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Homework Help: Mean Value Theorem Question

  1. Nov 12, 2006 #1
    Use the Intermediate Value Theorem and/or the Mean Value Theorem and/or properties of [tex]G'(x)[/tex] to show that the function [tex]G(x) = x^2 - e^{\frac{1}{1+x}}[/tex] assumes a value of 0 for exactly one real number x such that 0 < x < 2 . Hint: You may assume that [tex]e^{\frac{1}{3}} < 2 [/tex].

    So I'm completely lost.

    Here's the first thing I tried:

    [tex]G(x) = 0[/tex]

    [tex] 0 = x^2 - e^{\frac{1}{1+x}} [/tex]

    [tex]e^{\frac{1}{1+x}} = x^2[/tex]

    [tex] lne^{\frac{1}{1+x}} = lnx^2[/tex]

    [tex] \frac{1}{1+x} = 2lnx [/tex]

    [tex] 1 = 2(1+x)lnx [/tex]

    [tex] \frac{1}{2} = (1+x)lnx[/tex]

    Which isn't quite getting me anywhere.

    And I'm not sure how the mean value theorem is going to help me out much more since:

    [tex]G'(c) = \frac{G(2) - G(0)}{2-0}[/tex]

    [tex]2c + (1+c)e^{\frac{1}{1+c}} = \frac{[2^2-e^{\frac{1}{3}}] - [0^2 - e]}{2}[/tex]

    [tex] 4c + (2+2c)e^{\frac{1}{1+c}} = 4 - e^{\frac{1}{3}} + e [/tex]

    Which isn't going to simply easily.

    Oy vey...

    Any help or hints would be appreciated.

    Edit: There, I think I finally got it to render correctly.
    Last edited: Nov 12, 2006
  2. jcsd
  3. Nov 12, 2006 #2


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    Intermediate value theorem: G(0)<0, G(2)>0, so there is at least one root between. Now, how can you show it's going to be unique using G'(x)?
  4. Nov 12, 2006 #3
    Hrm, that's what I'm not sure of.

    I know the x value is within the interval, due to the intermediate value theorem, as you stated.

    I could suggest that I solve [tex]G'(x)=0[/tex], but I could imagine that would be useless as there's no reason to think that [tex]G(c)=0[/tex] is a critical point.

  5. Nov 12, 2006 #4
    Look at [tex] G'(x)[/tex] and decide what its properties are for 0<x<2. Then decide what that tells you about what [tex]G(x)[/tex] is doing on (0,2).
  6. Nov 13, 2006 #5
    Heh... I noticed I did my derivative entirely wrong. I guess that's why you shouldn't use the computer when you're tired.



    Which is < 0

    [tex]=4-e^{\frac{1}{3}} [/tex]

    Which is > 0

    Meaning [tex]G(x)=0[/tex] is contained somewhere between x=0 and x=2.

    [tex]G'(x)=2x + {e^{\frac{1}{1+x}}\over{(1+x)^2}}[/tex]

    But I'm not having an easy time trying to solve this equal to zero to find my critical numbers. It seems relatively safe to assume that [tex]G(x)[/tex] is increasing on this interval.
    Last edited: Nov 13, 2006
  7. Nov 16, 2006 #6
    Can you see whether G'(x) is positive or negative for positive x?
  8. Nov 18, 2006 #7
    I can. But how does that help me determine the value of X at which G(X) equals zero? :(
  9. Nov 18, 2006 #8


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    Homework Helper

    Try graphing it.
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