# 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 $0$.

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why do you say that $x=0$ is a cusp point of $f?$ To my knowledge, $x=0$ is just an angular point of $f,$ according to Giaquinta's book of Mathematical Analysis, Vol I, Functions of one variable, and see Page 98 of that book. And I am very happy if you give your definition of cusp point –  azc Oct 22 '14 at 2:59

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