Question

exam #1
score: 16.33/40 answered: 11/21
question 12
let ( f(x)=2 x^{3}-33 x^{2}+108 x + 3 ).
use a graphing calculator to find the local minimum and local maximum.
round your answer to the nearest whole number.
(a) local minimum at ( x= ) with output value of
(b) local maximum at ( x= ) with output value of
submit question

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 \(103\)