Absolute Value Graphing and Differentiability

If $f(x)= |x|$ and $g(x) = -|x|$, show that $(f+g)(x)$ is differentiable at $0$ while neither $f$ nor $g$ is.

I understand that neither $f$ or $g$ is not differentiable at $0$ because there is a cusp point at the origin. What I don't understand is that if I added both functions together wouldn't it just equal $0$.

-

2 Answers

$(f + g)(x) = |x| - |x| = 0$ which is just a constant function like $h(x) = 5$. Any constant function is differentiable, and differentiates to 0.

-
Oh I assumed that a horizontal line was not differentiable thanks –  EagleTamer Jan 13 at 0:46
Differentiation gives the rate of change of a function, if the function is constant everywhere then it doesn't change, hence the derivative is 0. –  Daniel Jan 13 at 0:47

The sum is in fact equal to the constant function with value zero and therefore it is differentiable..

-