为什么是f(a-a+1)≤f(3/4).求解析
a²-a+1
=(a-1/2)²-1/4+1
=(a-1/2)²+3/4
(a-1/2)²≥0
可知
(a-1/2)²+3/4≥3/4
因为是减函数,所以
f(a²-a+1)≤f(3/4)
f(a²-a+1)=f(a²-a+¼+¾)=f[(a-½)²+¾]∵(a-½)²≥0∴(a-½)²+¾≥¾
又∵f(x)为减函数
∴f(a²-a+1)≤f(¾)
a²-a+1=(a-1/2)²+3/4≥3/4
∵f(x)在(0,+∞)单调递减
∴f(a²-a+1)≤f(3/4)
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