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

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

Answer 1

$\text{f(x) is increasing at x=0}$

#### Explanation:

To determine if a function f(x) 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"

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

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) . g ' \left(x\right) - g \left(x\right) . h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$g \left(x\right) = - {x}^{3} + {x}^{2} - 3 x - 4 \Rightarrow g ' \left(x\right) = - 3 {x}^{2} + 2 x - 3$

$\text{and } h \left(x\right) = 4 x - 2 \Rightarrow h ' \left(x\right) = 4$

$\Rightarrow f ' \left(x\right)$

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

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

$= \frac{22}{4} = \frac{11}{2}$

Since f'(0) > 0 , then f(x) is increasing at x = 0
graph{(-x^3+x^2-3x-4)/(4x-2) [-20, 20, -10, 10]}

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