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

Apr 26, 2016

decreasing at x = 1

#### Explanation:

To determine if a function is increasing/decreasing at x = a , we evaluate f'(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

differentiate f(x) using the $\textcolor{b l u e}{\text{ quotient rule }}$

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

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

and h(x) $= {x}^{2} + x \Rightarrow h ' \left(x\right) = 2 x + 1$
$\text{---------------------------------------------------------------}$
Substitute these values into f'(x)

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

and f'(1) $= \frac{2. \left(- 7\right) - \left(- 2\right) . \left(3\right)}{2} ^ 2 = \frac{- 8}{4} = - 2$

Since f'(1) < 0 , then f(x) is decreasing at x = 1
graph{(-3x^2-x+2)/(x^2+x) [-10, 10, -5, 5]}