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

Jun 23, 2017

$\text{increasing at x = 1}$

#### Explanation:

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

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

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

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

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

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

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

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

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

$\rightarrow f ' \left(1\right) = \frac{2. \left(- 9\right) - \left(- 8\right) .3}{4}$

$\textcolor{w h i t e}{\Rightarrow f ' \left(1\right)} = \frac{6}{4} = \frac{3}{2}$

$\text{since " f'(1)>0" then f(x) is increasing at x = 1}$