作业帮函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?
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![函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?](/uploads/image/z/3697597-37-7.jpg?t=%E5%87%BD%E6%95%B0f%28x%29%3Dcx%2F2x%2B3%28x%E2%89%A0-3%2F2%29%2C%E6%BB%A1%E8%B6%B3f%5Bf%28x%29%5D%3Dx%2C%E5%88%99%E5%B8%B8%E6%95%B0c%3D%E5%B7%B2%E7%9F%A5g%28x%29%3D1-2x%2Cf%5Bg%28x%29%5D%3D%281-x%5E2%29%2Fx%5E2%28x%E2%89%A00%29%2C%E9%82%A3%E4%B9%88f%281%2F2%29%3D%3F)
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函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?
函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=
已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?
函数f(x)=cx/2x+3(x≠-3/2),满足f[f(x)]=x,则常数c=已知g(x)=1-2x,f[g(x)]=(1-x^2)/x^2(x≠0),那么f(1/2)=?
3
3
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