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

Oct 8, 2016

Increasing

#### Explanation:

To determine if the graph is increasing or decreasing at a certain point, we can use the first derivative.

• For values in which $f ' \left(x\right) > 0$, $f \left(x\right)$ is increasing as the gradient is positive.
• For values in which $f ' \left(x\right) < 0$, $f \left(x\right)$ is decreasing as the gradient is negative.

Differentiating $f \left(x\right)$,

We have to use quotient rule.
$f ' \left(x\right) = \frac{u ' v - v ' u}{v} ^ 2$
Let $u = {x}^{2} - 3 x - 2$ and $v = x + 1$
then $u ' = 2 x - 3$ and $v ' = 1$

So $f ' \left(x\right) = \frac{\left(2 x - 3\right) \left(x + 1\right) - \left({x}^{2} - 3 x - 2\right)}{x + 1} ^ 2 = \frac{{x}^{2} + 2 x - 1}{x + 1} ^ 2$

Subbing in $x = 1$,
$f ' \left(x\right) = \frac{{1}^{2} + 2 \left(1\right) - 1}{1 + 1} ^ 2 = \frac{1}{2} , \therefore f ' \left(x\right) > 0$

Since the $f ' \left(x\right) > 0$ for $x = 1$, $f \left(x\right)$ is increasing at $x = 1$