mean Value Theorem to prove that if p > 1 then $(1+x)^p$ $>$ $1+px$ for
$x$ $\in$ (−1, 0) U (0, $\infty$ ).

**Sudharaka** · February 17th 2011, 06:52 AM

Originally Posted by maximus101

Use the Mean Value Theorem to prove that if p > 1 then $(1+x)^p$ $>$ $1+px$ for
$x$ $\in$ (−1, 0) U (0, $\infty$ ).

Dear maximus101,

Using the Taylor's theorem you could write,

$(1+x)^p=1+px+p(p-1)(1+c)^{p-2}x^2~\text{where}~0<c<x~or~x<c<0$ ------(A)

$p>1\Rightarrow{p(p-1)>0}$ -----(1)

If $0<c<x~then~1<1+c$ -----(2)

If $x<c<0~then~{1+x<1+c<1}$ ----(3)

$x>-1\Rightarrow{1+x>0}$ ------(4)

By, (3) and (4); $If~x<c<0\Rightarrow{0<1+c<1}$ ----(5)

By (2) and (5); for both cases, $0<c<x~and~x<c<0\Rightarrow{0<1+c}$ ------(6)

Therefore, by (1) and (6); $p(p-1)(1+c)^{p-2}x^2>0~if~x\neq{0}~and~x>-1$ -----(6)

By (A);

$(1+x)^p-1-px=p(p-1)(1+c)^{p-2}x^2>0~if~x\neq{0}~and~x>-1$

$(1+x)^p>1+px~if~x>-1~and~x\neq{0}$

**maximus101** · February 19th 2011, 05:24 AM

Hey, thank you for this it was very helpful could you explain how I could use the mean value theorem to prove it?

**Sudharaka** · February 19th 2011, 05:50 AM

Originally Posted by maximus101

Hey, thank you for this it was very helpful could you explain how I could use the mean value theorem to prove it?

There are several mean value theorems; Rolle's mean value theorem, Cauchy's mean value theorem and the Taylor's mean value theorem. Please refer Mean value theorem - Wikipedia, the free encyclopedia. I have used the Taylor's mean value theorem in the above answer.

**maximus101** · February 19th 2011, 06:08 AM

Originally Posted by Sudharaka

There are several mean value theorems; Rolle's mean value theorem, Cauchy's mean value theorem and the Taylor's mean value theorem. Please refer Mean value theorem - Wikipedia, the free encyclopedia. I have used the Taylor's mean value theorem in the above answer.

Hey, sorry for my mistake, I meant the first one showing

$f'(c)$ = $\frac{f(b)-f(a)}{b-a}$

kind thanks

**Sudharaka** · February 19th 2011, 06:12 AM

Originally Posted by maximus101

Hey, sorry for my mistake, I meant the first one showing

$f'(c)$ = $\frac{f(b)-f(a)}{b-a}$

kind thanks

That is also know as Lagrange's mean value theorem. It is used to prove Taylor's theorem. Therefore it had been used in the answer.

**maximus101** · February 19th 2011, 06:34 AM

ok I worked it out

thank you

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