Global optimization

Find the absolute maximum and minimum values of f(x)=x^2 - 3 x + 1 over the interval [0,2].

f(x)=x^2-3x+1
f'(x)=2x+3
x=-3/2
You will finally plug in the x and the two values from the interval 0 and 2 into the original equation.
f(0)=(0)^2+3(0)+1=1
f(x)=(-3/2)^2+3(-3/2)+1=-1.25
f(x)=(2)^2+3(2)+1=11
The abs maximum is (2,11) and the abs minimum is (0,1)