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

Explanation:

Step1: Find the derivative of f(x)

First, find the derivative of $f(x)=\frac{8x^{3}}{3}+26x^{2}-28x$. Using the power - rule $(x^n)' = nx^{n - 1}$, we have $f'(x)=8x^{2}+52x - 28$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$, so $8x^{2}+52x - 28 = 0$. Divide through by 4 to simplify: $2x^{2}+13x - 7=0$. Factor the quadratic equation: $(2x - 1)(x + 7)=0$. Solving for x gives $x=\frac{1}{2}$ and $x=-7$.

Step3: Evaluate the function at critical points and endpoints

Evaluate $f(x)$ at the critical points $x=-7,\frac{1}{2}$ and the endpoints $x=-8,1$.

• $f(-8)=\frac{8(-8)^{3}}{3}+26(-8)^{2}-28(-8)=\frac{-4096}{3}+1664 + 224=\frac{-4096+4992 + 672}{3}=\frac{-4096 + 5664}{3}=\frac{1568}{3}$.
• $f(-7)=\frac{8(-7)^{3}}{3}+26(-7)^{2}-28(-7)=\frac{-2744}{3}+1274 + 196=\frac{-2744+3822+588}{3}=\frac{1666}{3}$.
• $f(\frac{1}{2})=\frac{8(\frac{1}{2})^{3}}{3}+26(\frac{1}{2})^{2}-28(\frac{1}{2})=\frac{8\times\frac{1}{8}}{3}+26\times\frac{1}{4}-14=\frac{1}{3}+\frac{13}{2}-14=\frac{2 + 39 - 84}{6}=-\frac{43}{6}$.
• $f(1)=\frac{8(1)^{3}}{3}+26(1)^{2}-28(1)=\frac{8}{3}+26 - 28=\frac{8 + 78 - 84}{3}=\frac{2}{3}$.

Step4: Determine the absolute minimum

Comparing the values $f(-8)=\frac{1568}{3}$, $f(-7)=\frac{1666}{3}$, $f(\frac{1}{2})=-\frac{43}{6}$, $f(1)=\frac{2}{3}$, the smallest value is $-\frac{43}{6}$ at $x = \frac{1}{2}$.

Answer:

A. The absolute minimum/minima is/are $-\frac{43}{6}$ at $x=\frac{1}{2}$