由題意
f(f(x+1))=(x+1)²-x-1+1=x²+x+1=f(f(x))+2x
得:f(f(1))=f(f(0))
求逆: f(1)=f(0) (1)
由原式:f(f(1))=1
故1為函數的不動點:f(1)=1
由式(1)得 f(0)=1
由題意
f(f(x+1))=(x+1)²-x-1+1=x²+x+1=f(f(x))+2x
得:f(f(1))=f(f(0))
求逆: f(1)=f(0) (1)
由原式:f(f(1))=1
故1為函數的不動點:f(1)=1
由式(1)得 f(0)=1
• 請教,由原式:f(f(1))=1 如何得到 “故1為函數的不動點:f(1)=1” -wxcfan123- ♂ (111 bytes) () 08/28/2024 postreply 11:40:29
• ,問得好 -大醬風度- ♂ (113 bytes) () 08/28/2024 postreply 12:23:41
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