exam #1\nscore: 16.33/40 answered: 11/21\nquestion 12\nlet ( f(x)=2 x^{3}-33 x^{2}+108 x + 3 ).\nuse a…

exam #1\nscore: 16.33/40 answered: 11/21\nquestion 12\nlet ( f(x)=2 x^{3}-33 x^{2}+108 x + 3 ).\nuse a graphing calculator to find the local minimum and local maximum.\nround your answer to the nearest whole number.\n(a) local minimum at ( x= ) with output value of \n(b) local maximum at ( x= ) with output value of \nsubmit question

Answer

Explanation:

Step1: Find the derivative of (f(x))

The derivative (f^\prime(x)=6x^{2}-66x + 108). Factor it: (f^\prime(x)=6(x^{2}-11x + 18)=6(x - 2)(x - 9))

Step2: Find critical points

Set (f^\prime(x)=0). Then (6(x - 2)(x - 9)=0), so (x = 2) and (x=9) are critical points.

Step3: Use the second - derivative test

The second - derivative (f^{\prime\prime}(x)=12x-66)

For (x = 2): (f^{\prime\prime}(2)=12\times2-66=-42<0), so (x = 2) is a local maximum.
For (x = 9): (f^{\prime\prime}(9)=12\times9-66 = 42>0), so (x = 9) is a local minimum.

Step4: Calculate function values

When (x = 9): (f(9)=2\times9^{3}-33\times9^{2}+108\times9 + 3=2\times729-33\times81+972 + 3=1458-2673+972 + 3=-240)
When (x = 2): (f(2)=2\times2^{3}-33\times2^{2}+108\times2 + 3=2\times8-33\times4 + 216+3=16-132 + 216+3=103)

Answer:

(a) Local Minimum at (x = 9) with output value of (-240) (b) Local Maximum at (x = 2) with output value of (103)