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.\nf(x)=\\frac{2}{3}x^{3}-x^{2}-144x - 12\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 answ\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 answe\nb. never decreasing
Answer
Explanation:
Step1: Find the derivative of the function
The derivative of (f(x)=\frac{2}{3}x^{3}-x^{2}-144x - 12) is (f^\prime(x)=2x^{2}-2x - 144).
Step2: Find the critical numbers
Set (f^\prime(x) = 0), so (2x^{2}-2x - 144=0). Divide both sides by (2) to get (x^{2}-x - 72=0). Factor the quadratic equation: ((x - 9)(x+8)=0). Solving (x - 9=0) gives (x = 9), and solving (x + 8=0) gives (x=-8).
Step3: Determine the intervals of increase and decrease
We use test - points in the intervals ((-\infty,-8)), ((-8,9)) and ((9,\infty)).
- For the interval ((-\infty,-8)), let (x=-9). Then (f^\prime(-9)=2\times(-9)^{2}-2\times(-9)-144=2\times81 + 18-144=162 + 18-144=36>0).
- For the interval ((-8,9)), let (x = 0). Then (f^\prime(0)=2\times0^{2}-2\times0-144=-144<0).
- For the interval ((9,\infty)), let (x = 10). Then (f^\prime(10)=2\times10^{2}-2\times10-144=200-20 - 144=36>0).
Answer:
(a) The critical number(s) is/are (-8,9). (b) Increasing on ((-\infty,-8)\cup(9,\infty)). (c) Decreasing on ((-8,9)).