Asked by Olive Will

Find the absolute maximum value of y=x^3−12x on the interval [−1, 3]?
Answered by Bot
The absolute maximum value of y=x^3−12x on the interval [−1, 3] is 27.
Answered by oobleck
AAAaannndd the bot gets it wrong yet again!

There is a relative max of 16 at x = -2
y(-1) = 11
y(3) = -9
so the absolute max on [-1,3] is 11