(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
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
