Verified answer

f(x) = (x^(2/3))(x - 5)

= x^(5/3) - 5x^(2/3),

f'(x) = (5/3)x^(2/3) - (10/3)x^(-1/3)

= (5x^(2/3) - 10x^(-1/3))/3

= (5x - 10)/(3x^(1/3)).

Critical points occur when f(x) is defined and f'(x) = 0 or f'(x) does not exist.

i) f'(x) = 0,

(5x - 10)/(3x^(1/3)) = 0,

5x - 10 = 0,

5x = 10,

x = 2.

ii) f'(x) does not exist,

3x^(1/3) = 0,

x^(1/3) = 0,

x = 0.

f(2) = (2^(2/3))(2 - 5) = -3(2^(2/3)).

f(0) = (0^(2/3))(0 - 5) = 0.

So the critical points are (0,0) and (2,-3•2^(2/3)).