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例设函数f(x)=ex-1-x-ax2.(1)若a=0,求f(x)的单调区间;(2)若当x≥0时,f(x)≥0,求a的取值范围.参考答案如下:(1)a=0时,f(x)=ex-1-x,f′(x)=ex-1.当x∈(-∞,0)时,f′(x)<0;当x∈(0,+∞)时,f′(x)>0.故f(x)在(-∞,0)上单调减少,在(0,+∞)上单调增加.
(1) If a = 0, find the monotone interval of f (x); (2) if f (x) ≥ 0 when x≥0, Find the range of a. The reference answers are as follows: (1) When a = 0, f (x) = ex-1-x and f ’(x) = ex- , f (x) <0; when x ∈ (0, + ∞), f ’(x)> 0. So f (x) decreases monotonically at (-∞, 0) ) Monotonically increase.