Question

Let f(x)=-x^2+3x on the interval [1,3]. Find the absolute maximum and absolute minimum of f(x) on this interval.

The absolute max occurs at x=.
The absolute min occurs at x=

Answer 1

at 1, y = 2
at 3, y = 0
dy/dx = -2x + 3 = 0 if x = 3/2
at x = 3/2, y = 2.25

so max at x = 3/2
and
min at x = 3