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

Jan 29, 2016

Increasing.

Explanation:

Find the value of the derivative at $x = - 3$. If it is positive, the function is increasing. If it's negative, the function is decreasing (at that point).

To find the derivative of the function, use the quotient rule.

$f ' \left(x\right) = \frac{\left({x}^{2} + 1\right) \frac{d}{\mathrm{dx}} \left[- {x}^{2} - 5 x - 2\right] - \left(- {x}^{2} - 5 x - 2\right) \frac{d}{\mathrm{dx}} \left[{x}^{2} + 1\right]}{{x}^{2} + 1} ^ 2$

Each of these derivatives can be found through the power rule:

$\frac{d}{\mathrm{dx}} \left[- {x}^{2} - 5 x - 2\right] = - 2 x - 5$

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

Plugging these back in yields

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

Distribution and simplification of the numerator gives a derivative of

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

We can now find the value of the derivative at $x = - 3 :$

$f ' \left(- 3\right) = \frac{5 \left(9\right) - 6 - 5}{9 + 1} ^ 2 = \frac{34}{100} = \frac{17}{50}$

Since this is $> 0$, the function is increasing at $x = - 3$.

We can check the graph of $f \left(x\right) :$

graph{(-x^2-5x-2)/(x^2+1) [-5, 5, -5.89, 3]}