Critical numbers

**GameTheory** · July 6th 2010, 11:39 PM

I'm asked to find the critical numbers of f(x)=2((e^-x)-(e^-2x))

I know you have to find the derivative first which I got:

f'(x)=2((-e^-x)+(2e^-2x))
f'(x)= (-2e^-x)+(4e^-2x)

But now, how do I go about finding one of the x? I believe you must rewrite it as a quadratic function but since the exponent is -2 indicating it is a reciprocal.

So:

(4(e^x)^-2)-(2(e^x)^-1)=0

Factoring: Divided by 4
(1/e^x)((1/e^x)-(1/2))=0

I can set ((1/e^x)-(1/2))=0 which will be x = ln 2

But when it comes to (1/e^x)=0 I'm at a COMPLETE LOSS. How do I do this?

Thank you in advance

**Prove It** · July 6th 2010, 11:44 PM

You have

$4e^{-2x} - 2e^{-x} = 0$

$4(e^{-x})^2 - 2e^{-x} = 0$

$4X^2 - 2X = 0$ where $X = e^{-x}$

$2X(2X - 1) = 0$

$2X = 0$ or $2X - 1 = 0$

$X = 0$ or $X = \frac{1}{2}$ .

Therefore $e^{-x} = 0$ or $e^{-x} = \frac{1}{2}$ .

But since $e^{-x} > 0$ for all $x$ , that means

$e^{-x} = \frac{1}{2}$

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

$e^{x} = 2$

$x = \ln{2}$ .

So there is a critical value at $x = \ln{2}$ . Use the second derivative test to determine its nature, and then substitute it back into the original function to find the corresponding $y$ value.

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