# Increasing and decreasing interval, natural exponential function

• Nov 14th 2008, 09:06 AM
angrynapkin
Increasing and decreasing interval, natural exponential function
Given: f(x) = e^(-2x^2)

f'(x) evaluated as -4xe^(-2x^2)

Now, every example I've seen thus far is a polynomial function, which makes it easy to find the roots and determine what interval it's increasing or decreasing, not so here. Any tips?
• Nov 14th 2008, 03:05 PM
mr fantastic
Quote:

Originally Posted by angrynapkin
Given: f(x) = e^(-2x^2)

f'(x) evaluated as -4xe^(-2x^2)

Now, every example I've seen thus far is a polynomial function, which makes it easy to find the roots and determine what interval it's increasing or decreasing, not so here. Any tips?

The solution to \$\displaystyle -4x e^{-2x^2} = 0\$ is x = 0 since \$\displaystyle e^{-2x^2} \neq 0\$ for real values of x.

Note also that \$\displaystyle e^{-2x^2} > 0\$ for all real values of x. So f'(x) > 0 => -4x > 0 and f'(x) < 0 => -4x < 0 ....
• Nov 14th 2008, 03:19 PM
angrynapkin
mr fantastic, thank you for your reply. I've worked through the problem and studied notes a little more carefully. The second derivative test returns concave upward for all real values, which is correct according to the graph of the original function.

f"(x) = 4e^(2x^2)[(1+4x^2)]

I set 1+4x^2 to zero as my two critical points and tested away.

Not too sure how you evaluated 4xe^(2x^2) = 0

From the graph, I know it has to be zero, but algebraically I'm not quite sure how to work that out (other than the assumption it HAS to equal zero)
• Nov 14th 2008, 03:39 PM
mr fantastic
Quote:

Originally Posted by angrynapkin
mr fantastic, thank you for your reply. I've worked through the problem and studied notes a little more carefully. The second derivative test returns concave upward for all real values, which is correct according to the graph of the original function.

f"(x) = 4e^(2x^2)[(1+4x^2)]

I set 1+4x^2 to zero as my two critical points and tested away.

Not too sure how you evaluated 4xe^(2x^2) = 0

From the graph, I know it has to be zero, but algebraically I'm not quite sure how to work that out (other than the assumption it HAS to equal zero)

If A B = 0 then either A = 0 or B = 0 .....
• Nov 14th 2008, 03:48 PM
angrynapkin
Ok thanks, if something similar pops up on the final I just wanted something like that to write out.