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

May 26, 2018

Decreasing in $\left(- \infty , - 3\right]$ , Increasing in $\left[- 3 , + \infty\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} {e}^{x}$ , $x$$\in$$\mathbb{R}$

We notice that $f \left(0\right) = 0$

$f ' \left(x\right) = \left({x}^{3} {e}^{x}\right) ' = 3 {x}^{2} {e}^{x} + {x}^{3} {e}^{x} = {x}^{2} {e}^{x} \left(3 + x\right)$

$f ' \left(x\right) = 0$ $\iff$ $\left(x = 0 , x = - 3\right)$

• When $x$$\in$$\left(- \infty , - 3\right)$ for example for $x = - 4$ we get

$f ' \left(- 4\right) = - \frac{16}{e} ^ 4 < 0$

• When $x$$\in$$\left(- 3 , 0\right)$ for example for $x = - 2$ we get

$f ' \left(- 2\right) = \frac{4}{e} ^ 2 > 0$

• When $x$$\in$$\left(0 , + \infty\right)$ for example for $x = 1$ we get

$f ' \left(1\right) = 4 e > 0$

$f$ is continuous in $\left(- \infty , - 3\right]$ and $f ' \left(x\right) < 0$ when $x$$\in$$\left(- \infty , - 3\right)$ so $f$ is strictly decreasing in $\left(- \infty , - 3\right]$

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

$f$ is continuous in $\left[0 , + \infty\right)$ and $f ' \left(x\right) > 0$ when $x$$\in$$\left(0 , + \infty\right)$ so $f$ is strictly increasing in $\left[0 , + \infty\right)$

$f$ is increasing in $\left[- 3 , 0\right) \cup \left(0 , + \infty\right)$ and $f$ is continuous at $x = 0$ , hence $f$ is strictly increasing in $\left[- 3 , + \infty\right)$

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]}