Find the absolute extrema of the function on the closed interval.

Question

f (x) = x^3 − 12x, [0, 4]

minimum (x, y) =( )
maximum (x, y) = ( )

I can't even give a setup cause I am lost please help!

Answer 1

f(x) = x^3 - 12x. You can find the maxima and minima by differentiating it (I'm assuming here that you can use calculus) to get f'(x) = 3x^2 - 12, and solving f'(x) = 0, so 3x^2 - 12 = 0, so 3x^2 = 12, so x^2 = 4, so x is either +2 or -2. But only one of those lies in the interval [0, 4], namely x = +2. The original function has at most two turning points (because that's the order of the derivative function), and only one of them lies within the interval, so any other maxima or minina must lie at the ends of the interval, i.e. either x=0 or x=4. At x=0, f(x) = 0. At x=4, f(x) = 64 - 48 = 16. At x=2, f(x) = 8 - 24 = -16. So the minimum must be -16 at x=2, and the maximum is +16 at x=4. Done.

Answer 2

Thank you :)