Advanved Mathematics

g(x) = 6x⁴ - 8x³ + 1
g'(x) = 24x³ - 24x²
g"(x) = 72x² - 48x

When g'(x) = 0 :
24x³ - 24x² = 0
x³ - x² = 0
x²(x - 1) = 0
x = 0 (double roots) or x = 1

When x = 0 :
g(x) = 6(0)⁴ - 8(0)³ + 1 = 1
g'(x) = 0
g"(x) = 72(0)² - 48(0) = 0
Hence, point of inflexion at (0, 1).

When x = 1 :
g(x) = 6(1)⁴ - 8(1)³ + 1 = -1
g'(x) = 0
g"(x) = 72(1)² - 48(1) = 24 > 0
Hence, minimum point at (1, -1).

Too 001：
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2012-10-19 17:38:52 補充：
g(x) = 6x⁴ - 8x³ + 1
g’(x) = 24x³ - 24x²
g”(x) = 72x² - 48x
g⁽³⁾(x) = 144x - 48


When g(x) is max. or min., g’(x) = 0.
24x³ - 24x² = 0
x²(x – 1) = 0
x = 0 or x = 1

When x = 0, g”(x) = 0. (rej.)
When x = 1, g”(x) = 72 – 48 = 24 > 0. (OK)
g(1) = 6 – 8 + 1 = -1
∴ (1, - 1) is min. pt.

2012-10-19 17:39:48 補充：
When (x, g(x)) is pt. of inflex., g”(x) = 0
72x² - 48x = 0
x² = 2/3
x(3x – 2) = 0
x = 0 or x = 2/3
g⁽³⁾(0) = - 48 < 0 ⇒ g’(x) is max. (OK)
or g⁽³⁾(2/3) = 144(2/3) – 48 = 48 > 0 ⇒ g’(x) is min. (OK)
g(0) = 1, g(2/3) = 6(2/3)⁴ - 8(2/3)³ + 1 = - 5/27
∴ (0, 1) & (2/3, - 5/27) are pt. of inflex.