设m,t为实数,函数f(x)=(mx+t)/(x^2+1),f(x)的图像在点M(0,f(0))处的切线斜率为1求m
问题描述:
设m,t为实数,函数f(x)=(mx+t)/(x^2+1),f(x)的图像在点M(0,f(0))处的切线斜率为1求m
答
f'(x)=[(mx+t)'*(x²+1)-(mx+t)*(x²+1)']/(x²+1)²=[m(x²+1)-2x(mx+t)]/(x²+1)²在点M(0,f(0))处的切线斜率为1即f'(0)=1所以(m-0)/(0+1)²=1m=1