# Is f(x)=(x^3-2x^2+5x-4)/(x-2) increasing or decreasing at x=0?

Nov 21, 2016

$f ' \left(0\right)$ is negative, therefore $f \left(x\right)$ is decreasing at $x = 0$.

#### Explanation:

This is the same as asking if $f ' \left(x\right)$ is positive or negative at $x = 0$.
We have to first find $f ' \left(x\right)$, then plug in 0 to get $f ' \left(0\right)$.

$f \left(x\right) = \frac{{x}^{3} - 2 {x}^{2} + 5 x - 4}{x - 2}$

$f ' \left(x\right) = \frac{\left(x - 2\right) \left(3 {x}^{2} - 4 x + 5\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 4\right) \left(1\right)}{{\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \frac{\left(3 {x}^{3} - 4 {x}^{2} + 5 x - 6 {x}^{2} + 8 x - 10\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 4\right)}{{\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \frac{\left(3 {x}^{3} - 10 {x}^{2} + 13 x - 10\right) - \left({x}^{3} - 2 {x}^{2} + 5 x - 4\right)}{{\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \frac{2 {x}^{3} - 8 {x}^{2} + 8 x - 6}{{\left(x - 2\right)}^{2}}$

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

$f ' \left(0\right) = \frac{\left(2\right) \left(- 3\right)}{{\left(- 2\right)}^{2}}$
$= \frac{- 6}{4}$
$= - \frac{3}{2}$

$f ' \left(0\right)$ is negative, therefore $f \left(x\right)$ is decreasing at $x = 0$.