# How do you find the critical numbers for f(x)= (x+3) / x^2 to determine the maximum and minimum?

Jul 19, 2017

$f \left(x\right) = \frac{x + 3}{x} ^ 2$ has a single critical point in $x = - 6$ where it has a local minimum.

#### Explanation:

Evaluate the derivative of the function:

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

$f ' \left(x\right) = - \frac{1}{x} ^ 2 - \frac{6}{x} ^ 3 = - \frac{x + 6}{x} ^ 3$

The critical points are the solutions of the equation:

$f ' \left(x\right) = 0$

$- \frac{x + 6}{x} ^ 3 = 0$

so the only critical point is for $x = - 6$.

Evaluate the second derivative:

$f ' ' \left(x\right) = \frac{2}{x} ^ 3 + \frac{18}{x} ^ 4 = \frac{2 x + 18}{x} ^ 4$

As $f ' ' \left(- 6\right) > 0$ the point is a local minimum.

graph{(x+3)/x^2 [-10, 2, -0.2, 0]}