What are the maximum and minimum values of the function #abs(2x-3# at closed intervals [1,2]?

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what are the maximum and minimum values of the function #abs(2x-3# at closed intervals [1,2]

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Answer 1 · 2018-03-28T16:22:51

Please see below for a non-calculus answer.

Non-calculus answer

The graph the absolute value function is #V# shaped.

The graph of #f(x) = abs(2x-3)# is a translation and shrinking/stretching of the graph of #absx#. So it also has a #V# shape.

Therefore, the maximum value on a closed, bounded interval occurs at an endpoint of the interval.

For #f(x) = abs(2x-3)# on #[1,2]# the value of #f# is #1# at both endpoints, so the maximum is #1#.

Clearly #f(x) = abs(2x-3) >= 0# for all #x# and it is #0# only at #x = 3/2#.

Since #3/2# is in #[1,2]#, the minimum value of #f# on the interval #[1,2]# is #0# (this value occurs at #x=3/2#)

Answer 2 · 2018-03-28T16:29:56

Please see below for a calculus answer.

Using the closed interval method for finding absolute extreme values:

#f(x) = abs(2x-3) = {(2x-3,"if", x >= 3/2),(-2x+3,"if", x < 3/2):}#

So #f'(x) = {(2,"if", x > 3/2),(-2,"if", x < 3/2):}#.

#f'(x)# is never #0# and #f'(3/2)# does not exist.

The only critical number for #f# is #3/2#.

Testing on the interval #{1,2}#

#{:(x," ",1," ",3/2," ",2),(f(x)," ",1," ",0," ",1):}#

The maximum is #1# and the minimum is #0#.