# Is f(x)=(x^3+5x^2-7x+2)/(x+1) increasing or decreasing at x=0?

May 5, 2018

Decreasing.

#### Explanation:

A function is increasing at $x$ when its derivative evaluated at $x$ is positive. That is, when $f ' \left(x\right) > 0$. Similarly, a function is decreasing at $x$ when $f ' \left(x\right) < 0$.

In order to test whether our function is increasing or decreasing at $x = 0$, we need to find it's derivative. Note that we use the quotient rule.

$f ' \left(x\right) = \frac{\left(x + 1\right) \left(3 {x}^{2} + 10 x - 7\right) - \left({x}^{3} + 5 {x}^{2} - 7 x + 1\right)}{x + 1} ^ 2$
$= \frac{3 {x}^{3} + 10 {x}^{2} - 7 x + 3 {x}^{2} + 10 x - 7 - {x}^{3} - 5 {x}^{2} + 7 x - 1}{x + 1} ^ 2$
$= \frac{2 {x}^{3} + 8 {x}^{2} + 10 x - 8}{x + 1} ^ 2$

Evaluating this at zero gives:

$f ' \left(0\right) = - 8$

Since $f ' \left(0\right) < 0$, $f \left(x\right)$ is decreasing at $x = 0$.