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Question
If f (x) = |x| + |x−1|, write the value of \[\frac{d}{dx}\left( f (x) \right)\]
Solution
\[f\left( x \right) = \left| x \right| + \left| x - 1 \right|\]
\[\text{ Case }1: x<0 (\therefore x-1<-1<0)\]
\[\left| x \right| = - x; \left| x - 1 \right| = - \left( x - 1 \right) = - x + 1\]
\[f\left( x \right) = - x + \left( - x + 1 \right) = - 2x\]
\[f'\left( x \right) = - 2\]
\[\text{ Case } 2: 0< x <1 (\therefore x>0 \text{ and } x-1<0)\]
\[\left| x \right| = x; \left| x - 1 \right| = - \left( x - 1 \right) = 1 - x\]
\[f\left( x \right) = x + 1 - x = 1\]
\[f'\left( x \right) = 0\]
\[\text{ Case } 3: x>1 \therefore x>1>0 \Rightarrow x>0)\]
\[\left| x \right| = x; \left| x - 1 \right| = x - 1\]
\[f\left( x \right) = x + x - 1 = 2x - 1\]
\[f'\left( x \right) = 2\]
\[f'(x)=\begin{cases}-2, \text{When } x < 0 \\0, \text{When }0 < x <1\\2, \text{When } x >1 \end{cases}\]
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