# Critical numbers

#### GameTheory

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?

#### Prove It

MHF Helper
You have

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

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

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

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

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

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

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

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

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

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

$$\displaystyle e^{x} = 2$$

$$\displaystyle x = \ln{2}$$.

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