# Lagrange's Mean Value Theorem

• Dec 15th 2008, 08:41 PM
Shivanand
Lagrange's Mean Value Theorem
Can u plzz solve this question... Using Lagrange mean value theorem show that e Rest to x > 1 + x, x > 0.
• Dec 15th 2008, 11:05 PM
I think you wrote
Quote:

$\displaystyle e^x>1+x, x>0$
Assume what you are trying to prove is not true ie. there exists y>0 such that $\displaystyle e^y<1+y$. Then the mean value theorem tells us that for some$\displaystyle x = c\in(0,y), \frac{d}{dx}e^x<\frac{1+y-1}{y}$. Find a contradiction.
• Dec 19th 2008, 04:09 AM
varunnayudu
Please is there any other way
Please sir is there any other weay because this is very confussing .If possible can u give me a reference from where i can get a complete explaination about hw u came to this conclusion....(Bow)
• Dec 20th 2008, 06:31 AM
bkarpuz
Quote:

Originally Posted by Shivanand
Can u plzz solve this question... Using Lagrange mean value theorem show that e Rest to x > 1 + x, x > 0.

Clearly, $\displaystyle \mathrm{e}^{x}$ is positive, increasing and convex for $\displaystyle x\geq0$.
Then, for $\displaystyle x\geq0$, the function $\displaystyle \mathrm{e}^{x}$ is always above its tangent lines; i.e., $\displaystyle \mathrm{e}^{x}\geq \mathrm{e}^{x_{0}}(x-x_{0})+\mathrm{e}^{x_{0}}$ for any $\displaystyle x,x_{0}\geq0$ (recall that a tangent line of a function $\displaystyle f$ at the point $\displaystyle x_{0}$ is given by $\displaystyle g(x):=f^{\prime}(x_{0})(x-x_{0})+f(x_{0})$).
Letting $\displaystyle x_{0}=0$, we get the desired result.