Determine the intervals where f(x)=e^-x^2/2 is increasing and decreasing

f(x) is increasing when f'(x)>0

ln(f)=-x^2/2 * e^(-x^2/2)

take the derivative.
f'/f(x)=-e^(-x^2/2)( -x)=xe^ ( )

then f'=xe^ ( ) * e^( )= xe^(-x^2)

check that.
so when x>0, f(x) is increasing.

To determine the intervals of increase and decrease for the function f(x) = e^(-x^2/2), we need to find the first derivative, f'(x), and analyze its sign.

Step 1: Find the first derivative of f(x)
Let's differentiate f(x) using the chain rule and the power rule:

f'(x) = d/dx [ e^(-x^2/2) ]
= -x * e^(-x^2/2)

Step 2: Analyze the sign of the first derivative
To determine where f(x) is increasing or decreasing, we need to find the intervals where f'(x) is positive or negative.

For f'(x) = -x * e^(-x^2/2) to be positive, two cases can occur:

Case 1: -x is positive and e^(-x^2/2) is positive.
In this case, x must be negative and larger than zero. However, since e^(-x^2/2) is always positive, it does not affect the sign. Therefore, we have x < 0.

Case 2: -x is negative and e^(-x^2/2) is negative.
In this case, x must be positive and smaller than zero. However, since e^(-x^2/2) is always positive, it cannot be negative. Therefore, there are no solutions for this case.

Step 3: Determine the intervals of increase and decrease
Based on the above analysis, f(x) is increasing for x < 0 and decreasing for x > 0.

In summary:
- f(x) is increasing for x < 0
- f(x) is decreasing for x > 0.