# Thread: Lagranges mean value Theorem

1. ## Lagranges mean value Theorem

State the Langrage mean value theorem .Hence show that
$\displaystyle e^x>1+x$ , x>0

2. Originally Posted by zorro
State the Langrage mean value theorem .Hence show that
$\displaystyle e^x>1+x$ , x>0
apply MVT to the function $\displaystyle f(x) = e^x$ on the interval $\displaystyle [0,x]$

3. ## Could u check if its right or no

Originally Posted by kalagota
apply MVT to the function $\displaystyle f(x) = e^x$ on the interval $\displaystyle [0,x]$

Use Langrange MVT

$\displaystyle f'(c) = \frac{e^x - 1}{x}$

What should i do next mate ......I am totally clueless from here on

4. Originally Posted by zorro
Use Langrange MVT

$\displaystyle f'(c) = \frac{e^x - 1}{x}$

What should i do next mate ......I am totally clueless from here on
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.

5. ## My work ,please let me know what i have done wrongly

Originally Posted by mr fantastic
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 .....

6. Originally Posted by zorro
$\displaystyle [a,b] = [0,x]$

$\displaystyle f(b) = e^x$

$\displaystyle f(a) = e^0 = 1$

$\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{f(x) - f(0)}{x-0} = \frac{e^x - 1}{x}$

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 .....
You want to arrive at an inequality $\displaystyle e^x> x+ 1$, not an equation. What is the smallest possible value for f(c) on $\displaystyle [0, \infty)$?

7. Originally Posted by zorro
$\displaystyle [a,b] = [0,x]$

$\displaystyle f(b) = e^x$

$\displaystyle f(a) = e^0 = 1$

$\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a} = \frac{f(x) - f(0)}{x-0} = \frac{e^x - 1}{x}$

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 .....
f(x) = e^x => f'(x) = e^x => f'(c) = e^c. Substitute this into then go back and read my previous post.

8. ## Is this correct

Originally Posted by HallsofIvy
You want to arrive at an inequality $\displaystyle e^x> x+ 1$, not an equation. What is the smallest possible value for f(c) on $\displaystyle [0, \infty)$?

The smallest possible value f(c) in $\displaystyle [0,\infty]$ is 0 and $\displaystyle e^0 = 1$. And if we go by that then $\displaystyle x + 1 = e^x$

I need help , could u please provide some material , this is getting confusing.

9. Originally Posted by zorro
The smallest possible value f(c) in $\displaystyle [0,\infty]$ is 0 and $\displaystyle e^0 = 1$. And if we go by that then $\displaystyle x + 1 = e^x$

I need help , could u please provide some material , this is getting confusing.
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 $\displaystyle e^x=1+e^c\cdot x>1+e^0\cdot x=1+x$.

10. Originally Posted by Drexel28
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 $\displaystyle e^x=1+e^c\cdot x>1+e^0\cdot x=1+x$.
Indeed. Which is what I said in posts #4 and #7 ....

11. Originally Posted by mr fantastic
Indeed. Which is what I said in posts #4 and #7 ....
Yes, I agree.

12. ## Here is what i have done

Originally Posted by Drexel28
Yes, I agree.

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

$\displaystyle e^x > x+1$

13. Originally Posted by zorro

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

$\displaystyle e^x > x+1$
You are sincerely missing the point. The above step is just...just I don't even know.You are done once you note that since $\displaystyle e^x$ is increasing on $\displaystyle [0,x]$ so that $\displaystyle e^x\geqslant e^0=1\implies x\cdot e^c+1\leqslant x+1$