设f(x)在x=3点可导,则lim{[f(3-h)-f(3)]/2h}=? h →0
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设f(x)在x=3点可导,则lim{[f(3-h)-f(3)]/2h}=? h →0
![设f(x)在x=3点可导,则lim{[f(3-h)-f(3)]/2h}=? h →0](/uploads/image/z/17392774-22-4.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8x%3D3%E7%82%B9%E5%8F%AF%E5%AF%BC%2C%E5%88%99lim%7B%5Bf%283-h%29-f%283%29%5D%2F2h%7D%3D%3F+h+%E2%86%920)
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)
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)
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