How do you find the absolute extrema of the function on the indicated interval by using the concept of the Extreme-Value Theorem f(x) = { |x| if -3 ≤ x ≤ 2 , 4-x if 2 < x ≤ 3 ; [ -3, 3]?

1 Answer
Jul 1, 2015

Answer:

The minimum is #0# (it occurs at #0#).
The maximum is #3# (it occurs at #-3#)

Explanation:

#f(x) = { ( absx," if ", -3 <= x <= 2), (4-x," if ", 2 < x <= 3) :}#

Note first that this function is continuous on #[-3,3]#.

The only possible "problem point" is #x=2# but we see that
#lim_(xrarr2^-)f(x) = lim_(xrarr2^+)f(x) = f(2)#. So #f# is continuous at #2#.

Thus, the Extreme Value Theorem guarantees that the function attains both a minimum and a maximum on the interval.

These extrema occur at values of #x# at which either #f'(x) =0#, or #f'(x)# does not exist, or #x# is an endpoint of the interval. (That is #x=-3# or #x=3#.)

(Values of #x# that are in the domain of #f# and at which either #f'(x) =0#, or #f'(x)# does not exist are called critical numbers for #f#.)

For the function in this question, we get the following piecewise defined derivative:

#f'(x) = { ( -1," if ", -3 < x < 0), ("DNE"," if ",x = 0), ( 1," if ", 0 < x < 2), ("DNE"," if ",x = 2), (-1," if ", 2 < x < 3) :}#

Note: #DNE# is used here to indicate that the derivative does not exist. We really could, and perhaps should, leave those lines out of the definition.

The only critical numbers are #0# and #2#.

The minimum and the maximum must occur at one of the values of #x#:
#x = -3, 0, 2, 3#

Evaluating #f# at each of these, we get:

#f(-3) =3#
#f(0)=0#
#f(2)=2#
#f(3) = 1#

The minimum is #0# (it occurs at #0#).
The maximum is #3# (it occurs at #-3#).

Some teachers and textbooks express this by writing:

Minimum: #f(0) =0#
Maximum: #f(-3)=3#