Question

求极值，高等数学

Answer 1

f(x)=(x-4)(x+1)^(2/3)
f'(x)=(x+1)^(2/3)+(2/3)*(x-4)/(x+1)^(1/3)
=(5/3)*(x-1)/(x+1)^(1/3)
当x<-1或x>1时，f'(x)>0
当-1<x<1时，f'(x)<0
当x=1时，f'(x)=0
所以x=1是f(x)的极小值点，f(1)=(-3)*4^(1/3)