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

determine the location and value of the absolute extreme values of f on the given interval, if they exist. f(x)=2x^3 - 30x^2 + 126x on 2,8 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 at x = (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.
Answer
Explanation:
Step1: Find the derivative of f(x)
$f'(x)=\frac{d}{dx}(2x^{3}-30x^{2}+126x)=6x^{2}-60x + 126$
Step2: Set the derivative equal to zero
$6x^{2}-60x + 126 = 0$ Divide through by 6: $x^{2}-10x + 21=0$ Factor: $(x - 3)(x - 7)=0$ Solve for x: $x=3$ or $x = 7$
Step3: Evaluate f(x) at critical points and endpoints
Evaluate at $x = 2$: $f(2)=2(2)^{3}-30(2)^{2}+126(2)=16-120 + 252=148$ Evaluate at $x = 3$: $f(3)=2(3)^{3}-30(3)^{2}+126(3)=54-270+378 = 162$ Evaluate at $x = 7$: $f(7)=2(7)^{3}-30(7)^{2}+126(7)=686-1470 + 882=98$ Evaluate at $x = 8$: $f(8)=2(8)^{3}-30(8)^{2}+126(8)=1024-1920+1008 = 112$
Step4: Determine the absolute maximum
The largest value among $f(2)=148$, $f(3)=162$, $f(7)=98$, $f(8)=112$ is 162 at $x = 3$
Answer:
A. The absolute maximum/maxima is/are 162 at x = 3