设t=-h,h →0 ,则t →0
h →0 lim{[f(3-h)-f(3)]/2h}
= t →0 lim{[f(3+t)-f(3)]/(-2t)}
=(-1/2) t →0lim{[f(3+t)-f(3)]/t}
f(x)在x=3点可导
则按导数定义
t →0 lim{[f(3+t)-f(3)]/t}=f'(3)
所以答案为(-1/2)f'(3)
设t=-h,h →0 ,则t →0
h →0 lim{[f(3-h)-f(3)]/2h}
= t →0 lim{[f(3+t)-f(3)]/(-2t)}
=(-1/2) t →0lim{[f(3+t)-f(3)]/t}
f(x)在x=3点可导
则按导数定义
t →0 lim{[f(3+t)-f(3)]/t}=f'(3)
所以答案为(-1/2)f'(3)