Question

On what interval isthe function f(x)=x^3.e^x increasing and diecreasing?

Answer 1

Decreasing in `(-oo,-3]` , Increasing in `[-3,+oo)`

Explanation:

`f(x)=x^3e^x` , `x` `in` `RR`

We notice that `f(0)=0`

`f'(x)=(x^3e^x)'=3x^2e^x+x^3e^x=x^2e^x(3+x)`

`f'(x)=0` `<=>` `(x=0,x=-3)`

• When `x` `in` `(-oo,-3)` for example for `x=-4` we get

`f'(-4)=-16/e^4<0`

• When `x` `in` `(-3,0)` for example for `x=-2` we get

`f'(-2)=4/e^2>0`

• When `x` `in` `(0,+oo)` for example for `x=1` we get

`f'(1)=4e>0`

`f` is continuous in `(-oo,-3]` and `f'(x)<0` when `x` `in` `(-oo,-3)` so `f` is strictly decreasing in `(-oo,-3]`

`f` is continuous in `[-3,0]` and `f'(x)>0` when `x` `in` `(-3,0)` so `f` is strictly increasing in `[-3,0]`

`f` is continuous in `[0,+oo)` and `f'(x)>0` when `x` `in` `(0,+oo)` so `f` is strictly increasing in `[0,+oo)`

`f` is increasing in `[-3,0)uu(0,+oo)` and `f` is continuous at `x=0` , hence `f` is strictly increasing in `[-3,+oo)`

Here is a graph which will help you see how this function behaves

graph{x^3e^x [-4.237, 1.922, -1.736, 1.34]}