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已知函数f（X）=ax+b,若f(0)=1,且f(x+1)=f(x+1)=f(x)+3,求f(x).
已知函数f（X）=ax+b,若f(0)=1,且f(x+1)=f(x+1)=f(x)+3,求f(x).
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将f（0）带进去得到b=1 ,则f（x+1）=a（x+1）+1 ,f（x）+3=ax+4
又f(x+1)=f(x)+3,即a（x+1）+1 =ax+4 化简得到a=3
所以f(x）=3x+1
又f(x+1)=f(x)+3,即a（x+1）+1 =ax+4 化简得到a=3
所以f(x）=3x+1
1年前
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