Critical numbers

Jun 2010
23
0
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

MHF Helper
Aug 2008
12,897
5,001
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.