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January 已知f[f(x)=2x-1,则一次函数f(x)=_____?
已知f[f(x)=2x-1,则一次函数f(x)=_____?
sqrt(2)*x+1-sqrt(2),or -sqrt(2)*x+sqrt(2)+1
最佳答案 - 由投票者2007-10-05 23:22:00选出
解:设f(x)=ax+b,则f[f(x)]=a(ax+b)+b=a^2x+(a+1)b=2x-1.于是
a^2=2........(1)
(a+1)b=-1.....(2).
由(1)(2)得a=√2,b=1-√2,或a=-√2,b=1+√2. 故所求的一次函数为:
y=(√2)x-1+√2,或y=(-√2)x-1-√2.
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