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)=4x³ - 39x² - 144x + 9\na) find the critical number(s). select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○ a. the critical number(s) is/are\n(type an integer or a simplified fraction. use a comma to separate answers as needed.)\n○ b. there are no critical numbers.\nb) list any interval(s) on which the function is increasing. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○ a. the function is increasing on the interval(s)\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.)\n○ b. the function is never increasing.\nc) list any interval(s) on which the function is decreasing. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\n○ a. the function is decreasing on the interval(s)\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.)\n○ b. the function is never decreasing.
Answer
Explanation:
Step1: Find the derivative of the function
Given ( f(x)=4x^{3}-39x^{2}-144x + 9). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (f^\prime(x)=12x^{2}-78x-144). Factor out (6): (f^\prime(x)=6(2x^{2}-13x - 24)). Factor the quadratic (2x^{2}-13x - 24=(2x + 3)(x-8)). So (f^\prime(x)=6(2x + 3)(x - 8)).
Step2: Find the critical numbers
Set (f^\prime(x)=0). (6(2x + 3)(x - 8)=0). Since (6\neq0), then (2x+3 = 0) or (x - 8=0). Solving (2x+3=0) gives (x=-\frac{3}{2}), and solving (x - 8=0) gives (x = 8).
Step3: Determine the intervals of increase and decrease
We use test - points in the intervals ((-\infty,-\frac{3}{2})), ((-\frac{3}{2},8)) and ((8,\infty)).
- For the interval ((-\infty,-\frac{3}{2})), let (x=-2). Then (f^\prime(-2)=6[2(-2)+3][-2 - 8]=6(-1)(-10)=60>0). So the function is increasing on ((-\infty,-\frac{3}{2})).
- For the interval ((-\frac{3}{2},8)), let (x = 0). Then (f^\prime(0)=6(0 + 3)(0 - 8)=6\times3\times(-8)=-144<0). So the function is decreasing on ((-\frac{3}{2},8)).
- For the interval ((8,\infty)), let (x=9). Then (f^\prime(9)=6(2\times9+3)(9 - 8)=6(21)(1)=126>0). So the function is increasing on ((8,\infty)).
Answer:
a) The critical numbers are (x =-\frac{3}{2}) and (x = 8). b) The function is increasing on ((-\infty,-\frac{3}{2})\cup(8,\infty)). c) The function is decreasing on ((-\frac{3}{2},8)).