Question

let f(x)=2x³ - 12x² - 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 \(f(x)\)

Differentiate \(f(x)=2x^{3}-12x^{2}-30x - 6\) using the power - rule \((x^n)^\prime=nx^{n - 1}\).
\(f^\prime(x)=6x^{2}-24x - 30=6(x^{2}-4x - 5)=6(x - 5)(x+1)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(6(x - 5)(x + 1)=0\).
Solving \(x - 5=0\) and \(x+1=0\) gives \(x=-1\) and \(x = 5\).

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

Evaluate \(f(x)\) at \(x=-6,x=-1,x = 5,x=6\).
\(f(-6)=2(-6)^{3}-12(-6)^{2}-30(-6)-6=2(-216)-12(36)+180 - 6=-432-432 + 180-6=-708\)
\(f(-1)=2(-1)^{3}-12(-1)^{2}-30(-1)-6=-2-12 + 30-6=10\)
\(f(5)=2(5)^{3}-12(5)^{2}-30(5)-6=2(125)-12(25)-150-6=250-300-150-6=-206\)
\(f(6)=2(6)^{3}-12(6)^{2}-30(6)-6=2(216)-12(36)-180-6=432-432-180-6=-186\)

Step4: Determine the 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 \(-708\) which occurs at \(x=-6\), so the ordered - pair is \((-6,-708)\)

Answer:

a. \((-1,10)\)
b. \((-6,-708)\)