f(x)为二次函数且f(0)=3,f(x+2)-f(x)=4x+2
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设f(x) = ax² + bx + c
f(0) = 3,代入
c = 3
f(x + 2) - f(x) = 4x + 2
f(- 2 + 2) - f(- 2) = 4(- 2) + 2
3 - f(- 2) = - 6
f(- 2) = 9,代入
9 = 4a - 2b + 3 ==> 4a - 2b = 6
f(- 4 + 2) - f(- 4) = 4(- 4) + 2
9 - f(- 4) = - 14
f(- 4) = 23,代入
23 = 16a - 4b + 3 ==> 4a - b = 5
(4a - 2b) - 2(4a - b) = 6 - 2(5)
- 4a = - 4 ==> a = 1
b = 4(1) - 5 = - 1
所以f(x) = x² - x + 3
f(0) = 3,代入
c = 3
f(x + 2) - f(x) = 4x + 2
f(- 2 + 2) - f(- 2) = 4(- 2) + 2
3 - f(- 2) = - 6
f(- 2) = 9,代入
9 = 4a - 2b + 3 ==> 4a - 2b = 6
f(- 4 + 2) - f(- 4) = 4(- 4) + 2
9 - f(- 4) = - 14
f(- 4) = 23,代入
23 = 16a - 4b + 3 ==> 4a - b = 5
(4a - 2b) - 2(4a - b) = 6 - 2(5)
- 4a = - 4 ==> a = 1
b = 4(1) - 5 = - 1
所以f(x) = x² - x + 3
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