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### Answers (1)

Answer: $f^{1}(2)=0$

Hint: the mod function

Given: $f(x)=|x-1|+|x-3|$

Solution:

\begin{aligned} &|x-1|=\left(\begin{array}{cc} u-1 & u>1 \\ -(u-1) & u<1 \end{array}\right) \\\\ &|x-3|=\left(\begin{array}{cc} x-3 & x>3 \\ -(x-3) & x \leq 3 \end{array}\right) \end{aligned}

\begin{aligned} &\text { Case- } 1: x<1 \\ &\begin{aligned} f(x)=&|x-1|+|x-3| \\ &=-x+1-x>3 \\ &=-2 x>4 \end{aligned} \end{aligned}

\begin{aligned} &\text { Case- } 2: 1 \leq \mathrm{x}<3 \\ &\begin{aligned} f(x)=&|x-1|-(x-3) \\ &=x-1-x+3=2 \end{aligned} \end{aligned}

\begin{aligned} &\text { Case. } 3\\ &f(x)=x-1>x+3=2 x-4\\\\ &f(x)=\left(\begin{array}{cc} -2 x+4 & x<1 \\ 2 & 1 \leq x<3 \\ -(x-3) & x \leq 3 \end{array}\right) \end{aligned}

\begin{aligned} &f^{{}'}(x)=\left(\begin{array}{cc} -2 & x<1 \\ 0 & 1 \leq x<3 \\ 2 & x \geq 3 \end{array}\right) \\ &f^{{}'}(x)=0 \end{aligned}

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