Question

for the function below, find (a) the critical numbers; (b) the open intervals where the function is increasing; and (c) the open intervals where it is d

f(x)=\frac{x + 3}{x + 2}

b. there are no critical numbers.

(b) determine the interval(s) on which the function is increasing. select the correct choice below and, if necessary, fill in the answer box to complete choice.

a. the function is increasing on the interval(s) (type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression. use a comma to se answers as needed.)

b. the function is never increasing.

(c) determine the interval(s) on which the function is decreasing. select the correct choice below and, if necessary, fill in the answer box to complete y choice.

a. the function is decreasing on the interval(s) (type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression. use a comma to sepa answers as needed.)

b. the function is never decreasing.

Explanation:

Step1: Compute derivative of f(x)

Use quotient rule: $f'(x)=\frac{(1)(x+2)-(x+3)(1)}{(x+2)^2}=\frac{x+2-x-3}{(x+2)^2}=\frac{-1}{(x+2)^2}$

Step2: Find critical numbers

Critical numbers where $f'(x)=0$ or undefined. $f'(x)=\frac{-1}{(x+2)^2}$ never equals 0; undefined at $x=-2$ (not in domain of f(x)). So no critical numbers.

Step3: Analyze increasing intervals

$f'(x)=\frac{-1}{(x+2)^2}<0$ for all $x≠-2$. So function never increasing.

Step4: Analyze decreasing intervals

$f'(x)<0$ for all $x≠-2$. So function decreasing on $(-\infty,-2)\cup(-2,\infty)$.

Answer:

(a) B. There are no critical numbers.
(b) B. The function is never increasing.
(c) A. The function is decreasing on the interval(s) $(-\infty,-2)\cup(-2,\infty)$