对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值
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![对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值](/uploads/image/z/10365837-69-7.jpg?t=%E5%AF%B9%E4%BA%8E%E5%87%BD%E6%95%B0f%28x%29%3Dlg1%2Bx%2F1-x%2C%E8%8B%A5f%28y%2Bz%2F1%2Byz%29%3D1%2Cf%28y-z%2F1-yz%29%3D2%2C%E5%85%B6%E4%B8%AD-1%26lt%3By%26lt%3B1%2C-1%26lt%3Bz%26lt%3B1%2C%E6%B1%82f%28yf%28Y%29%E5%92%8Cf%28z%29%E7%9A%84%E5%80%BC)
对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值
对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(y
f(Y)和f(z)的值
对于函数f(x)=lg1+x/1-x,若f(y+z/1+yz)=1,f(y-z/1-yz)=2,其中-1<y<1,-1<z<1,求f(yf(Y)和f(z)的值
注意到:
Ka=1+(y+z)/(1+yz)=(1+y+z+yz)/(1+yz)=(1+y)(1+z)/(1+yz)
Kb=1-(y+z)/(1+yz)=(1-y-z+yz)/(1+yz)=(1-y)(1-z)/(1+yz)
Kc=1+(y-z)/(1-yz)=(1+y-z-yz)/(1-yz)=(1+y)(1-z)/(1-yz)
Kd=1-(y-z)/(1-yz)=(1-y+z-yz)/(1-yz)=(1-y)(1+z)/(1-yz)
所以:
Ka/Kb=[(1+y)/(1-y)][(1+z)/(1-z)]
Kc/Kd=[(1+y)/(1-y)][(1-z)/(1+z)]
因为f(x)=lg[(1+x)/(1-x)]
所以f((y+z)/(1+yz))=lg(Ka/Kb)=lg[(1+y)/(1-y)][(1+z)/(1-z)]
=lg[(1+y)/(1-y)]+lg[(1+z)/(1-z)]
=f(y)+f(z)
同理f((y-z)/(1-yz))=f(y)-f(z)
解方程组:
f(y)+f(z)=1
f(y)-f(z)=2
可得:
f(y)=1.5
f(y)=-0.5