Locate the absolute extrema of the function on the closed interval

f(x) = 2(3-x) [-1,2]

2 answers

6 - 2x is a straight line with a negative slope. In the interval [-1,2], it is highest when x is smallest (-1) and lowest when x is highest (2).

You don't need to use calculus to answer this question.
drwls is right no need for calculus here but if youd like to use it

f(x)=2(3-x) [-1,2]
f'(x)=-2
no critical values therefore only test endpoints
f(-1)=8 <--- abs. max (-1,8)
f(2)=2 <---- abs. min (2,2)