Question

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

f(x)=\frac{12x^{3}}{3}+10x^{2}-8x on -3,1

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

a. the absolute maximum/maxima is/are \square at x = \square.
(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: 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 \(x=-2\)