Question

已知二次函数f(x)满足f(2x+1)=4x的平方-6x+5，求f(x) 求高人会达

已知二次函数f(x)满足f(2x+1)=4x的平方-6x+5，求f(x) 求高人会达详细一点 可以的话请说明是用什么方法做题的

Answer 1

解：f(2x+1)=4x^2-6x+5
=(2x+1)^2-（4x+1)-6x+5
=(2x+1)^2-5(2x+1)+5+4
=(2x+1)^2-5(2x+1)+9
∴f(x)=x^2-5x+9

Answer 2

令2x+1=t
x=(t-1)/2
f(2x+1)
=f(t)=4[(t-1)/2]^2-6[(t-1)/2]+5
=4(t-1)^2/4-3(t-1)+5
=t^2-2t+1-3t+3+5
=t^2-5t+9

∴f(x)=x^2-5x+9

Answer 3

令2x+1=u, x=(u-1)/2.
f(u)=4*[(u-1)/2]^2-6(u-1)/2+5.
f(u)=(u-1)^2-6(u-1)+5.
=(u-1-5)(u-1-1).
=(u-6)(u-2)
=u^2-8u+12.
∴f(x)=x^2-8x+12.