determine the location and value of the absolute extreme values of f on the given interval, if they…

determine the location and value of the absolute extreme values of f on the given interval, if they exist.\n\nf(x)=\\frac{12x^{3}}{3}+10x^{2}-8x on -3,1\n\nwhat is/are the absolute maximum/maxima of f on the given interval? select the correct choice below and, if necessary, fill in the answer \n\na. the absolute maximum/maxima is/are \\square at x = \\square.\n(use a comma to separate answers as needed. type exact answers, using radicals as needed.)\nb. there is no absolute maximum of f on the given interval.

Answer

Explanation:

Step1: Simplify the function and find its derivative

Simplify (f(x)=\frac{12x^{3}}{3}+10x^{2}-8x) to (f(x) = 4x^{3}+10x^{2}-8x). The derivative (f^{\prime}(x)) using the power rule ((x^{n})^\prime=nx^{n - 1}) is (f^{\prime}(x)=12x^{2}+20x - 8). Factor (f^{\prime}(x)): (f^{\prime}(x)=4(3x^{2}+5x - 2)=4(3x - 1)(x + 2)).

Step2: Find the critical points

Set (f^{\prime}(x)=0), so (4(3x - 1)(x + 2)=0). Solving (3x-1 = 0) gives (x=\frac{1}{3}), and solving (x + 2=0) gives (x=-2). Both (x=-2) and (x=\frac{1}{3}) are in the interval ([-3,1]).

Step3: Evaluate the function at critical points and endpoints

Evaluate (f(x)) at (x=-3), (x=-2), (x=\frac{1}{3}), and (x = 1).

For (x=-3): (f(-3)=4(-3)^{3}+10(-3)^{2}-8(-3)=4\times(-27)+10\times9 + 24=-108 + 90+24 = 6).
For (x=-2): (f(-2)=4(-2)^{3}+10(-2)^{2}-8(-2)=4\times(-8)+10\times4+16=-32 + 40+16=24).
For (x=\frac{1}{3}): (f(\frac{1}{3})=4(\frac{1}{3})^{3}+10(\frac{1}{3})^{2}-8(\frac{1}{3})=4\times\frac{1}{27}+10\times\frac{1}{9}-\frac{8}{3}=\frac{4 + 30-72}{27}=-\frac{38}{27}).
For (x = 1): (f(1)=4(1)^{3}+10(1)^{2}-8(1)=4 + 10-8 = 6).

Answer:

A. The absolute maximum/maxima is/are (24) at (x=-2)