Playing with Graphs - 8

Calculus Level 4

Given that

{f(x)=xg(x)=∣1−f(x)∣h(x)=2−g(x)L(x)=h(∣x∣)+∣h(x)∣\begin{cases} f (x) = x \\ g (x) = | 1 - f (x) | \\ h (x) = 2 - g (x) \\ L (x) = h (|x|) + | h (x) | \end{cases}

What is the number of points where L(x)L (x) is non-differentiable?

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Aniket Sanghi by Aniket Sanghi
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