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

Jan 28, 2016

f(x) is decreasing at x = 1

#### Explanation:

To find if the function is increasing / decreasing at x = 1

Require to check the value of f'(x) at x = 1.

• If f')x) > 0 then increasing

• If f'(x) < 0 then decreasing

differentiate using $\textcolor{b l u e}{\text{ quotient rule }}$

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

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

$= \frac{- 3 {x}^{4} - 2 {x}^{3} - {x}^{2} - 9 {x}^{3} - 6 {x}^{2} - 3 x - \left(- 2 {x}^{4} - 3 {x}^{3} - 2 {x}^{3} - 3 {x}^{2} - 2 {x}^{2} - 3 x + 4 x + 6\right)}{{x}^{2} + 3 x} ^ 2$

$= \frac{- 3 {x}^{4} - 11 {x}^{3} - 7 {x}^{2} - 3 x + 2 {x}^{4} + 5 {x}^{3} + 5 {x}^{2} - 4 x - 6}{{x}^{2} + 3 x} ^ 2$

$= \frac{- {x}^{4} - 6 {x}^{3} - 2 {x}^{2} - 7 x - 6}{{x}^{2} + 3 x} ^ 2$

$f ' \left(1\right) = \frac{- 1 - 6 - 2 - 7 - 6}{1 + 3} ^ 2 = - \frac{22}{16} = - \frac{11}{8}$

now f'(1) < 0 hence f(x) is decreasing at x = 1

The graph of f(x) illustrates decreasing at x = 1

graph{(-x3-x^2-x+2)/(x^2+3x) [-10, 10, -5, 5]}