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
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                    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
                    
                    
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                    ,問得好
                     -大醬風度- 
                    ♂                    
                        
                    
                                                            
                        
                    
                                        
                    
                    
                                        
                        (113 bytes)
                        ()
                        08/28/2024 postreply
                                                12:23:41
                    
                    
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