State the Langrage mean value theorem .Hence show that
$\displaystyle e^x>1+x$ , x>0
next, you should state completely what the mean Value Theorem says since you have stated/used only part of it.
Then, from the equation you posted it follows that $\displaystyle e^x = 1 + e^c x$ (the steps should be basic at this level and are left for you to fill in). Since c > 0 the required result easily follows.
$\displaystyle [a,b] = [0,x]$
$\displaystyle f(b) = e^x $
$\displaystyle f(a) = e^0 = 1$
$\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{f(x) - f(0)}{x-0} = \frac{e^x - 1}{x}$
is this right ?
between [0,x] what should i put as the c
could u please explain me or provide me with a link through which i could understand .....I am a little weak with the Lagrange MVT .....
Friend, what we have is that $\displaystyle e^x=1+e^c\cdot x$ for $\displaystyle c\in(0,x)$. But! $\displaystyle e^z$ is an increasing function on that interval so that $\displaystyle e^c>e^0=1\quad \forall c\in(0,x)$ so then $\displaystyle e^x=1+e^c\cdot x>1+e^0\cdot x=1+x$.
Solution :
Here $\displaystyle f(x) = e^x$
$\displaystyle f'(x) = \ \frac{d}{dx} e^x = \ e^x$
Using Lagrange Mean value theorem
$\displaystyle f'(c) \ = \ \frac{f(b) - f(a)}{b - a} \qquad [a,b]$
$\displaystyle f(a) = f(0) = e^0 = 1$ and$\displaystyle f(b) = f(x) = e^x since [a,b] = [0,x]$
$\displaystyle f'(c) = \ \frac{f(b) - f(a)}{b - a} = \ \frac{e^x - 1}{x - 0} = \frac{e^x - 1}{x }$.............(1)
$\displaystyle f(x) = e^x$
Differentiating wrt x weget
$\displaystyle f'(x) = e^x$..........................(2)
Now comparing eq(1) and (2) weget
$\displaystyle f'(c) = \frac{e^x - 1}{x }$
$\displaystyle e^c = \frac{e^x - 1}{x }$ ..........[since $\displaystyle f'(x) = e^x \therefore f'(c) = e^c$]
$\displaystyle x. e^c + 1 = e^x$.............................(3)
Taking c = 0 eq (3) becomes
$\displaystyle x. e^c + 1 = e^x$
$\displaystyle x+1 = e^x$......................................(4)
Taking c = x in eq(3) weget
$\displaystyle x. e^x + 1 = e^x$..............................(5)
Therefore u can see that the 2 term is less that the first term
$\displaystyle \therefore x. e^x + 1 > e^x $............................(6)
Now from (4) and (5) we can write the same eq (6) as
$\displaystyle e^x > x+1$