# Is f(x)=(x^2e^x)/(x+2) increasing or decreasing at x=-1?

##### 1 Answer
Jan 22, 2016

Thus, $f \left(x\right)$ is decreasing at $x = - 1$.

#### Explanation:

To determine if $f \left(x\right)$ is increasing or decreasing at a specific $x$, first of all, you need to compute the derivative.

Here, you can use the quotient rule:

if $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, then the derivative can be computed as follows:

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$

For you, $g \left(x\right) = {x}^{2} {e}^{x}$ and $h \left(x\right) = x + 2$.

The derivatives of $g \left(x\right)$ and $h \left(x\right)$ are:

$g ' \left(x\right) = {x}^{2} {e}^{x} + 2 x {e}^{x}$ (with the product rule)

$h ' \left(x\right) = 1$

Thus, your derivative is:

$f ' \left(x\right) = \frac{\left({x}^{2} {e}^{x} + 2 x {e}^{x}\right) \left(x + 2\right) - {x}^{2} {e}^{x} \cdot 1}{x + 2} ^ 2$

$= \frac{{e}^{x} \left({x}^{2} + 2 x\right) \left(x + 2\right) - {e}^{x} {x}^{2}}{x + 2} ^ 2$

$= \frac{{e}^{x} \left[\left({x}^{2} + 2 x\right) \left(x + 2\right) - {x}^{2}\right]}{x + 2} ^ 2$

$= \frac{{e}^{x} \left({x}^{3} - 3 {x}^{2} + 4 x\right)}{x + 2} ^ 2$

Now, let's evaluate the derivative at $x = - 1$:

$f ' \left(- 1\right) = \frac{{e}^{- 1} \left({\left(- 1\right)}^{3} - 3 {\left(- 1\right)}^{2} + 4 \cdot \left(- 1\right)\right)}{- 1 + 2} ^ 2$

$= {e}^{- 1} \left(- 1 - 3 - 4\right)$

$= - 8 {e}^{- 1}$

As ${e}^{- 1} > 0$,

$f ' \left(- 1\right) = - 8 {e}^{- 1} < 0$.

Thus, $f \left(x\right)$ is decreasing at $x = - 1$.