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, Question 1
Given:
𝑓(𝑥) = 𝑥 4 − 2𝑥 3 + 12𝑥 − 5
Matlab/Octave code:
p = [1 -2 0 12 -5];
roots(p)
Explanation:
• Coefficients correspond to: 𝑥 4 − 2𝑥 3 + 0𝑥 2 + 12𝑥 − 5
• roots(p) computes all solutions numerically.
Question 2
We need extrema of 𝑓 ′ (𝑥)on [−2,2]
Differentiate
𝑓 ′ (𝑥) = 4𝑥 3 − 6𝑥 2 + 12
Find critical points of 𝒇′ (𝒙)
Differentiate again:
𝑓 ′′ (𝑥) = 12𝑥 2 − 12𝑥
Set:
12𝑥(𝑥 − 1) = 0 ⇒ 𝑥 = 0,1
Evaluate 𝒇′ (𝒙)at endpoints + critical points
𝑓 ′ (−2) = 4(−8) − 6(4) + 12 = −32 − 24 + 12 = −44
𝑓 ′ (0) = 12
𝑓 ′ (1) = 4 − 6 + 12 = 10
𝑓 ′ (2) = 32 − 24 + 12 = 20
Final Answer: