Another interval prob: decr/incr for f(x) = xe^(x^2)

[hank]

Find where the intervals are decreasing/increasing for f(x) = xe^(x^2)

Step one: Find derivative using product rule.
= x*e^(x^2) + (1) * e^(x^2) //e raised to x squared.
= xe^(x^2) + e^(x^2)
= e^(x^2) (x + 1)

Step two: Set that mess to zero and find critical points.
e^(x^2) (x + 1) = 0
(x + 1) = 0 //divide both sides by e^(x^2) to get rid of it.
x = -1

Step three: Plug in test values into f'(x) to find intervals.
f'(-2) = ---, therefore interval is decreasing on (-inf, -1]
f'(2) = +++, therefore interval is increasing on [-1, inf)

I don't think it's right, and I think I screwed up finding the derivative, or setting it to zero.

I'm thinking that I should have set e^(x^2) to zero and solved for x, but that would be impossible as ln0 is undefined.

Help?

 

[galactus]

Re: Another interval problem

Find where the intervals are decreasing/increasing for f(x) = xe^(x^2)

Step one: Find derivative using product rule.
= x*e^(x^2) + (1) * e^(x^2) //e raised to x squared.
= xe^(x^2) + e^(x^2)
= e^(x^2) (x + 1)

Product rule:

\(\displaystyle \L\\x(2xe^{x^{2}})+(e^{x^{2}})(1)=(2x^{2}+1)e^{x^{2}}\)

Note the graph. Do you see critical points?. What does the derivative say about that?.

Check the limit of your function.

\(\displaystyle \L\\\lim_{x\to{-}\infty}(xe^{x^{2}})={-}\infty\)

\(\displaystyle \L\\\lim_{x\to\infty}(xe^{x^{2}})=+\infty\)

If these hold, then the function has neither a max nor a min. on \(\displaystyle (-\infty,+\infty)\)

Now, try them again.

 

[hank]

Ok, I figured out how you got to the result you did.
Now, my question is does this mean there is no increasing/decreasing intervals?

If I set e^x^2 to 0, it is undefined because I would have to take ln of both sides and ln0 is undefined.

Also, 2x^2 = 1 becomes the sqrt of -1/2 which is also undefined.

 

[skeeter]

f'(x) = (2x² + 1)e^x²

2x² + 1 > 0 for all x

e^x² > 0 for all x

so ... f'(x) > 0 for all x

f(x) is increasing over its entire domain. have a look at galactus' graph.