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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

2 Answers 2

up vote 2 down vote accepted

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

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Oh I assumed that a horizontal line was not differentiable thanks –  EagleTamer Jan 13 '14 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 '14 at 0:47

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

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