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已知函数y=f(x),f(1)=2 f(x+3)≤f(x)+3 f(x+2)≥f(x)+2,求f(2009)
人气:344 ℃ 时间:2020-05-21 14:12:31
解答
f(2009)=2010
f(x+3)<=f(x)+3.(1)
f(x+2)>=f(x)+2.(2)
将x=x+1代入(2)
f(x+3)>=f(x+1)+2又f(x+3)<=f(x)+3
即f(x+1)+2<=f(x+3)<=f(x)+3
f(x+1)<=f(x)+1.(3)
将x=x+2代入(2)
f(x+4)>=f(x+2)+2>=f(x)+4>=f(x)+3+1>=f(x+3)+1
即f(x+4)>=f(x+3)+1
即f(x+1)>=f(x)+1.(4)
(3)(4)连立f(x+1)=f(x)+1
f(1)=2,f(2)=3.,f(2009)=2010
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