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 decreasing.
$f(x)=\frac{4}{3}x^3 - 4x^2 - 12x - 36$
(a) determine the critical numbers. select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the critical number(s) is/are $-1, 3$.
(type an integer or a simplified fraction. use a comma to separate answers as needed.)
b. there are no critical numbers.
(b) on which intervals is the function increasing? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. increasing on
(type your answer in interval notation. simplify your answer. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)
b. never increasing

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x)=\frac{4}{3}x^{3}-4x^{2}-12x - 36 \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), the derivative \( f^\prime(x) \) is:
\( f^\prime(x)=4x^{2}-8x - 12 \) (since \( \frac{d}{dx}(\frac{4}{3}x^{3}) = 4x^{2} \), \( \frac{d}{dx}(-4x^{2})=-8x \), \( \frac{d}{dx}(-12x)=-12 \), and derivative of constant - 36 is 0).

Step2: Factor the derivative

Factor \( f^\prime(x)=4x^{2}-8x - 12 \). First, factor out a 4: \( f^\prime(x)=4(x^{2}-2x - 3) \). Then factor the quadratic: \( x^{2}-2x - 3=(x - 3)(x + 1) \). So \( f^\prime(x)=4(x - 3)(x + 1) \).

Step3: Find critical numbers (already given as - 1, 3, we can verify by setting \( f^\prime(x)=0 \))

Set \( f^\prime(x)=0 \): \( 4(x - 3)(x + 1)=0 \). Since 4≠0, we have \( x - 3 = 0 \) or \( x + 1 = 0 \), so \( x = 3 \) or \( x=-1 \), which matches the given critical numbers.

Step4: Test intervals for increasing/decreasing

The critical numbers - 1 and 3 divide the real line into three intervals: \( (-\infty, - 1) \), \( (-1, 3) \), and \( (3, \infty) \).

• For \( x < - 1 \) (e.g., \( x=-2 \)): \( f^\prime(-2)=4(-2 - 3)(-2 + 1)=4(-5)(-1)=20>0 \). So \( f(x) \) is increasing on \( (-\infty, - 1) \).
• For \( -1 < x < 3 \) (e.g., \( x = 0 \)): \( f^\prime(0)=4(0 - 3)(0 + 1)=4(-3)(1)=-12<0 \). So \( f(x) \) is decreasing on \( (-1, 3) \).
• For \( x > 3 \) (e.g., \( x = 4 \)): \( f^\prime(4)=4(4 - 3)(4 + 1)=4(1)(5)=20>0 \). So \( f(x) \) is increasing on \( (3, \infty) \).

Combining the increasing intervals: \( (-\infty, - 1)\cup(3, \infty) \).

Answer:

(a) The critical number(s) is/are \(-1, 3\) (matching option A).
(b) Increasing on \((-\infty, - 1)\cup(3, \infty)\) (so for part (b), the correct choice is A with interval \((-\infty, - 1)\cup(3, \infty)\))