# Is f(x)=(3x^3-5x^2+13x-11)/(x-3) increasing or decreasing at x=2?

Jan 11, 2016

Increasing

#### Explanation:

Use the quotient rule of differentiation to find the gradient. Then if the gradient is negative the function is decreasing, and if it is positive the function is increasing.
If $f \left(x\right) = g \frac{x}{h} \left(x\right)$
$f ' \left(x\right) = \frac{g \left(x\right) h ' \left(x\right) - g ' \left(x\right) h \left(x\right)}{{h}^{2} \left(x\right)}$

$f ' \left(x\right) = \frac{\left(3 {x}^{3} - 5 {x}^{2} + 13 x - 11\right) \left(1\right) - \left(9 {x}^{2} - 10 x + 13\right) \left(x - 3\right)}{x - 3} ^ 2$
$= \frac{\left(3 {x}^{3} - 5 {x}^{2} + 13 x - 11\right) - \left(9 {x}^{3} - 10 {x}^{2} + 13 x - 27 {x}^{2} + 30 x - 39\right)}{x - 3} ^ 2$
$= \frac{- 6 {x}^{3} + 32 {x}^{2} - 30 x + 28}{x - 3} ^ 2$
$= \frac{- 2 \left(3 {x}^{3} - 16 {x}^{2} + 15 x - 14\right)}{x - 3} ^ 2$
Substituting in $x = 2$ gives $\frac{- 2 \left(24 - 64 + 30 - 14\right)}{1}$
$= - 2 \left(54 - 78\right) = - 2 \cdot \left(- 24\right) = 48$

The gradient (slope) of $48$ is positive and the function is therefore increasing.