The function f(x) = x^3 - 48x has a local minimum at x = and a local maximum at x = ? f'(x) = 3x^2 - 48 Set this equal to 0: 3x^2 - 48 = 0 Add 48 to each side: 3x^2 = 48 Divide each side by 3: x^2 = 16 Therefore, x = -4, 4 Test f(4) f(4) = 4^3 - 48(4) f(4) = 64 - 192 f(4) = -128 <-- Local minimum Test f(-4) f(-4) = -4^3 - 48(-4) f(-4) = -64 + 192 f(-4) = 128 <-- Local maximum