How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for f(x) = x^2e^x - 3?

1 Answer
Jul 17, 2017

f(x) is strictly increasing in (-oo,-2), has a local maximum in x=-2, it is strictly decreasing in (-2,0) reaching a local minimum in x=0 and again increasing for x in (0,+oo)

Explanation:

Given:

f(x) = x^2e^x-3

consider the derivative of the function:

f'(x) = d/dx (x^2e^x-3) = (d/dx x^2)e^x + x^2 (d/dx e^x)

f'(x) = 2xe^x+x^2e^x

f'(x) = x(2+x)e^x

Solve now the inequality:

f'(x) > 0

As e^x > 0 AA x in RR, the sign of f'(x) is the sign of x(2+x), then we have:

f'(x) > 0 for x in (-oo,-2) uu (0,+oo)

f'(x) < 0 for x in (-2,0)

and the critical points where f'(x) = 0 are x_1 =-2 and x_2 =0

We can conclude that f(x) is strictly increasing in (-oo,-2), has a local maximum in x=-2, it is strictly decreasing in (-2,0) reaching a local minimum in x=0 and again increasing for x in (0,+oo)

graph{x^2e^x-3 [-10, 10, -5, 5]}