# Is f(x)=(x^3+3x^2-x-9)/(x+1) increasing or decreasing at x=-2?

Feb 25, 2016

Increasing.

#### Explanation:

To determine if a function is increasing or decreasing at a point, examine the sign of the first derivative at that point.

• If $f ' \left(- 2\right) < 0$, then $f \left(x\right)$ is decreasing at $x = - 2$.
• If $f ' \left(- 2\right) > 0$, then $f \left(x\right)$ is increasing at $x = - 2$.

To find $f ' \left(x\right)$, use the quotient rule.

This gives us:

$f ' \left(x\right) = \frac{\left(x + 1\right) \frac{d}{\mathrm{dx}} \left({x}^{3} + 3 {x}^{2} - x - 9\right) - \left({x}^{3} + 3 {x}^{2} - x - 9\right) \frac{d}{\mathrm{dx}} \left(x + 1\right)}{x + 1} ^ 2$

Find each derivative through the power rule.

$f ' \left(x\right) = \frac{\left(x + 1\right) \left(3 {x}^{2} + 6 x - 1\right) - \left({x}^{3} + 3 {x}^{2} - x - 9\right)}{x + 1} ^ 2$

Simplification of the numerator yields:

$f ' \left(x\right) = \frac{2 {x}^{3} + 6 {x}^{2} + 6 x + 8}{x + 1} ^ 2$

Find the sign of the derivative at $x = - 2$:

$f ' \left(- 2\right) = \frac{2 \left(- 8\right) + 6 \left(4\right) + 6 \left(- 2\right) + 8}{- 2 + 1} ^ 2 = 4$

Since $f ' \left(2\right) = 4 > 0$, the function is increasing at $x = - 2$. We can check the graph of $f \left(x\right)$:

graph{(x^3+3x^2-x-9)/(x+1) [-7.87, 6.18, -1.176, 5.85]}