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Mean Value Theorem Questionby calcnd
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#1
Nov1206, 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)}{20}[/tex] [tex]2c + (1+c)e^{\frac{1}{1+c}} = \frac{[2^2e^{\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. 


#2
Nov1206, 04:05 PM

Emeritus
Sci Advisor
PF Gold
P: 4,500

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)?



#3
Nov1206, 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 


#4
Nov1206, 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).



#5
Nov1306, 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)=0^2e^(\frac{1}{1+0})[/tex] [tex]=e[/tex] Which is < 0 [tex]G(2)=2^2e^{\frac{1}{1+2}}[/tex] [tex]=4e^{\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. 


#6
Nov1606, 09:29 PM

P: 7

Can you see whether G'(x) is positive or negative for positive x?



#7
Nov1806, 02:26 PM

P: 20

I can. But how does that help me determine the value of X at which G(X) equals zero? :(



#8
Nov1806, 02:36 PM

HW Helper
P: 2,567

Try graphing it.



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