let ( f(x)=2x^{3}-12x^{2}-30x - 6) be defined on (-6,6). find: a. the absolute maximum. write your answer as…

Explanation:

Step1: Find the derivative of (f(x))

First, find (f'(x)) for (f(x)=2x^{3}-12x^{2}-30x - 6). Using the power - rule ((x^n)'=nx^{n - 1}), we have (f'(x)=6x^{2}-24x - 30).

Step2: Set the derivative equal to zero

Set (f'(x)=0), so (6x^{2}-24x - 30 = 0). Divide through by 6: (x^{2}-4x - 5=0). Factor the quadratic equation: ((x - 5)(x+1)=0). Solving for (x) gives (x = 5) or (x=-1).

Step3: Evaluate (f(x)) at critical points and endpoints

The endpoints of the interval ([-6,6]) are (x=-6) and (x = 6), and the critical points are (x=-1) and (x = 5).

• (f(-6)=2(-6)^{3}-12(-6)^{2}-30(-6)-6=2(-216)-12(36)+180 - 6=-432-432 + 180-6=-700).
• (f(-1)=2(-1)^{3}-12(-1)^{2}-30(-1)-6=-2-12 + 30-6=10).
• (f(5)=2(5)^{3}-12(5)^{2}-30(5)-6=2(125)-12(25)-150-6=250-300-150-6=-206).
• (f(6)=2(6)^{3}-12(6)^{2}-30(6)-6=2(216)-12(36)-180-6=432-432-180-6=-186).

Step4: Determine absolute maximum and minimum

a. The absolute maximum value of (f(x)) on ([-6,6]) is (10) which occurs at (x=-1). So the ordered - pair is ((-1,10)). b. The absolute minimum value of (f(x)) on ([-6,6]) is (-700) which occurs at (x=-6). So the ordered - pair is ((-6,-700)).

Answer:

a. ((-1,10)) b. ((-6,-700))