lim[f(1)-f(1-2x)]/2x
=lim[f(1)-f(1-2x)]/(0-2x)
=f'(1)
=-1
∴ 曲线y=f(x)在点(1,f(1))处的斜率是-1f'(1)=-1怎么来的?f'(1)=lim[f(1)-f(1-2x)]/2x不是-f(x)'吗lim[f(1)-f(1+【-2x】)]/(-2x)=-f(1)'=-1那不就是1了吗lim[f(1)-f(1-2x)]/2x=lim[f(1+【-2x】)-f(1)]/(-2x)=f'(1)
设f(x)为可导函数,且满足lim[f(1)+f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率
设f(x)为可导函数,且满足lim[f(1)+f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率
设f(x)为可导函数,且满足lim[f(1)-f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率
设f(x)为可导函数,且满足lim[f(1)-f(1-2x)]/2x=-1,x趋于0时,求曲线y=f(x)在点(1,f(1))处的斜率
数学人气:987 ℃时间:2019-08-14 01:20:58
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