Expert: Sherry Wallin - 7/22/2010

Question
Hello Sir
I was confused by this problem;

I know the function |X-1|(absolute Value) is not differentiable at one because it has sharp corner in that point
but is the function differentiable in (-∞,1] including one
and on [1,∞) including one?
I am confused since it is not differentiable at 1 how can those inclusive braces be used?

Answer
Michael~

The absolute value function is really a piecewise function:
f(x) = |x - 1| is the same as saying:

             x - 1  for x in [1,∞)
f(x) =
            -(x - 1)  for x in (-∞,1] which is the same as saying 1 - x  in (-∞,1]

If you use the definition of the derivative: A function f is said to be differentiable at 'a' if the limit of the difference quotient exists or if
f'(a) = lim (h->0) [f(a + h) - f(a)]/h exists

Since the absolute function is defined piecewise you need to check it on both pieces:

f'(x) = lim(h->0) [(x + h) - 1 - (x - 1)]/h = lim(h->0)[x + h - 1 - x + 1]/h
= lim(h->0) h/h = lim(h->0) 1 = 1

and

f'(x) = lim(h->0) [1 - (x + h) - (1 - x)]/h = lim(h->0)[1 - x - h - 1 + x]/h
= lim(h->0) -h/h = lim(h->0) -1 = -1

Hopefully this clears up your questions, if not please write more.

Math Prof

** In a sense you are examining each piece as if they are separate functions. On the interval [1,∞) you are actually examining what happens as h approaches 0 from the right and on the interval (-∞,1] you are examining as h approaches 0 from the left. Just because the limit of |x - 1| doesn't exist doesn't mean it's individual pieces don't have a limit.
If you look at the graph (from left to right) the slope of the tangent line is -1 for the first piece and the slope of the second piece is 1, these both exist but are not the same and this is why we say the whole function is not differentiable because in order to be differentiable the limits must both exist and be the same, and in this case they fail because the limits  are not equal.

!!! Sorry I made a couple of typos on the first draft but I believe the typos are corrected in this version, sorry...