# Is f(x)=(-2x^3+9x^2-5x+6)/(x-2) increasing or decreasing at x=0?

Aug 24, 2016

$f$ is $\uparrow$ at $x = 0$.

#### Explanation:

We know that, $f$ is $\uparrow \mathmr{and} \downarrow a t x = 0$ according as f'(0) >, or, < 0"#.

So, we need to check $f ' \left(0\right)$.

Now, $f \left(x\right) = \frac{- 2 {x}^{3} + 9 {x}^{2} - 5 x + 6}{x - 2}$

$= \frac{\underline{- 2 {x}^{3} + 4 {x}^{2}} + \underline{5 {x}^{2} - 10 x} + \underline{5 x - 10} + 16}{x - 2}$

$= \frac{- 2 {x}^{2} \left(x - 2\right) + 5 x \left(x - 2\right) + 5 \left(x - 2\right) + 16}{x - 2}$

$= \left\{\frac{\cancel{\left(x - 2\right)} \left(- 2 {x}^{2} + 5 x + 5\right)}{\cancel{\left(x - 2\right)}} + \frac{16}{x - 2}\right\}$

$\therefore f \left(x\right) = - 2 {x}^{2} + 5 x + 5 + \frac{16}{x - 2} \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

Knowing that, $\frac{d}{\mathrm{dt}} \left(\frac{1}{t}\right) = - \frac{1}{t} ^ 2 , \text{we have, by} \left(1\right)$,

$f ' \left(x\right) = - 4 x + 5 - \frac{16}{x - 2} ^ 2$

$\Rightarrow f ' \left(0\right) = 5 - \frac{16}{- 2} ^ 2 = 5 - 4 = 1 > 0$

Hence, $f$ is $\uparrow$ at $x = 0$.

Enjoy Maths.!