Question

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

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. \((-6,-700)\)