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 \boxed{}.
(type an integer or a simplified fraction. use a comma to separate answers as needed.)

b. there are no critical numbers.

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:
\( f^{\prime}(x)=\frac{4}{3}\times3x^{2}-4\times2x-12\times1 \)
Simplify each term: \( f^{\prime}(x) = 4x^{2}-8x - 12 \)

Step2: Solve \( f^{\prime}(x)=0 \) for critical numbers

Set \( 4x^{2}-8x - 12 = 0 \). First, divide both sides by 4:
\( x^{2}-2x - 3=0 \)
Factor the quadratic equation: \( (x - 3)(x + 1)=0 \)
Set each factor equal to zero: \( x - 3 = 0 \) or \( x + 1 = 0 \)
Solving these, we get \( x = 3 \) or \( x=-1 \)

Answer:

A. The critical number(s) is/are \(-1, 3\)