题目

已知函数f（x）=ax+（a＞1） （1）证明：函数f（x）在（﹣1，+∞）上为增函数； （2）用反证法证明f（x）=0没有负数根． 答案：【考点】反证法与放缩法；函数单调性的判断与证明． 【分析】（1）由于函数f（x）=ax+1﹣，而函数 y=ax（a＞1）和函数y=﹣在（﹣1，+∞）上都为增函数，可得函数f（x）在（﹣1，+∞）上为增函数． （2）假设f（x）=0有负数根为x=x0＜0，则有+1= ①．分当x0∈（﹣1，0）时、当x0∈（﹣∞，﹣1）两种情况 —Mary’s got crazy and has been sent to the mental hospital. Did you tell her boss about that?  —Yes, but I _______her husband first.    A. can have told    B. should have told   C. need have told     D. must have told