- #1

calcnd

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- 0

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.

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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