# Mean Value Theorem Question

by calcnd
Tags: theorem
 P: 20 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.
 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 $$G'(x)=0$$, but I could imagine that would be useless as there's no reason to think that $$G(c)=0$$ is a critical point. :S
 P: 18 Mean Value Theorem Question Look at $$G'(x)$$ and decide what its properties are for 0
 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, $$G(0)=0^2-e^(\frac{1}{1+0})$$ $$=-e$$ 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.