Question

find the absolute maximum and minimum, if either exists, for the function on the indicated interval.

$f(x)=(x - 3)(x - 7)^{3}+3$
(a) $0,5$
(b) $2,9$
(c) $5,8$

(a) find the absolute maximum. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

a. the absolute maximum is at $x=$
(use a comma to separate answers as needed.)

b. there is no absolute maximum.

Explanation:

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

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\), where \(u=(x - 3)\) and \(v=(x - 7)^3\).
\(u^\prime=1\), \(v^\prime = 3(x - 7)^2\).
So \(t^\prime(x)=(x - 7)^3+3(x - 3)(x - 7)^2=(x - 7)^2[(x - 7)+3(x - 3)]=(x - 7)^2(4x-16)=4(x - 7)^2(x - 4)\)

Step2: Find the critical points

Set \(t^\prime(x)=0\), then \(4(x - 7)^2(x - 4)=0\).
The critical points are \(x = 4\) and \(x = 7\). But for interval \([0,5]\), \(x = 7\) is not in the interval.

Step3: Evaluate the function at critical points and endpoints

For \(x = 0\): \(t(0)=(0 - 3)(0 - 7)^3+3=(-3)\times(-343)+3 = 1032\)
For \(x = 4\): \(t(4)=(4 - 3)(4 - 7)^3+3=(1)\times(-27)+3=-24\)
For \(x = 5\): \(t(5)=(5 - 3)(5 - 7)^3+3=(2)\times(-8)+3=-13\)

Answer:

A. The absolute maximum is \(1032\) at \(x = 0\)