Does G(x) = x^2 - e^{1/(1+x)} Assume a Value of 0 for 0 < x < 2?

[calcnd]

Use the Intermediate Value Theorem and/or the Mean Value Theorem and/or properties of G&#039;(x) to show that the function G(x) = x^2 - e^{\frac{1}{1+x}} assumes a value of 0 for exactly one real number x such that 0 < x < 2 . Hint: You may assume that e^{\frac{1}{3}} &lt; 2.

So I'm completely lost.

Here's the first thing I tried:

G(x) = 0

0 = x^2 - e^{\frac{1}{1+x}}

e^{\frac{1}{1+x}} = x^2

lne^{\frac{1}{1+x}} = lnx^2

\frac{1}{1+x} = 2lnx

1 = 2(1+x)lnx

\frac{1}{2} = (1+x)lnx

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:

G&#039;(c) = \frac{G(2) - G(0)}{2-0}

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

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

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.

 

[Office_Shredder]

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]

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 G&#039;(x)=0, but I could imagine that would be useless as there's no reason to think that G(c)=0 is a critical point.

:S

 

[slearch]

Look at G&#039;(x) and decide what its properties are for 0<x<2. Then decide what that tells you about what G(x) is doing on (0,2).

 

[calcnd]

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,

G(0)=0^2-e^(\frac{1}{1+0})
=-e

Which is < 0

G(2)=2^2-e^{\frac{1}{1+2}}
=4-e^{\frac{1}{3}}

Which is > 0

Meaning G(x)=0 is contained somewhere between x=0 and x=2.

G&#039;(x)=2x + {e^{\frac{1}{1+x}}\over{(1+x)^2}}

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 G(x) is increasing on this interval.

 

[dustball]

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

 

[calcnd]

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

 

[StatusX]

Try graphing it.