Question

函数的解析式设函数y=f(x)对任意x属于R,都有f(x+1)=af(x)(x>0).若当x属于(0,1]时f(x)=2

Answer 1

f(x)=a^n*(2^x+2^-x)

当n=1时,显然f(t+1)=a*f(t)=a(2^x+2^-x),0