Question

函数f(x)=(x+1)^(1/2)+b/x在x=3时有拐点,求b

Answer 1

就是二阶导数在x=3时为0

f'=1/[2倍根号下（x+1)]-b/x^2

f''=-1/4(x+1）^(-3/2)=2b x^(-3)

当x=3时,f''(3)=0.即得b=27/64,自己仔细再算算吧