Question

On what interval is the function f (x) = x^3 e^x increasing?

Answer 1

`f(x)` is increasing in `(-3,0)`, `(0, oo)`.

Explanation:

STEP 1: Remember that our function, `f(x)`, is increasing when the derivative, `f'(x) > 0`. So our first step is to find `f'(x)`.

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

STEP 2: Find the zeros of the function.

`x = 0, x = -3`

STEP 3: Use a sign chart to determine when `f'(x) > 0` at each interval.

STEP 3a: Plug any value less than -3 into the `f'(x)`.

When `x < -3`, `f'(x) < 0`

STEP 3b: Plug in any value between -3 and 0 into `f'(x)`.

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

STEP 3c: Plug any value greater than 0 into the `f'(x)`.

when `x > 0`, `f'(x) > 0`.

ANSWER: `f(x)` is increasing when `f'(x) > 0`, which happens on the intervals `(-3,0)`, `(0, oo)`.

—Note that `f(x)` is not increasing at `x = 0`, because `f'(x)=0` there, which is not `> 0`

Answer 2

The interval of increasing of `f(x)` is `x in (-3,0) uu (0,+oo)`

Explanation:

We need

`(uv)'=u'v+uv'`

The function is

`f(x)=x^3e^x`

This is a product of two functions

`u(x)=x^3`, `=>`, `u'(x)=3x^2`

`v(x)=e^x`, `=>`, `v'(x)=e^x`

Therefore,

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

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

The critical points are when `f'(x)=0`

`e^x(3x^2+x^3)=0`

`AA x in RR, e^x>0`

`3x^2+x^3=0`, `=>`, `x^2(3+x)=0`

Therefore, the critical points are when

`x=0` and `x=-3`

We can make a sign chart

`color(white)(aaaa)` `Interval` `color(white)(aaaa)` `(-oo,-3)` `color(white)(aaaa)` `(-3,0)` `color(white)(aaaa)` `(0,+oo)`

`color(white)(aaaa)` `f'(x)` `color(white)(aaaaaaaaaaa)` `-` `color(white)(aaaaaaaaaa)` `+` `color(white)(aaaaaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaaaaa)` `↘` `color(white)(aaaaaaaaaa)` `↗` `color(white)(aaaaaaaa)` `↗`

The interval of increasing of `f(x)` is `x in (-3,0) uu (0,+oo)`

graph{x^3e^x [-12.66, 12.65, -6.33, 6.33]}