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)

$f'(x)=\frac{d}{dx}(\frac{8x^{3}}{3}+26x^{2}-28x)=8x^{2}+52x - 28$

Step2: Set the derivative equal to zero

$8x^{2}+52x - 28 = 0$ Divide through by 4: $2x^{2}+13x - 7=0$ Factor: $(2x - 1)(x + 7)=0$ Solve for x: $x=\frac{1}{2},x=-7$

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

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

Step4: Compare the values

$\frac{1666}{3}>\frac{1568}{3}>\frac{2}{3}>-\frac{43}{6}$

Answer:

A. The absolute maximum/maxima is/are $\frac{1666}{3}$ at $x=-7$