Question

determine the location and value of the absolute extreme values of f on the given interval, if they exist
f(x)=\frac{12 x^{3}}{3}+10 x^{2}-8 x \text { 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 boxes to complete your choice
a. the absolute maximum/maxima is/are 24 at ( x=-2 )
(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: Simplify the function

Simplify \( f(x)=\frac{12x^{3}}{3}+10x^{2}-8x \) to \( f(x) = 4x^{3}+10x^{2}-8x \).

Step2: Find the derivative

Differentiate \( f(x) \) using the power rule \( (x^{n})^\prime=nx^{n - 1} \).
\( f^\prime(x)=12x^{2}+20x - 8 \).
Factor \( f^\prime(x) \): \( f^\prime(x)=4(3x^{2}+5x - 2)=4(3x - 1)(x + 2) \).

Step3: Find critical points

Set \( f^\prime(x)=0 \), then \( 4(3x - 1)(x + 2)=0 \).
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 \([-3,1]\).

Step4: Evaluate the function at critical points and endpoints

• Evaluate \( f(x) \) at \( x=-3\):

\( f(-3)=4\times(-3)^{3}+10\times(-3)^{2}-8\times(-3)=4\times(-27)+10\times9 + 24=-108 + 90+24 = 6 \).

• Evaluate \( f(x) \) at \( x=-2\):

\( f(-2)=4\times(-2)^{3}+10\times(-2)^{2}-8\times(-2)=4\times(-8)+10\times4+16=-32 + 40+16=24 \).

• Evaluate \( f(x) \) at \( x=\frac{1}{3}\):

\( f(\frac{1}{3})=4\times(\frac{1}{3})^{3}+10\times(\frac{1}{3})^{2}-8\times\frac{1}{3}=4\times\frac{1}{27}+10\times\frac{1}{9}-\frac{8}{3}=\frac{4 + 30-72}{27}=-\frac{38}{27}\approx - 1.41 \).

• Evaluate \( f(x) \) at \( x = 1\):

\( f(1)=4\times1^{3}+10\times1^{2}-8\times1=4 + 10-8 = 6 \).

Answer:

The absolute maximum is \(24\) at \(x=-2\).
The absolute minimum is \(-\frac{38}{27}\) at \(x=\frac{1}{3}\). So for the first question, the answer is A. The absolute maximum/maxima is/are \(24\) at \(x=-2\). For the second question, the answer is A. The absolute minimum/minima is/are \(-\frac{38}{27}\) at \(x = \frac{1}{3}\).