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 $(-\infty, -1), (3, \infty)$
(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

(c) on which intervals is the function decreasing? select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. decreasing on $\square$
(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 decreasing

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 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: Determine where \( f^{\prime}(x)<0 \) (function decreasing)

To find where the function is decreasing, we need to find the intervals where \( f^{\prime}(x)<0 \). The critical numbers are \( x=-1 \) and \( x = 3 \) (from part (a)). These divide the number line into three intervals: \( (-\infty,-1) \), \( (-1,3) \), and \( (3,\infty) \)

• Test a value in \( (-\infty,-1) \), say \( x=-2 \): \( f^{\prime}(-2)=4(-2 - 3)(-2 + 1)=4(-5)(-1)=20>0 \)
• Test a value in \( (-1,3) \), say \( x = 0 \): \( f^{\prime}(0)=4(0 - 3)(0 + 1)=4(-3)(1)=-12<0 \)
• Test a value in \( (3,\infty) \), say \( x = 4 \): \( f^{\prime}(4)=4(4 - 3)(4 + 1)=4(1)(5)=20>0 \)

So the function is decreasing on the interval \( (-1,3) \)

Answer:

A. Decreasing on \((-1, 3)\)