Question

求解數學

求解這一題的解答過程🙏🙏#多項式的運算與運用

(5) 11
+12)=1 +13)=1;
5.設f(x)為3次多項式且f(1)+1=f(2)-1=f(3)-13=f(4)-41=0,則
(4) 7
①
ƒ(0)
=
(1)1.則(10)=?
(4)60
h = -17
(x fi 1) = -1
f(2) = 1 = a(z-h)² + k
且其所小值】
(2) 2
(3)4.
(5) 10
f(3)=13=a(3-h)+K1=
錢更個或兩個以上不給分)

Answer 1

✨ ベストアンサー ✨

qn

<解法一> 假設一般式
設 f(x) = ax³+bx²+cx+d
f(1) = a + b + c + d = -1 ...(1)
f(2) = 8a + 4b + 2c + d = 1 ...(2)
f(3) = 27a + 9b + 3c + d = 13 ...(3)
f(4) = 64a + 16b + 4c + d = 41 ...(4)

第2、3、4式各別減掉第1式
7a + 3b + c = 2 ...(5)
26a + 8b + 2c = 14 → 13a + 4b + c = 7 ...(6)
63a + 15b + 3c = 42 → 21a + 5b + c = 14 ...(7)

第6、7式各別減掉第5式
6a + b = 5
14a + 2b = 12 → 7a + b = 6

兩式解得 a = 1 , b = -1
代入第5式得 c = 2 - 7 + 3 = -2
再代入第1式得 d = -1 - 1 + 1 + 2 = 1

故 f(0) = d = 1

<解法二> 牛頓插值法
設 f(x) = a(x-3)(x-2)(x-1) + b(x-2)(x-1) + c(x-1) + d
f(1) = d = -1
f(2) = c + d = 1 → c = 2
f(3) = 2b + 2c + d = 13 → b = 5
f(4) = 6a + 6b + 3c + d = 41 → a = 1

⇒ f(x) = (x-3)(x-2)(x-1) + 5(x-2)(x-1) + 2(x-1) - 1
⇒ f(0) = (-3)(-2)(-1) + 5(-2)(-1) + 2(-1) - 1
= -6 + 10 - 2 - 1
= 1

qn

<解法三> 階差
x 0 1 2 3 4
f(x) y -1 1 13 41
+z +2 +12 +28
+w +10 +16
+s +6
+0
→ s=6 → w=4 → z=-2 → y=1