Question

如何用函数方程法解题

f(x)+2f(2-x)=3x²-8x+8这道题怎么解

Answer 1

f(x)+2f(2-x)=3x²-8x+8,①
以2-x代x,得f(2-x)+2f(x)=3(2-x)^2-8(2-x)+8,②
②*2-①，得3f(x)=6(2-x)^2-3x^2-32+24x+8,

所以f(x)=2(2-x)^2-x^2+8x-8
=x^2.

Answer 2

有待定系数法，设f(x)=ax^2+bx+c
f(x)+2f(2-x)
=ax^2+bx+c+2[a(2-x)^2+b(2-x)+c]
=ax^2+bx+c+2(4a-4ax+ax^2+2b-bx+c)
=ax^2+bx+c+8a-8ax+2ax^2+4b-2bx+2c
=3ax^2-(b+8a)x+3c+8a+4b
=3x^2-8x+8
所以3a=3，8a+b=8，8a+4b+3c=8
a=1，b=0，c=0
所以f(x)=x^2

Answer 3

f(x)+2f(2-x)=3x²-8x+8 ①
f(2-x)+2f(x)=3(2-x)²-8(2-x)+8 ②
②*2-① f(x)=2*[3(2-x)²-8(2-x)+8]-3x²-8x+8