Asked by Jenn
Locate the absolute extrema of the function on the closed interval
f(x) = 2(3-x) [-1,2]
f(x) = 2(3-x) [-1,2]
Answers
Answered by
drwls
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.
You don't need to use calculus to answer this question.
Answered by
Anonymous
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)
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)