Question

The minimum value of $|1-x|+|3+x|+|5-x|$ is

• A. 8
• B. 12
• C. 9
• D. 7

Answer 1

Answer: (A)

8

$f\left(x\right)=\{\begin{array}{cc}1-x-(3+x)+5-x=3-3x& x\le -3\\ 1-x+3+x+5-x=9-x,& -3<x<1\\ x-1+3+x+5-x=7+x& 1\le x<5\\ x-1+3+x+x-5=3x-3& x\ge 5\end{array}$
$f^{{}^{\prime }}\left(x\right)<0$ in $(-\infty ,-3)$ and $(-3,1)$ and
$f^{{}^{\prime }}\left(x\right)>0in[1,5)$ and $[5,\infty )$
$\therefore f^{{}^{\prime }}\left(x\right)$ is decreasing in $[-\infty ,1)$ and increasing in $[1,\infty ).$
So the minimum value in $[-\infty ,1)$ is $f\left(1\right)=7+1=8.$