for the function below, find (a) the critical numbers; (b) the open intervals where the function is…

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.\n$f(x)=\\frac{4}{3}x^{3}-6x^{2}-72x - 16$\n(a) determine the critical numbers. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the critical number(s) is/are\n(type an integer or a simplified fraction. use a comma to separate answers as needed.)\nb. there are no critical numbers.\n(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.\na. increasing on\n(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.)\nb. never increasing\n(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.\na. decreasing on\n(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.)\nb. never decreasing
Answer
Explanation:
Step1: Find the derivative of (f(x))
The function is (f(x)=\frac{4}{3}x^{3}-6x^{2}-72x - 16). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (f^\prime(x)=4x^{2}-12x - 72). Factor out (4): (f^\prime(x)=4(x^{2}-3x - 18)). Factor the quadratic: (f^\prime(x)=4(x - 6)(x+3)).
Step2: Find the critical numbers (a)
Set (f^\prime(x) = 0). (4(x - 6)(x + 3)=0). Since (4\neq0), then (x-6=0) or (x + 3=0). Solving (x-6=0) gives (x = 6), solving (x + 3=0) gives (x=-3).
Step3: Determine intervals of increase and decrease (b and c)
We use test - points in the intervals ((-\infty,-3)), ((-3,6)), and ((6,\infty)).
- For the interval ((-\infty,-3)), let (x=-4). Then (f^\prime(-4)=4((-4)-6)((-4)+3)=4\times(-10)\times(-1)=40>0). So the function is increasing on ((-\infty,-3)).
- For the interval ((-3,6)), let (x = 0). Then (f^\prime(0)=4(0 - 6)(0+3)=4\times(-6)\times3=-72<0). So the function is decreasing on ((-3,6)).
- For the interval ((6,\infty)), let (x = 7). Then (f^\prime(7)=4(7 - 6)(7+3)=4\times1\times10 = 40>0). So the function is increasing on ((6,\infty)).
Answer:
(a) The critical numbers are (-3) and (6). (b) The function is increasing on ((-\infty,-3)\cup(6,\infty)). (c) The function is decreasing on ((-3,6)).