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.

Answer 1

take f ' (x)
set it equal to zero and solve for x
(I see x = 3/2)

plug that back into f(x) and evaluate
Also plug in the end values of your domain

pick the smallest and largest values

Answer 2

I got my maximum as 3/2 but cannot find the minimum that is in the closed interval (1,3)

Answer 3

3/2 is not the maximum f(3/2) is the local maximum

f(3/2) = -(9/4) + 3(3/2) = 9/4 or 2.25

checking the endvalues of your domain
f(1) = -1+3 = 2
f(3) = -9 + 9 = 0

so within your domain, the absolute min is 0 and the absolute max is 2.25

look at the graph from Wolfram
http://www.wolframalpha.com/input/?i=plot+-x%5E2%2B3x+%2C+1%3Cx%3C3