Mean Value Theorem Question


by calcnd
Tags: theorem
calcnd
calcnd is offline
#1
Nov12-06, 03:22 PM
P: 20
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.
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Office_Shredder
Office_Shredder is offline
#2
Nov12-06, 04:05 PM
Mentor
P: 4,499
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)?
calcnd
calcnd is offline
#3
Nov12-06, 04:25 PM
P: 20
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.

:S

slearch
slearch is offline
#4
Nov12-06, 04:53 PM
P: 18

Mean Value Theorem Question


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).
calcnd
calcnd is offline
#5
Nov13-06, 08:23 PM
P: 20
Heh... I noticed I did my derivative entirely wrong. I guess that's why you shouldn't use the computer when you're tired.

Anyways,

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

Which is < 0

[tex]G(2)=2^2-e^{\frac{1}{1+2}}[/tex]
[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.
dustball
dustball is offline
#6
Nov16-06, 09:29 PM
P: 7
Can you see whether G'(x) is positive or negative for positive x?
calcnd
calcnd is offline
#7
Nov18-06, 02:26 PM
P: 20
I can. But how does that help me determine the value of X at which G(X) equals zero? :(
StatusX
StatusX is offline
#8
Nov18-06, 02:36 PM
HW Helper
P: 2,566
Try graphing it.


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