函数f(x)=x^2+lnx,若曲线 y=f(x)在点(1,f(1))处的切线方程为y=ax+b,a+b=

2025-05-15 16:34:25
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回答1:

f(x)=x^2+lnx => f '(x) = 2x + 1/x
f(1) = 1, f '(1) = 3
点(1,f(1))处的切线方程: y-1 = 3(x-1)
即 y = 3x - 2
=> a=3, b= -2
a+b = 1