Question

If $f\left(x\right)=|x-2|+|x-5|$ then find $f^{\prime }\left(4\right)$.

Answer 1

$\begin{array}{c}{f}^{\prime }\left(x\right)=\left|x-2\right|+\left|x-5\right|\\ =\{\begin{array}{c}-\left(x-2\right)-\left(x-5\right),x<2\\ x-2-\left(x-5\right),2<x<5\\ x-2+x-5,x<5\end{array}\\ =\{\begin{array}{c}-x+2-x+5,x<2\\ x-2-x+5,2<x<5\\ 2x-7,x>5\end{array}\\ =\{\begin{array}{c}-2x+7,x<2\\ 3,2<x<5\\ 2x-7,x>5\end{array}\\ f^{\prime }\left(x\right)=\{\begin{array}{c}-2,x<2\\ 0,2<x<5\\ 2,x>5\end{array}\\ So,f^{\prime }\left(4\right)=0\end{array}$