# Is f(x)=(-x^3+2x^2-3x+7)/(x-2) increasing or decreasing at x=-1?

Oct 24, 2016

So first we have to find the derivative of the function,

$y = f \frac{x}{g} \left(x\right)$

$y ' = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g} {\left(x\right)}^{2}$

so with,
$y = \frac{- {x}^{3} + 2 {x}^{2} - 3 x + 7}{x - 2}$

$f \left(x\right) = - {x}^{3} + 2 {x}^{2} - 3 x + 7$

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

and,

$g \left(x\right) = x - 2$

$g ' \left(x\right) = 1$

so subbing into the formula,

$y ' = \frac{\left(- 3 {x}^{2} + 4 x - 3\right) \left(x - 2\right) - \left(- {x}^{3} + 2 {x}^{2} - 3 x + 7\right)}{x - 2} ^ 2$

$y ' = \frac{- 2 {x}^{3} - 8 {x}^{2} + 8 x + 1}{x - 2} ^ 2$

now to find what the gradient is at $x = - 1$
we need to sub in $x = - 1$ into $y '$

$y ' = \frac{- 2 {\left(- 1\right)}^{3} - 8 {\left(- 1\right)}^{2} + 8 \left(- 1\right) + 1}{\left(- 1\right) - 2} ^ 2$

$y ' = \frac{17}{9}$

this means that the gradient is positive at $x = - 1$ and as a result the function in increasing at this value.