# Is f(x)=(x^3-4x^2-4x+5)/(x+2) increasing or decreasing at x=3?

Jul 27, 2017

$\text{increasing at } x = 3$

#### Explanation:

$\text{to determine if f(x) is increasing/decreasing at x = a}$
$\text{differentiate and evaluate at x = a}$

• " if "f'(a)>0" then f(x) is increasing at x = a"

• " if " f'(a)<0" then f(x) is decreasing at x = a"

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

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

$g \left(x\right) = {x}^{3} - 4 {x}^{2} - 4 x + 5 \Rightarrow g ' \left(x\right) = 3 {x}^{2} - 8 x - 4$

$h \left(x\right) = x + 2 \Rightarrow h ' \left(x\right) = 1$

$\Rightarrow f ' \left(x\right) = \frac{\left(x + 2\right) \left(3 {x}^{2} - 8 x - 4\right) - \left({x}^{3} - 4 {x}^{2} - 4 x + 5\right)}{x + 2} ^ 2$

$\Rightarrow f ' \left(3\right) = \frac{5 \left(- 1\right) - \left(- 16\right)}{25} = \frac{11}{25}$

$f ' \left(3\right) > 0 \text{ hence f(x) is increasing at x = 3}$
graph{(x^3-4x^2-4x+5)/(x+2) [-10, 10, -5, 5]}