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

Feb 24, 2016

increasing at x = 1

#### Explanation:

To check if function is increasing / decreasing , require to find the value of f'(1)

• If f'(1) > 0 then f(x) is increasing at x = 1

• If f'(1) < 0 then f(x) is decreasing at x = 1

Differentiate using the $\textcolor{b l u e}{\text{ Quotient rule }}$

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

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

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

$f ' \left(1\right) = \frac{\left(1 - 3\right) \left(2 - 2\right) - \left(1 - 2 - 2\right)}{1 - 3} ^ 2 = \frac{0 - \left(- 3\right)}{4} = \frac{3}{4}$

$\Rightarrow \text{ since " f'(1) > 0 " then f(x) is increasing at x = 1 }$
This can be seen on graph of function.
graph{(-x^2-2x-2)/(x-3) [-10, 10, -5, 5]}