Q.If $f(x)=\min\{|x|,|x-1|,|x+1|\}$ for $-1\le x\le 2$ , number of points of non-differentiability is

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.If f(x)=min⁡{∣x∣,∣x−1∣,∣x+1∣}f(x)=\min\{|x|,|x-1|,|x+1|\}f(x)=min{∣x∣,∣x−1∣,∣x+1∣} for −1≤x≤2-1\le x\le 2−1≤x≤2, number of points of non-differentiability is