# Mean Value Theorem Question

1. Nov 12, 2006

### calcnd

Use the Intermediate Value Theorem and/or the Mean Value Theorem and/or properties of $$G'(x)$$ 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 $$e^{\frac{1}{3}} < 2$$.

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:

$$G'(c) = \frac{G(2) - G(0)}{2-0}$$

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

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

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.

Last edited: Nov 12, 2006
2. Nov 12, 2006

### Office_Shredder

Staff Emeritus
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. Nov 12, 2006

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

:S

4. Nov 12, 2006

### slearch

Look at $$G'(x)$$ 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).

5. Nov 13, 2006

### 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.

$$G'(x)=2x + {e^{\frac{1}{1+x}}\over{(1+x)^2}}$$

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.

Last edited: Nov 13, 2006
6. Nov 16, 2006

### dustball

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

7. Nov 18, 2006

### calcnd

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

8. Nov 18, 2006

### StatusX

Try graphing it.