Question

critical numbers of f(x)=(x^2-4) ^ 2/3 find relative max and min and point of inflection

Answer 1

f'(x)=(2x).(2/3)(x^2-4)^-1/3=0
this gives x=0,2,-2
now put these value in f''(x). and find max(-ve) , min(=ve) , and point of inflection(0)