Is f(x)=(3x^3+x^2-2x-14)/(x-1) increasing or decreasing at x=2?

Oct 13, 2016

At $x = 2$, $f \left(x\right) = \frac{3 {x}^{3} + {x}^{2} - 2 x - 14}{x - 1}$ is increasing.

Explanation:

For any function, if $\frac{\mathrm{df}}{\mathrm{dx}}$ is positive, it is increasing, if $\frac{\mathrm{df}}{\mathrm{dx}}$ is negative, it is decreasing and if if $\frac{\mathrm{df}}{\mathrm{dx}} = 0$, it is flat.

in the case of $f \left(x\right) = \frac{3 {x}^{3} + {x}^{2} - 2 x - 14}{x - 1}$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\left(x - 1\right) \left(9 {x}^{2} + 2 x - 2\right) - \left(3 {x}^{3} + {x}^{2} - 2 x - 14\right) \times 1}{x - 1} ^ 2$

= $\frac{9 {x}^{3} + 2 {x}^{2} - 2 x - 9 {x}^{2} - 2 x + 2 - 3 {x}^{3} - {x}^{2} + 2 x + 14}{x - 1} ^ 2$

= $\frac{6 {x}^{3} - 8 {x}^{2} - 2 x + 16}{x - 1} ^ 2$

And at $x = 2$,

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{6 \cdot {2}^{3} - 8 \cdot {2}^{2} - 2 \cdot 2 + 16}{2 - 1} ^ 2$

= $\frac{48 - 32 - 4 + 16}{1} = 28$

Hence, at $x = 2$, $\frac{3 {x}^{3} + {x}^{2} - 2 x - 14}{x - 1}$ is increasing.