# Is f(x)=(-2x^3+x^2-2x-4)/(4x-2) increasing or decreasing at x=0?

Jan 7, 2018

$\text{increasing at x = 0}$

#### Explanation:

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

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

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

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

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

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

$g \left(x\right) = - 2 {x}^{3} + {x}^{2} - 2 x - 4 \Rightarrow g ' \left(x\right) = - 6 {x}^{2} + 2 x - 2$

$h \left(x\right) = 4 x - 2 \Rightarrow h ' \left(x\right) = 4$

$f ' \left(x\right) = \frac{\left(4 x - 2\right) \left(- 6 {x}^{2} + 2 x - 2\right) - 4 \left(- 2 {x}^{3} + {x}^{2} - 2 x - 4\right)}{4 x - 2} ^ 2$

$\Rightarrow f ' \left(0\right) = \frac{\left(- 2\right) \left(- 2\right) - 4 \left(- 4\right)}{4} = 5 > 0$

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