# Lagrange's Mean Value Theorem

• December 15th 2008, 09:41 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.
• December 16th 2008, 12:05 AM
I think you wrote
Quote:

$e^x>1+x, x>0$
Assume what you are trying to prove is not true ie. there exists y>0 such that $e^y<1+y$. Then the mean value theorem tells us that for some $x = c\in(0,y), \frac{d}{dx}e^x<\frac{1+y-1}{y}$. Find a contradiction.
• December 19th 2008, 05:09 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)
• December 20th 2008, 07:31 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, $\mathrm{e}^{x}$ is positive, increasing and convex for $x\geq0$.
Then, for $x\geq0$, the function $\mathrm{e}^{x}$ is always above its tangent lines; i.e., $\mathrm{e}^{x}\geq \mathrm{e}^{x_{0}}(x-x_{0})+\mathrm{e}^{x_{0}}$ for any $x,x_{0}\geq0$ (recall that a tangent line of a function $f$ at the point $x_{0}$ is given by $g(x):=f^{\prime}(x_{0})(x-x_{0})+f(x_{0})$).
Letting $x_{0}=0$, we get the desired result.