f(x)=xe^-x^2. What interval is this function increasing.

To determine the interval on which a function is increasing, we need to find the values of x where the derivative is positive.

Let's start by finding the derivative of the given function, f(x)=xe^(-x^2). We can use the product rule and chain rule for this calculation.

Using the product rule: (uv)' = u'v + uv'

Let u = x and v = e^(-x^2), we have:
f'(x) = (x)'(e^(-x^2)) + (x)(e^(-x^2))'

Now, let's find the derivatives of u and v:
u' = 1 (since the derivative of x is 1)
v' = (e^(-x^2))' = -2x * e^(-x^2) (by the chain rule)

Substituting these derivatives into the expression for f'(x), we get:
f'(x) = 1 * e^(-x^2) + x * (-2x * e^(-x^2))

Simplifying further:
f'(x) = e^(-x^2) - 2x^2 * e^(-x^2)

To find the interval on which the function is increasing, we need to look for values of x where f'(x) > 0.

Setting f'(x) > 0:
e^(-x^2) - 2x^2 * e^(-x^2) > 0

We can factor out e^(-x^2):
e^(-x^2) (1 - 2x^2) > 0

Since e^(-x^2) is always positive, we can focus on the inequality:
1 - 2x^2 > 0

Solving for x, we get:
-2x^2 + 1 > 0

Rearranging the inequality, we have:
-2x^2 > -1

Dividing both sides by -2 (and reversing the inequality since we're dividing by a negative number), we get:
x^2 < 1/2

Taking the square root of both sides, we find:
|x| < sqrt(1/2)

Since we're looking for the interval, we need to consider both positive and negative values of x:
- sqrt(1/2) < x < sqrt(1/2)

Therefore, the function f(x)=xe^(-x^2) is increasing on the interval (-sqrt(1/2), sqrt(1/2)).