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

Dec 5, 2017

Decreasing; see explanation

#### Explanation:

In order to determine the solution, we must find the derivative $f ' \left(3\right)$

The Quotient Rule states that, for $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)} , f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$

In this case, we have via the power rule: $g \left(x\right) = 3 {x}^{3} + 3 {x}^{2} + 5 x + 2 , g ' \left(x\right) = 9 {x}^{2} + 6 x + 5 , h \left(x\right) = x - 2 , h ' \left(x\right) = 1$

Thus:

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

We can simplify further if we wish, but for the purposes of this problem it's unnecessary.

Now find $f ' \left(c\right)$ by plugging in c...

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

$= \frac{\left(81 + 18 + 5\right) \left(1\right) - \left(81 + 27 + 15 + 2\right)}{{1}^{2}} = - \frac{21}{1} = - 21$

Because the derivative is negative at this point, $f \left(x\right)$ is decreasing at $x = 3$

Viewing the graph below, we can verify this is correct.

graph{(3x^3 + 3x^2 + 5x + 2)/(x-2) [-2.947, 9.2, 121.787, 127.86]}