SOLUTION: Please help me with this problem: Let f be the function f(x) = {{{xe^(-x^(2))}}} where x is a set of real numbers. Find the critical numbers of f. (It is helpful to note that {{{e

Algebra ->  Test -> SOLUTION: Please help me with this problem: Let f be the function f(x) = {{{xe^(-x^(2))}}} where x is a set of real numbers. Find the critical numbers of f. (It is helpful to note that {{{e      Log On


   



Question 1033806: Please help me with this problem:
Let f be the function f(x) = xe%5E%28-x%5E%282%29%29 where x is a set of real numbers. Find the critical numbers of f. (It is helpful to note that e%5E%28-x%5E%282%29%29is nonzero for any value of x.) Find the intervals on which f is increasing and on which f is decreasing. Use the information found to tell where f attains local maximum and minimum values.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+xe%5E%28-x%5E2%29 ==> f'(x) = %281-2x%5E2%29e%5E%28-x%5E2%29.
Let f'(x) be equal to 0.
==> 1-2x%5E2=0 ==> x+=+sqrt%282%29%2F2 or x+=+-sqrt%282%29%2F2, the critical numbers of f(x).
As you mentioned e%5E%28-x%5E2%29%3E0, so no solution comes from this.
In (-infinity, -sqrt%282%29%2F2), f'(x) < 0, so f(x) is decreasing there.
In (-sqrt%282%29%2F2,sqrt%282%29%2F2), f'(x) > 0, so f(x) is increasing there.
In (sqrt%282%29%2F2,infinity), f'(x) < 0, so f(x) is decreasing there.
Therefore f(x) attains local min at x+=+-sqrt%282%29%2F2, while it attains a local max at x+=+sqrt%282%29%2F2.