# Is f(x)=(x^3+3x^2-4x-9)/(x+1) increasing or decreasing at x=0?

Jan 11, 2017

It's increasing, because the slope is positive at $x = 0$.

graph{(x^3 + 3x^2 - 4x - 9)/(x+1) [-5, 5, -12, 5]}

The first derivative tells you whether the function is decreasing ($\frac{\mathrm{df}}{\mathrm{dx}} < 0$), increasing ($\frac{\mathrm{df}}{\mathrm{dx}} > 0$), or at an extremum (a turning point, $\frac{\mathrm{df}}{\mathrm{dx}} = 0$).

For this, I would actually try to do some quick synthetic division to simplify it if possible. I would guess $x = - 1$ in hopes of cancelling out the denominator, but if it doesn't work, that's OK. It'll at least become simpler.

$\underline{- 1} | \text{ "1" "3" "-4" } - 9$
$\text{-----------------------------------}$
$\text{ "" "" } 1$

Multiply the divisor (the factor you guessed, $- 1$) by the term you just brought down ($1$), and place it under the next column ($- 1 \times 1 = - 1$).

$\underline{- 1} | \text{ "1" "3" "-4" } - 9$
$\text{ "" "" "" } - 1$
$\text{-----------------------------------}$
$\text{ "" "" } 1$

Add the current column, and repeat the previous steps:

$\underline{- 1} | \text{ "1" "3" "-4" } - 9$
$\text{ "" "" "" } - 1$
$\text{-----------------------------------}$
$\text{ "" "" "1" } 2$

$\underline{- 1} | \text{ "1" "3" "-4" } - 9$
$\text{ "" "" "" "-1" } - 2$
$\text{-----------------------------------}$
$\text{ "" "" "1" "2" } - 6$

$\underline{- 1} | \text{ "1" "3" "-4" } - 9$
$\text{ "" "" "" "-1" "-2" "" } 6$
$\text{-----------------------------------}$
$\text{ "" "" "1" "2" "-6" } - 3$

You started with a cubic, so you end up with a quadratic. The remainder is $- 3$, and one way of writing it is to leave it divided by $x + 1$ (your denominator):

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

This is indeed easier to differentiate, so now:

$\frac{d}{\mathrm{dx}} \left[{x}^{2} + 2 x - 6 - \frac{3}{x + 1}\right]$

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

Now, plug in $x = 0$ to find out what is going on with this function at $x = 0$:

$f ' \left(0\right) = | \left[\frac{\mathrm{dy}}{\mathrm{dx}}\right] {|}_{x = 0} = 2 \left(0\right) + 2 + \frac{3}{0 + 1} ^ 2 = 5 > 0$

Therefore, the function is increasing at $x = 0$.