關於Differentiation的A.Maths.問題2

y = x(x – 5)2
dy/dx = (x – 5)2 + 2x(x – 5)
dy/dx = (x – 5)(x – 5 + 2x) = (x – 5)(3x – 5)
d2y/dx2 = 3x – 5 + 3(x – 5) = 6x – 20
Increasing portion is determined by dy/dx > 0
(x – 5)(3x – 5) > 0
x > 5 or x < 5/3
So there are two portions, from negative infinity to 5/3 AND from 5 to positive infinity
Decreasing portion is determined by dy/dx < 0
(x – 5)(3x – 5) < 0
5/3 < x < 5
Maximum and minimum points are determined by dy/dx = 0
(x – 5)(3x – 5) = 0
x = 5 or x = 5/3
d2y/dx2 at x = 5 is 6(5) – 20 = 10 => minimum point; y = 0
Alternatively since y decreases from the point x= 5/3 to the point x = 5, and then changes to increasing from x = 5 onwards, so x = 5 represents a minimum.
d2y/dx2 at x = 5/3 is 6(5/3) – 20 = -10 => maximum point; y = (5/3)(5/3 – 5)2 = 500/27
Alternatively since y increases from the point x= negative infinity to the point x = 5/3, and then changes to decreasing from x = 5/3 onwards, so x = 5/3 represents a maximum.
Minimum point is (5,0)
Maximum point is (5/3, 500/27)

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