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 = absolute minimum =

Explanation:

Step1: Expand the function

First, expand \(f(x)=(x - 2)(x - 5)^3+8\). Let \(u=x - 5\), then \(x=u + 5\), and \(f(u)=(u + 3)u^3+8=u^4+3u^3+8\). Then \(f(x)=x^4-17x^3 + 110x^2-325x + 383\). Now, find the derivative \(f'(x)\) using the power - rule \((x^n)'=nx^{n - 1}\). \(f'(x)=4x^3-51x^2 + 220x-325\).

Step2: Find critical points

Set \(f'(x)=0\). We can try to find the roots of \(4x^3-51x^2 + 220x-325 = 0\) by trial - and - error with values in the interval \([1,4]\). We can also use a graphing utility or numerical methods. Another way is to check the values of \(f(x)\) at the endpoints of the interval \(x = 1\) and \(x = 4\).

• When \(x = 1\): \(f(1)=(1 - 2)(1 - 5)^3+8=(-1)\times(-64)+8=64 + 8=72\).
• When \(x = 4\): \(f(4)=(4 - 2)(4 - 5)^3+8=2\times(-1)+8=6\).

Step3: Determine absolute extrema

Compare the function values at the endpoints. Since \(72>6\), the absolute maximum of \(f(x)\) on the interval \([1,4]\) is \(72\) and the absolute minimum is \(6\).

Answer:

Absolute maximum = \(72\)
Absolute minimum = \(6\)