# How do you find the intervals of increasing and decreasing using the first derivative given y=(x+4)/x^2?

Jul 25, 2017

$y \left(x\right)$ is strictly decreasing in $\left(- \infty , - 8\right)$ and reaches a local minimum in $x = 8$, then it increases in $\left(- 8 , 0\right)$, and it is decreasing again in $\left(0 , + \infty\right)$

#### Explanation:

Evaluate the first derivative of the function:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\frac{x + 4}{x} ^ 2\right) = \frac{{x}^{2} - 2 x \left(x + 4\right)}{x} ^ 4 = - \frac{x + 8}{x} ^ 3$

and solve the inequality:

$\frac{\mathrm{dy}}{\mathrm{dx}} < 0$

The inequality is satisfied when numerator and denominator are both positive or both negative, so:

$\frac{\mathrm{dy}}{\mathrm{dx}} < 0$ for $x \in \left(- \infty , - 8\right) \cup \left(0 , + \infty\right)$

and:

$\frac{\mathrm{dy}}{\mathrm{dx}} > 0$ for $x \in \left(- 8 , 0\right)$

while the only stationary point is for $x = - 8$.

So $y \left(x\right)$ is strictly decreasing in $\left(- \infty , - 8\right)$ and reaches a local minimum in $x = 8$, then it increases in $\left(- 8 , 0\right)$, and it is decreasing again in $\left(0 , + \infty\right)$

graph{(x+4)/x^2 [-1.1181, -6.118, -0.86, 1.64]}

graph{(x+4)/x^2 [-9, -7, -0.08, -0.04]}