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题 8 设函数ｆ(ｘ) =2ｘ -ｘ2 - 1 (ｘ≥ 1 ) ,试解答下列问题 :1 )解不等式ｆ(ｘ) ≥ 2 ;2 )求出使ｆ(ｘ) 在Ｉ上是递增函数的最大的区间Ｉ ;3 )求出最大的实数ａ ,使得ｆ(ｘ) ≥ａ·ｘ恒成立 .解 1 )把 ｆ(ｘ) ≥ 2写为 2 (ｘ - 1 )≥(ｘ - 1 ) (ｘ +1 ) ,显然ｘ =1是该不
Problem 8 Let the function f(x) = 2x -x2 - 1 (x≥ 1) to answer the following questions: 1) Solve the inequality f(x) ≥ 2 ;2) Find that f(x) is incremented on I. The largest interval of the function I 3) Find the maximum real number a such that f(x) ≥ a · x is constant. Solution 1) Write f(x) ≥ 2 as 2 (x - 1 )≥(x - 1) (x +1), obviously x =1 is not