設f(x)=x^4+ax^3+bx^2+cx+d 證明若3a^2<=8b 則f沒有反曲點 若3a^2>8b 則f有兩個反曲點?

f(x) = x⁴ + ax³ + bx² + cx + d
f'(x) = 4x³ + 3ax² + 2bx + c
f''(x) = 12x² + 6ax + 2b = 12(x + a/4)² + 2b - 3a²/4 = 12(x + a/4)² - (3a² - 8b)/4

f''(x) = 0
12x² + 6ax + 2b = 0
6x² + 3ax + b = 0
x = [-3a ± √(9a² - 24b)]/12

考慮 Δ < 0
9a² - 24b < 0
3a² < 8b
沒有反曲點

考慮 Δ = 0
9a² - 24b = 0
3a² - 8b = 0
x = -a/4
f''(x < -a/4) > 0
f''(x > -a/4) > 0
沒有反曲點

考慮 Δ > 0
9a² - 24b > 0
3a² > 8b

f''(x) = 12(x + a/4)² - (3a² - 8b)/4 ...... [ - (3a² - 8b)/4 < 0 ]
x = [-3a ± √(9a² - 24b)]/12

f''( x < [-3a - √(9a² - 24b)]/12 ) > 0
f''( [-3a - √(9a² - 24b)]/12 < x < [-3a + √(9a² - 24b)]/12 ) < 0
f''( x > [-3a - √(9a² - 24b)]/12 ) > 0
有兩個反曲點

∴ 若 3a² ≤ 8b 則 f 沒有反曲點
若 3a² > 8b 則 f 有兩個反曲點