# Is f(x)=(x^3+2x^2-x-2)/(x+3) increasing or decreasing at x=-2?

Dec 8, 2015

Increasing.

#### Explanation:

Find the first derivative at the given point. If the value is positive, the function is increasing. If the value is negative the function is decreasing.

Finding $f ' \left(x\right)$ will require the quotient rule:

$f ' \left(x\right) = \frac{\left(x + 3\right) \frac{d}{\mathrm{dx}} \left[{x}^{3} + 2 {x}^{2} - x + 2\right] - \left({x}^{3} + 2 {x}^{2} - x - 2\right) \frac{d}{\mathrm{dx}} \left[x + 3\right]}{x + 3} ^ 2$

$= \frac{\left(x + 3\right) \left(3 {x}^{2} + 4 x - 1\right) - \left({x}^{3} + 2 {x}^{2} - x - 2\right)}{x + 3} ^ 2$

$= \frac{3 {x}^{3} + 13 {x}^{2} + 11 x - 3 - \left({x}^{3} + 2 {x}^{2} - x - 2\right)}{x + 3} ^ 2$

$= \frac{2 {x}^{3} + 11 {x}^{2} + 12 x - 1}{x + 3} ^ 2$

Now, calculate the derivative at the point.

$f ' \left(- 2\right) = \frac{2 \left(- 8\right) + 11 \left(4\right) + 12 \left(- 2\right) - 1}{- 2 + 3} ^ 2$

$= \frac{- 16 + 44 - 24 - 1}{1}$

$= 3$

Since $3 > 0$, the function is increasing when $x = - 2$.