已知a>0,函数f(x)=x 3 -a,x∈(0,+∞),设x 1 >0,记曲线y=f(x)在点(x 1 ,f(x 1 ))处的切线为l, (1)求l的方程; (2)设l与x轴交点为(x 2 ,0)证明: ① x 2 ≥ a 1 3 ; ②若 x 2 > a 1 3 则 a 1 3 < x 2 < x 1 .
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