Question

find the absolute maximum and minimum, if either exists, for the function on the indicated interval.
$f(x)=(x - 3)(x - 7)^{2}+3$
(a) $0,5$ (b) $2,9$ (c) $5,8$
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 $-24$ at $x = 4$
(use a comma to separate answers as needed.)
b. there is no absolute minimum.
(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 $\square$ at $x = \square$
(use a comma to separate answers as needed.)
b. there is no absolute maximum.

Explanation:

Step1: Expand the function

$$\begin{align*} f(x)&=(x - 3)(x - 7)^{2}+3\\ &=(x - 3)(x^{2}-14x + 49)+3\\ &=x^{3}-14x^{2}+49x-3x^{2}+42x-147 + 3\\ &=x^{3}-17x^{2}+91x-144 \end{align*}$$

Step2: Find the derivative

Using the power rule \((x^n)^\prime=nx^{n - 1}\), we have \(f^\prime(x)=3x^{2}-34x + 91\)

Step3: Solve \(f^\prime(x)=0\)

Factor \(3x^{2}-34x + 91=(3x - 13)(x - 7)=0\). So \(x=\frac{13}{3}\approx4.33\) and \(x = 7\)

Step4: Evaluate the function on the interval \([2,9]\)

• \(f(2)=(2 - 3)(2 - 7)^{2}+3=(-1)\times25 + 3=-22\)
• \(f(\frac{13}{3})=(\frac{13}{3}-3)(\frac{13}{3}-7)^{2}+3=(\frac{4}{3})(-\frac{8}{3})^{2}+3=\frac{4}{3}\times\frac{64}{9}+3=\frac{256}{27}+3=\frac{256 + 81}{27}=\frac{337}{27}\approx12.48\)
• \(f(7)=(7 - 3)(7 - 7)^{2}+3=3\)
• \(f(9)=(9 - 3)(9 - 7)^{2}+3=6\times4+3=27\)

Answer:

The absolute maximum is \(27\) at \(x = 9\)