Find the absolute maximum on [0,3]: g(x)=2x^3+3x^2-36x+2

Question

Find the absolute maximum on [0,3]:

Answer 1

g'(x) = 6x^2 + 6x - 36 = 0

        = x^2 + x - 6 = 0

        = (x + 3)(x - 2) = 0

so you have critical points at x = -3 and x = 2

since x = -3 we can ignore this one. We take the second derivative to get

g''(x) = 2x + 1

g''(2) = 5 > 0. Since this is a minimum, we check the endpoints x = 0 and x = 3 and plug them into g(x) to get

g(0) = 2

g(3) = 2 (3)^3 + 3 (3)^2 - 36(3) + 2

        = 2(27) + 27 - 108 + 2

        = 54 + 27 - 108 + 2

        = 81 - 108 + 2 = -25

Therefore the absolute maximum of the function in the interval [0,3] is at x = 0 where g(0) = 2.