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 your choice
a. the absolute maximum/maxima is/are - 108 at $x = 6$
(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
what is/are the absolute minimum/minima of f on the given interval? select the correct choice below and, if necessary, fill in the answer boxes to complete your choice
a. the absolute minimum/minima is/are at $x =$
(use a comma to separate answers as needed. type exact answers, using radicals as needed.)
b. there is no absolute minimum of f on the given interval

Explanation:

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

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 \(-6(x - 3)(x - 6)=0\).
The critical points are \(x = 3\) and \(x=6\). Both \(x = 3\) and \(x = 6\) are in the interval \([2,7]\)

Step3: Evaluate \(f(x)\) at the critical points and endpoints

• 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:

For the absolute maximum:
A. The absolute maximum/maxima is/are \(-108\) at \(x = 6\)
For the absolute minimum:
A. The absolute minimum/minima is/are \(-135\) at \(x = 3\)