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

May 31, 2016

increasing at x = 0

#### Explanation:

To determine if a function is increasing/decreasing at x = a ,consider the following.

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

$f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} \text{ then } f ' \left(x\right) = \frac{h \left(x\right) . g ' \left(x\right) - g \left(x\right) . h ' \left(x\right)}{h \left(x\right)} ^ 2$
$\text{--------------------------------------------------------------------}$
$g \left(x\right) = {x}^{3} - 4 {x}^{2} + 3 x - 4 \Rightarrow g ' \left(x\right) = 3 {x}^{2} - 8 x + 3$

$h \left(x\right) = 4 x - 2 \Rightarrow h ' \left(x\right) = 4$
$\text{-------------------------------------------------------------------}$
Substitute these values into f'(x)

$f ' \left(x\right) = \frac{\left(4 x - 2\right) \left(3 {x}^{2} - 8 x + 3\right) - \left({x}^{3} - 4 {x}^{2} + 3 x - 4\right) .4}{4 x - 2} ^ 2$

and $f ' \left(0\right) = \frac{\left(- 2\right) \left(+ 3\right) - \left(- 4\right) .4}{- 2} ^ 2 = \frac{5}{2}$

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