设f(x)= (I)求f'(x); (Ⅱ)证明:x=0是f(x)的极大值点; (Ⅲ)令x n ,考察f'(x n )是正的还是负的,n为非零整数; (Ⅳ)证明:对
【正确答案】正确答案:(I)当x≠0时按求导法则得 当x=0时按导数定义得 (II)由于f(x)一f(0)=一x 2 (2+sin )<0(x≠0),即f(x)<f(0),于是由极值的定义可知x=0是f(x)的极大值点. (III)令x N (N=±1,±2,±3,…),则sin =0,cos =(一1) n ,于是 (Ⅳ)对 δ>0,当n为 负奇数且|n|充分大时x n ∈(一δ,0),f'(x n )<0 f(x)在(一δ,0)不单调上升;当n为正偶数且n充分大时x n ∈(0,δ),f'(x n )>0
【答案解析】