Find the extrema for
f(x)=0.5e^(-x^2)

Question

Find the extrema for 
f(x)=0.5e^(-x^2)

anonymous · Answer

$$f(x)=0.5e ^{-x ^{2}}$$

anonymous · Answer

I got the derivative

$$f'(x)=-e ^{-x ^{2}}$$

But I never took college algebra and never learned about e. I don't know how to solve for x.

jhannybean · Answer

$$\large f(x) = 0.5e^{-x^2}$$$$\large f'(x)=(0.5)(-2x)e^{-x^2}$$$$\large f'(x) = -xe^{-x^2}$$

anonymous · Answer

Oh yeah. I missed an x typing.

jhannybean · Answer

Now you have to set $f'(x) = -xe^{-x^2} =0$ and solve for x to get your critical points. Look up the extreme value theorem for further assistance. http://tutorial.math.lamar.edu/Classes/CalcI/AbsExtrema.aspx

anonymous · Answer

Yeah, I've read everything I can find on the extreme value theorem online and in my textbook. I understand it. I don't understand HOW to solve for x in that equation. I even know the answer. I just don't know how to do it. Thanks anyway.

jhannybean · Answer

Alright. So you've gotten your first derivative, and you've set it equal to 0, but now you see that theyre multiplied to eachother. 

Evaluate them separately. 
$$\large -x=0$$$$\large e^{-x^2} = 0$$
solving for x = 0, divide both sides by -1, you'll find that x=0.

Once you understand the domain of $e^x$ you'll find that the function never passes through 0, it comes really close to it. (If you've got a graphing calculator you can graph the function $e^{-x^2}$. 

So your only extrema in this case would be 0.

anonymous · Answer

Thank you!

jhannybean · Answer

I'd suggest reading up on your natural log functions to better understand the function e.

anonymous · Answer

I know. I've done that. My Calc teacher and my stats teacher have both tried to help me with that. I have a college algebra book sitting open next to me with that section open. It doesn't help. I'm old and it doesn't stick in my brain. :)

jhannybean · Answer

:) I had to practice a ton of problems with derivatives and integrals of e to see how the function worked.