1.f(a+3h)-f(a-h)=f(a+3h)-f(a)+f(a)-f(a-h),
limh→0[f(a+3h)-f(a)]/3h=f'(a),
limh→0[f(a-h)-f(a)]/(-h)=f'(a),
所以 limh→0[f(a+3h)-f(a-h)]/2h
=limh→0[f(a+3h)-f(a)+f(a)-f(a-h)]/2h
=limh→0[f(a+3h)-f(a)]/2h+limh→0[f(a)-f(a-h)]/2h
=limh→03/2[f(a+3h)-f(a)]/3h+1/2limh→0[f(a-h)-f(a)]/(-h)
=3/2f'(a)+1/2f'(a)=2f'(a)
2.与上题类似,lim x->0[f(1)-f(1-2x)]/2x=f'(1),
所以f'(1)=-1,因为函数在点x0处的导数几何意义就是函数在此点的切线的斜率,所以,曲线y=f(x)上的点（1,f(1)）处 的切线的斜率为-1