(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)-f(0)=-x 2 <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 =(-1) n ,于是 f'(x n )= (Ⅳ)对 >0,当n为 负奇数且|n|充分大时x n ∈(-δ,0),f'(x n )<0 f(x)在(-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】