# Finding stationary points

• Jul 21st 2010, 07:01 AM
Glitch
Finding stationary points
The question:
Find and identify the stationary points for \$\displaystyle y = e^{2x}(1 - x)\$

My attempt:
I differentiated the function as per here.

But I'm not sure how to solve for x when \$\displaystyle y^{'} = e^{2x}(3 - 2x) = 0\$. I tried using Wolfram Alpha, but it said there are no solutions. My text states that there's a minimum point at (1/2, e/2), however.

How can I solve this? Assistance is appreciated.
• Jul 21st 2010, 07:08 AM
Quote:

Originally Posted by Glitch
The question:
Find and identify the stationary points for \$\displaystyle y = e^{2x}(1 - x)\$

My attempt:
I differentiated the function as per here.

But I'm not sure how to solve for x when \$\displaystyle y^{'} = e^{2x}(3 - 2x) = 0\$. I tried using Wolfram Alpha, but it said there are no solutions. My text states that there's a minimum point at (1/2, e/2), however.

How can I solve this? Assistance is appreciated.

dy/dx=e^(2x)(-1)+(1-x)(2)(e^(2x))=e^(2x)[1-2x]

So when you set that to be 0, either e^(2x)=0 or 1-2x=0

but e^(2x) cannot be 0, so the only solution is 1/2.
• Jul 21st 2010, 07:09 AM
Ackbeet
With your corrected derivative, and using this command, I was able to get WolframAlpha to produce the x value that solves the equation (although, really, it's quite straight-forward to solve by inspection: the exponential is never zero, and the linear multiplier term (1-2x) is zero when x = 1/2. Done.) What command did you use to try to find the root?

[EDIT] This is the same as mathaddict's post, essentially.
• Jul 21st 2010, 07:14 AM
Glitch
Aha! Thanks guys!