Lagranges mean value Theorem

**zorro** · December 31st 2008, 09:10 PM

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

**kalagota** · December 31st 2008, 11:36 PM

Originally Posted by zorro

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

apply MVT to the function $f(x) = e^x$ on the interval $[0,x]$

**zorro** · December 1st 2009, 09:59 PM

Originally Posted by kalagota

apply MVT to the function $f(x) = e^x$ on the interval $[0,x]$

Use Langrange MVT

$<br /> f'(c) = \frac{e^x - 1}{x}<br />$

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

**mr fantastic** · December 2nd 2009, 01:25 AM

Originally Posted by zorro

Use Langrange MVT

$<br /> f'(c) = \frac{e^x - 1}{x}<br />$

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 $e^x = 1 + e^c x$ (the steps should be basic at this level and are left for you to fill in). Since c > 0 the required result easily follows.

**zorro** · December 3rd 2009, 01:27 AM

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 $e^x = 1 + e^c x$ (the steps should be basic at this level and are left for you to fill in). Since c > 0 the required result easily follows.

$[a,b] = [0,x]$

$f(b) = e^x$

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

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

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 .....

**HallsofIvy** · December 3rd 2009, 02:25 AM

Originally Posted by zorro

$[a,b] = [0,x]$

$f(b) = e^x$

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

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

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

**mr fantastic** · December 3rd 2009, 02:35 AM

Originally Posted by zorro

$[a,b] = [0,x]$

$f(b) = e^x$

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

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

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.

**zorro** · December 3rd 2009, 01:24 PM

Originally Posted by HallsofIvy

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

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

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

**Drexel28** · December 3rd 2009, 01:27 PM

Originally Posted by zorro

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

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

Friend, what we have is that $e^x=1+e^c\cdot x$ for $c\in(0,x)$ . But! $e^z$ is an increasing function on that interval so that $e^c>e^0=1\quad \forall c\in(0,x)$ so then $e^x=1+e^c\cdot x>1+e^0\cdot x=1+x$ .

**mr fantastic** · December 3rd 2009, 04:12 PM

Originally Posted by Drexel28

Friend, what we have is that $e^x=1+e^c\cdot x$ for $c\in(0,x)$ . But! $e^z$ is an increasing function on that interval so that $e^c>e^0=1\quad \forall c\in(0,x)$ so then $e^x=1+e^c\cdot x>1+e^0\cdot x=1+x$ .

Indeed. Which is what I said in posts #4 and #7 ....

**Drexel28** · December 3rd 2009, 04:16 PM

Originally Posted by mr fantastic

Indeed. Which is what I said in posts #4 and #7 ....

Yes, I agree.

**zorro** · December 27th 2009, 07:17 PM

Originally Posted by Drexel28

Yes, I agree.

Solution :

Here $f(x) = e^x$

$f'(x) = \ \frac{d}{dx} e^x = \ e^x$

Using Lagrange Mean value theorem

$f'(c) \ = \ \frac{f(b) - f(a)}{b - a} \qquad [a,b]$

$f(a) = f(0) = e^0 = 1$ and $f(b) = f(x) = e^x since [a,b] = [0,x]$

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

$f(x) = e^x$

Differentiating wrt x weget

$f'(x) = e^x$ ..........................(2)

Now comparing eq(1) and (2) weget

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

$e^c = \frac{e^x - 1}{x }$ ..........[since $f'(x) = e^x \therefore f'(c) = e^c$ ]

$x. e^c + 1 = e^x$ .............................(3)

Taking c = 0 eq (3) becomes

$x. e^c + 1 = e^x$

$x+1 = e^x$ ......................................(4)

Taking c = x in eq(3) weget

$x. e^x + 1 = e^x$ ..............................(5)

Therefore u can see that the 2 term is less that the first term
$\therefore x. e^x + 1 > e^x$ ............................(6)

Now from (4) and (5) we can write the same eq (6) as

$e^x > x+1$

**Drexel28** · December 27th 2009, 11:54 PM

Originally Posted by zorro

Taking c = x in eq(3) weget

$x. e^x + 1 = e^x$ ..............................(5)

Therefore u can see that the 2 term is less that the first term
$\therefore x. e^x + 1 > e^x$ ............................(6)

Now from (4) and (5) we can write the same eq (6) as

$e^x > x+1$

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 $e^x$ is increasing on $[0,x]$ so that $e^x\geqslant e^0=1\implies x\cdot e^c+1\leqslant x+1$

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