Question

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`?

Answer 1

`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]}