Question

find the absolute maximum and absolute minimum values of the function ( f(x)=(x - 2)(x - 6)^{3}+11 ) on each of the indicated intervals. enter dne if a value does not exist.
a. interval: (1,4)
absolute maximum:
absolute minimum:
b. interval: (1,8)
absolute maximum:
absolute minimum:
a. interval: (4,9)
absolute maximum:
absolute minimum:

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=(x - 2)\) and \(v=(x - 6)^3\).
\(u^\prime=1\), \(v^\prime = 3(x - 6)^2\)
\(f^\prime(x)=(x - 6)^3+3(x - 2)(x - 6)^2=(x - 6)^2[(x - 6)+3(x - 2)]=(x - 6)^2(4x-12)=4(x - 6)^2(x - 3)\)

Step2: Find the critical points

Set \(f^\prime(x)=0\), then \(x = 3\) or \(x = 6\)

For interval \([1,4]\)

• Evaluate \(f(x)\) at critical points \(x = 3\) and endpoints \(x = 1,x = 4\)
• \(f(1)=(1 - 2)(1 - 6)^3+11=(-1)\times(-125)+11 = 136\)
• \(f(3)=(3 - 2)(3 - 6)^3+11=(1)\times(-27)+11=-16\)
• \(f(4)=(4 - 2)(4 - 6)^3+11=(2)\times(-8)+11=-5\)

For interval \([1,8]\)

• Evaluate \(f(x)\) at critical points \(x = 3,x = 6\) and endpoints \(x = 1,x = 8\)
• \(f(1)=136\)
• \(f(3)=-16\)
• \(f(6)=(6 - 2)(6 - 6)^3+11 = 11\)
• \(f(8)=(8 - 2)(8 - 6)^3+11=(6)\times8+11=59\)

For interval \([4,9]\)

• Evaluate \(f(x)\) at critical points \(x = 6\) and endpoints \(x = 4,x = 9\)
• \(f(4)=-5\)
• \(f(6)=11\)
• \(f(9)=(9 - 2)(9 - 6)^3+11=(7)\times27+11=200\)

Answer:

a. Interval \([1,4]\):
Absolute maximum: \(136\)
Absolute minimum: \(-16\)

b. Interval \([1,8]\):
Absolute maximum: \(136\)
Absolute minimum: \(-16\)

c. Interval \([4,9]\):
Absolute maximum: \(200\)
Absolute minimum: \(-5\)