已知函数f(x)=e^x-e^-x,g(x)=e^x+e^-x 设f(x)f(y)=4,g(x)g (y)=8,求g(x

f(x)=e^x-e^x,g(x)=e^x+e^(-x)
f(x)f(y)=8
即(e^x-e^(-x)][e^y-e^(-y)]=4
e^(x+y)-e^(-x-y)-e^(-x+y)+e^(x-y)=4
∴ g(x+y)-g(x-y)=4 ①
g(x)g(y)=8
即[e^x+e^(-x)][e^y+e^(-y)]=8
e^(x+y)+e^(-x-y)+e^(-x+y)+e^(x-y)=8
g(x+y)+g(x-y)=8 ②
①②解得:
g(x+y)=6 g(x-y)=2
因此:
g(x+y)/g(x-y)=3