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

Mar 31, 2018

$f \left(x\right)$ is increasing at $x = - 3.$

#### Explanation:

If $f ' \left(- 3\right) > 0 , f \left(x\right)$ is increasing at that point.

If $f ' \left(- 3\right) < 0 , f \left(x\right)$ is decreasing at that point.

So, first, let's determine $f ' \left(x\right)$ using the Quotient Rule:

$f ' \left(x\right) = \frac{\left({x}^{2} + 1\right) \frac{d}{\mathrm{dx}} \left({x}^{2} - 4 x - 2\right) - \left({x}^{2} - 4 x - 2\right) \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 2$

$f ' \left(x\right) = \frac{\left({x}^{2} + 1\right) \left(2 x - 4\right) - \left({x}^{4} - 4 x - 2\right) \left(2 x\right)}{{x}^{2} + 1} ^ 2$

There's not really much of a need for simplification since we're evaluating the derivative at a point only to determine whether it is positive or negative.

$f ' \left(- 3\right) = \frac{\left(9 + 1\right) \left(2 \left(- 3\right) - 4\right) - \left(81 + 12 - 2\right) \left(- 6\right)}{{10}^{2}} = \frac{7}{50} > 0$

$f \left(x\right)$ is increasing at $x = - 3.$