Question

find the absolute maximum and absolute minimum values of the function f(x)=(x - 2)(x - 5)^3+8 on each of the indicated intervals. enter none for any absolute extrema that does not exist. (a) interval = 1,4. absolute maximum = 72 absolute minimum = -\frac{139}{256} (b) interval = 1,8. absolute maximum = 170 absolute minimum = -\frac{139}{256} (c) interval = 4,9. absolute maximum = 465 absolute minimum =

Explanation:

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

Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u=(x - 2)$ and $v=(x - 5)^3$.
$u^\prime=1$ and $v^\prime = 3(x - 5)^2$.
So $f^\prime(x)=(x - 5)^3+3(x - 2)(x - 5)^2=(x - 5)^2[(x - 5)+3(x - 2)]=(x - 5)^2(4x-11)$.

Step2: Find the critical points

Set $f^\prime(x)=0$.
$(x - 5)^2(4x - 11)=0$.
The critical points are $x = 5$ and $x=\frac{11}{4}=2.75$.

Step3: Evaluate the function at critical points and endpoints for each interval

(A) Interval $[1,4]$

• $f(1)=(1 - 2)(1 - 5)^3+8=(-1)(-64)+8=72$.
• $f(2.75)=(2.75 - 2)(2.75 - 5)^3+8=(0.75)(-2.25)^3+8=-\frac{139}{256}$.
• $f(4)=(4 - 2)(4 - 5)^3+8=2(-1)+8 = 6$.

Answer:

Absolute maximum = $72$
Absolute minimum = $-\frac{139}{256}$

(B) Interval $[1,8]$

• $f(1)=72$.
• $f(2.75)=-\frac{139}{256}$.
• $f(5)=(5 - 2)(5 - 5)^3+8 = 8$.
• $f(8)=(8 - 2)(8 - 5)^3+8=6\times27 + 8=170$.