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数学
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函数f(x)=cx/2x+3,(x不等于-3/2)满足f[f(x)]=x,则常数c等于几?为什么?
人气:377 ℃ 时间:2019-10-17 00:57:42
解答
设y=f(x)
则y=f(x)=cx/(2x+3)
y=cx/(2x+3)
x=f[f(x)]=f(y)=cy/(2y+3)
所以
cx=2xy+3y
cy=2xy+3x
两式相减得:
c(x-y)=3(y-x)
所以c=-3
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