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) $5,8$

(b) 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 128 at $x = 2$
(use a comma to separate answers as needed.)
b. there is no absolute maximum.
find the absolute minimum. select the correct choice below and, if necessary, fill in the answer boxes to complete your choice
a. the absolute minimum is $\square$ at $x = \square$
(use a comma to separate answers as needed.)
b. there is no absolute minimum.

Explanation:

Step1: Expand the function

First, expand \(f(x)=(x - 3)(x - 7)^{3}+3\).
Using the formula \((a - b)(c - d)^{n}\) and the binomial expansion \((x - 7)^{3}=x^{3}-21x^{2}+147x - 343\), then \(f(x)=(x - 3)(x^{3}-21x^{2}+147x - 343)+3=x^{4}-24x^{3}+210x^{2}-786x + 1032\).

Step2: Find the derivative

Differentiate \(f(x)\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(f^\prime(x)=4x^{3}-72x^{2}+420x - 786=2(2x^{3}-36x^{2}+210x - 393)\).
Set \(f^\prime(x) = 0\). We can also use the product - rule: if \(y=(x - 3)(x - 7)^{3}+3\), by \(y=uv + 3\) where \(u=x - 3\), \(v=(x - 7)^{3}\), \(u^\prime=1\), \(v^\prime=3(x - 7)^{2}\), then \(y^\prime=(x - 7)^{3}+3(x - 3)(x - 7)^{2}=(x - 7)^{2}(x - 7+3x - 9)=(x - 7)^{2}(4x - 16)=4(x - 7)^{2}(x - 4)\).
Set \(y^\prime = 0\), we get \(x = 4\) or \(x = 7\).

Step3: Evaluate the function at critical points and endpoints for interval \([2,9]\)

• For \(x = 2\): \(f(2)=(2 - 3)(2 - 7)^{3}+3=(-1)\times(-125)+3=128\).
• For \(x = 4\): \(f(4)=(4 - 3)(4 - 7)^{3}+3=1\times(-27)+3=-24\).
• For \(x = 7\): \(f(7)=(7 - 3)(7 - 7)^{3}+3=3\).
• For \(x = 9\): \(f(9)=(9 - 3)(9 - 7)^{3}+3=6\times8 + 3=51\).

Answer:

The absolute minimum is \(-24\) at \(x = 4\).