Function f(x) = | x | − | x − 1 | is monotonically increasing when

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0 < x < 1 f\left( x ight) = \left| x ight| - \left| x - 1 ight| ext { Case 1: Let }x < 0 ext { If x < 0 , then }\left| x ight| = - x \Rightarrow \left| x - 1 ight| = - \left( x - 1 ight) ext { Now,} f\left( x ight) = \left| x ight| - \left| x - 1 ight| = - x - \left( - x + 1 ight) = - x + x - 1 = - 1 f'\left( x ight) = 0 ext { So,f }\left( x ight) ext { is not monotonically increasing when x< 0.} ext { Case 2: Let }0 < x < 1 ext { Here,} \left| x ight| = x \Rightarrow \left| x - 1 ight| = - \left( x - 1 ight) ext { Now,} f\left( x ight) = \left| x ight| - \left| x - 1 ight| = x + x - 1 = 2x - 1 f'\left( x ight) = 2 > 0 ext { So },f\left( x ight) ext { is monotonically increasing when }0 < x < 1 . ext { Case 3: Let x > 1} ext { Ifx > 0, then }\left| x ight| = x \Rightarrow \left| x - 1 ight| = \left( x - 1 ight) ext { Now,} f\left( x ight) = \left| x ight| - \left| x - 1 ight| = x - x + 1 = 1 f'\left( x ight) = 0 ext { So },f\left( x ight) ext { is not monotonically increasing when x >1 }. ext { Thus },f\left( x ight) ext { is monotonically increasing when 0 < x < 1} .

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