Question

determine the location and value of the absolute extreme values of f on the given interval, if they exist.
f(x)=-2x^{3}+27x^{2}-108x on 2,7
what is/are the absolute maximum/maxima of f on the given interval? select the correct choice below and, if necessary, fill in the answer boxes to complete you
a. the absolute maximum/maxima is/are at x=
(use a comma to separate answers as needed. type exact answers, using radicals as needed.)
b. there is no absolute maximum of f on the given interval.

Explanation:

Step1: Find the derivative of the function

The derivative of \(f(x)=-2x^{3}+27x^{2}-108x\) is \(f^{\prime}(x)=-6x^{2}+54x - 108\).
Factor \(f^{\prime}(x)\): \(f^{\prime}(x)=-6(x^{2}-9x + 18)=-6(x - 3)(x - 6)\).

Step2: Find the critical points

Set \(f^{\prime}(x)=0\), then \(-6(x - 3)(x - 6)=0\).
Solving \(x - 3=0\) gives \(x = 3\), and solving \(x - 6=0\) gives \(x = 6\). Both \(x = 3\) and \(x = 6\) are in the interval \([2,7]\).

Step3: Evaluate the function at the critical points and endpoints

Evaluate \(f(x)\) at \(x = 2\), \(x = 3\), \(x = 6\), and \(x = 7\).

• When \(x = 2\): \(f(2)=-2\times2^{3}+27\times2^{2}-108\times2=-16 + 108-216=-124\).
• When \(x = 3\): \(f(3)=-2\times3^{3}+27\times3^{2}-108\times3=-54 + 243-324=-135\).
• When \(x = 6\): \(f(6)=-2\times6^{3}+27\times6^{2}-108\times6=-432+972 - 648=-108\).
• When \(x = 7\): \(f(7)=-2\times7^{3}+27\times7^{2}-108\times7=-686+1323-756=-119\).

Answer:

A. The absolute maximum/maxima is/are \(-108\) at \(x = 6\).