Find the critical points of the function f(x) = (x^2-4)^2/5

Answer 1

f'(x) = 4x(x² - 4)/5 = 0; x = 0 or x = - 2 or x = 2

f''(x) = 4/5(3x² - 4); f''(0) = -16/5 < 0; at x = 0 we have max (0, f(0)) = (0, 16/5)

f''(2) = f''(- 2) = 32/5 > 0. So we have 2 min: (- 2, f(- 2)) = (- 2, 0) and (2, f(2)0 = (2, 0)