f(x) = (x^4)(x-1)^3

=> f ' (x) = (4x^3)(x - 1)^3 + 3(x^4)(x - 1)^2

=> f ' (x) = (x^3)(x - 1)^2 * [4(x - 1) + 3x]

=> f ' (x) = (x^3)(x - 1)^2 * [7x - 4]

for critical numbers we have f ' (x) = 0

=> (x^3)(x - 1)^2 * [7x - 4] = 0

=> (x^3) = 0

or (x - 1)^2 = 0

or 7x - 4 = 0

=> x = 0

or x = 1

or x = 4/7