# Thread: How do you Use the Mean Value Theorem to prove this

1. ## How do you Use the Mean Value Theorem to prove this

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

2. Originally Posted by maximus101
Use the Mean Value Theorem to prove that if p > 1 then $\displaystyle (1+x)^p$ $\displaystyle >$ $\displaystyle 1+px$ for
$\displaystyle x$ $\displaystyle \in$ (−1, 0) U (0,$\displaystyle \infty$).
Dear maximus101,

Using the Taylor's theorem you could write,

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

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

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

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

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

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

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

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

By (A);

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

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

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

4. 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.

5. 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

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

kind thanks

6. Originally Posted by maximus101
Hey, sorry for my mistake, I meant the first one showing

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

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.

7. ok I worked it out thank you