Question

If f (x) = |x| + |x−1|, write the value of $\frac{d}{dx}\left(f\left(x\right)\right)$.

Answer 1

$$\begin{gathered} f\left(x\right)=\left|x\right|+\left|x-1\right| \\ \text{Case 1: x<0 (∴ x-1<-1<0)} \\ \left|x\right|=-x;\left|x-1\right|=-\left(x-1\right)=-x+1 \\ f\left(x\right)=-x+\left(-x+1\right)=-2x \\ f\text{'}\left(x\right)=-2 \\ \text{Case 2: 0< x <1 (∴ x>0 and x-1<0)} \\ \left|x\right|=x;\left|x-1\right|=-\left(x-1\right)=1-x \\ f\left(x\right)=x+1-x=1 \\ f\text{'}\left(x\right)=0 \\ \text{Case 3: x>1 ∴ x>1>0 ⇒ x>0)} \\ \left|x\right|=x;\left|x-1\right|=x-1 \\ f\left(x\right)=x+x-1=2x-1 \\ f\text{'}\left(x\right)=2 \\ \text{From case 1, case 2 and case 3, we have:} \\ f\text{'}\left(x\right)=\left\{\begin{array}{l}\begin{array}{c}-2,\mathrm{when}x<0\\ 0,\mathrm{when}0<x<1\\ 2,\mathrm{when}x>1\end{array}\\ \end{array}\right. \end{gathered}$$