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Thread: Lagranges mean value Theorem

  1. #1
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    Lagranges mean value Theorem

    State the Langrage mean value theorem .Hence show that
    $\displaystyle e^x>1+x$ , x>0
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by zorro View Post
    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 $\displaystyle [0,x]$
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  3. #3
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    Could u check if its right or no

    Quote Originally Posted by kalagota View Post
    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
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  4. #4
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    Quote Originally Posted by zorro View Post
    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.
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    My work ,please let me know what i have done wrongly

    Quote Originally Posted by mr fantastic View Post
    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 .....
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    Quote Originally Posted by zorro View Post
    $\displaystyle [a,b] = [0,x]$

    $\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 $\displaystyle [0, \infty)$?
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    Quote Originally Posted by zorro View Post
    $\displaystyle [a,b] = [0,x]$

    $\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.
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    Is this correct

    Quote Originally Posted by HallsofIvy View Post
    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zorro View Post
    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 $\displaystyle e^x=1+e^c\cdot x>1+e^0\cdot x=1+x$.
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  10. #10
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    Quote Originally Posted by Drexel28 View Post
    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 ....
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Indeed. Which is what I said in posts #4 and #7 ....
    Yes, I agree.
    Last edited by mr fantastic; Dec 3rd 2009 at 04:19 PM. Reason: Quite right ;) But I don't want anyone getting upset.
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  12. #12
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    Here is what i have done

    Quote Originally Posted by Drexel28 View Post
    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

    $\displaystyle e^x > x+1$
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  13. #13
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by zorro View Post

    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 $\displaystyle e^x\geqslant e^0=1\implies x\cdot e^c+1\leqslant x+1$
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