大學微積分，請問圈起來的題目怎麼解?

f '
= 3ax^2 + 2bx + c
= 3a*( x - 2 )( x - 4 ) , 因為相對極值發生在 x = 2 , 4
= 3a*( x^2 - 6x + 8 )

f '' = 3a( 2x - 6 ) = 0
x = 3 , 故反曲點確實在此

f
= ∫ f ' dx
= ∫ 3a*( x^2 - 6x + 8 ) dx
= 3a*[ (1/3)x^3 - 3x^2 + 8x + K1 ]
= ax^3 - 9ax^2 + 24ax + K

f(2) = 8a - 36a + 48a + K = 20a + K = 4 ..... (1)
f(4) = 64a - 144a + 96a + K = 16a + K = 2 ..... (2)
(1) - (2) 得:
4a = 2
a = 1/2
K = 4 - 20a = 4 - 10 = - 6

f(x)
= ax^3 - 9ax^2 + 24ax + K
= (1/2)x^3 - (9/2)x^2 + 12x - 6

Ans: a = 1/2 , b = - 9/2 , c = 12 , d = - 6