Let # f(x) = |x-1|. # 1) Verify that # f(x) # is neither even nor odd. 2) Can # f(x) # be written as the sum of an even function and an odd function ? a) If so, exhibit a solution. Are there more solutions ? b) If not, prove that it is impossible.

1 Answer
Max G.
Feb 6, 2018

Let #f(x) = | x -1 |#.
If f were even, then #f(-x)# would equal #f(x)# for all x.
If f were odd, then #f(-x)# would equal #-f(x)# for all x.
Observe that for x = 1
#f(1) = | 0 | = 0#
#f(-1) = | -2 | = 2#
Since 0 is not equal to 2 or to -2, f is neither even nor odd.

Might f be written as #g(x) + h(x)#, where g is even and h is odd?

If that were true then #g(x) + h(x) = | x - 1|#. Call this statement 1.
Replace x by -x.
#g(-x) + h(-x) = | -x - 1|#
Since g is even and h is odd, we have:
#g(x) - h(x) = | -x - 1|# Call this statement 2.

Putting statements 1 and 2 together, we see that
#g(x) + h(x) = | x - 1|#
#g(x) - h(x) = | -x - 1|#
ADD THESE to obtain
#2g(x) = | x - 1| + | -x - 1|#
#g(x) = (| x - 1| + | -x - 1|)/2#

This is indeed even, since #g(-x) = (| -x - 1| + | x - 1|)/2 = g(x)#

From statement 1
#(| -x - 1| + | x - 1|)/2 + h(x) = | x - 1|#
#| -x - 1|/2 + | x - 1|/2 + h(x) = | x - 1|#
#h(x) = | x - 1|/2 - | -x - 1|/2#

This is indeed odd, since
#h(-x) = | -x - 1|/2 - | x - 1|/2 = -h(x)#.

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