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
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).