find the absolute maximum and absolute minimum values of f on the given interval.\n\n$f(x)=6x^{3}-18x^{2}-54x…

find the absolute maximum and absolute minimum values of f on the given interval.\n\n$f(x)=6x^{3}-18x^{2}-54x + 8,-2,4$\n\nstep 1\n\nthe absolute maximum and minimum values of f occur either at a critical point inside the interval or at an endpoint of the interval. recall that a critical point is a point where $f(x)=0$ or is undefined. we begin by finding the derivative of f.\n\n$f(x)=18x^{2}-36x - 54$\n\nstep 2\n\nwe now solve $f(x)=0$ for x, which gives the following critical numbers. (enter your answers as a comma - separated list.)\n\n$x=-1,3$\n\nstep 3\n\nwe must now find the function values at the critical numbers we just found and at the endpoints of the interval $-2,4$.\n\n$f(-1)=$\n$f(3)=$\n$f(-2)=$\n$f(4)=$

find the absolute maximum and absolute minimum values of f on the given interval.\n\n$f(x)=6x^{3}-18x^{2}-54x + 8,-2,4$\n\nstep 1\n\nthe absolute maximum and minimum values of f occur either at a critical point inside the interval or at an endpoint of the interval. recall that a critical point is a point where $f(x)=0$ or is undefined. we begin by finding the derivative of f.\n\n$f(x)=18x^{2}-36x - 54$\n\nstep 2\n\nwe now solve $f(x)=0$ for x, which gives the following critical numbers. (enter your answers as a comma - separated list.)\n\n$x=-1,3$\n\nstep 3\n\nwe must now find the function values at the critical numbers we just found and at the endpoints of the interval $-2,4$.\n\n$f(-1)=$\n$f(3)=$\n$f(-2)=$\n$f(4)=$

Answer

Explanation:

Step1: Calculate (f(-1))

Substitute (x = - 1) into (f(x)=6x^{3}-18x^{2}-54x + 8). [ \begin{align*} f(-1)&=6\times(-1)^{3}-18\times(-1)^{2}-54\times(-1)+8\ &=6\times(-1)-18\times1 + 54+8\ &=-6-18 + 54+8\ &=38 \end{align*} ]

Step2: Calculate (f(3))

Substitute (x = 3) into (f(x)=6x^{3}-18x^{2}-54x + 8). [ \begin{align*} f(3)&=6\times3^{3}-18\times3^{2}-54\times3+8\ &=6\times27-18\times9-162 + 8\ &=162-162-162 + 8\ &=-154 \end{align*} ]

Step3: Calculate (f(-2))

Substitute (x=-2) into (f(x)=6x^{3}-18x^{2}-54x + 8). [ \begin{align*} f(-2)&=6\times(-2)^{3}-18\times(-2)^{2}-54\times(-2)+8\ &=6\times(-8)-18\times4 + 108+8\ &=-48-72 + 108+8\ &=-4 \end{align*} ]

Step4: Calculate (f(4))

Substitute (x = 4) into (f(x)=6x^{3}-18x^{2}-54x + 8). [ \begin{align*} f(4)&=6\times4^{3}-18\times4^{2}-54\times4+8\ &=6\times64-18\times16-216 + 8\ &=384-288-216 + 8\ &=-112 \end{align*} ]

Answer:

(f(-1)=38), (f(3)=-154), (f(-2)=-4), (f(4)=-112)