Question

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.

Answer 1

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)`.