# How do you find the critical numbers for f(x) = (x - 1 )/( x + 3) to determine the maximum and minimum?

May 5, 2017

There are no critical numbers (maximums or minimums)

#### Explanation:

Critical numbers are found when $f ' \left(x\right) = 0$.

Find $f ' \left(x\right)$ using the Quotient Rule: $\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

Given: $f \left(x\right) = \frac{x - 1}{x + 3}$

Let u = x-1; " " u' = 1

Let v = x + 3; " " v' = 1

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

Simplify:

$f ' \left(x\right) = \frac{x + 3 - x + 1}{x + 3} ^ 2 = \frac{4}{x + 3} ^ 2$

Find critical numbers (f'(x) = 0:

$\frac{4}{x + 3} ^ 2 = 0$

Multiply both sides by the denominator: $4 = 0$

There are no critical numbers (maximums or minimums).

This can be seen by graphing the function:

graph{(x-1)/(x+3) [-11.66, 8.34, -5.12, 4.88]}