calculus

a) To find the maximum of the function in the interval [0,5], we need to determine the turning points of the function. To find the turning points, we need to solve the derivative. We have f'(x) = 6x^2 - 42x + 72 =0 By solving this equation, we have x=0 and x=6 Since the interval [0,5] only includes 0 and 5, the turning point will only occur at x=0. Therefore, the maximum will be when x=5, which is f(5) = 2(5)^3 - 21(5)^2 + 72(5) = 240. b) To find the maximum of the function in the interval [0,4], we need to determine the turning points of the function. To find the turning points, we need to solve the derivative. We have f'(x) = 6x^2 - 42x + 72 =0 By solving this equation, we have x=0 and x=6 Since the interval [0,4] only includes 0, the turning point will only occur at x=0. Therefore, the maximum will be when x=4, which is f(4) = 2(4)^3 - 21(4)^2 + 72(4) = 192.