Question

The minimum value of$f\left(x\right)=|3-x|+|2+x|+|5-x|$

• A. $7$
• B. $10$
• C. $8$
• D. $0$

Answer 1

The correct option is A $7$

Let us redefine $f\left(x\right)$ for considering values less than or greater than -2, 3 and 5 because of different mods appearing in it.Consider the following case $x\le -2,-2\le x<3,3\le x<5,x\ge 5$
$f\left(x\right)=y=\{\begin{array}{cc}6-3x& x\le -2\\ 10-x& -2\le x<3\\ x+4& 3\le x<5\\ 3x-6& x\ge 5\end{array}$
$\frac{dy}{dx}=-ive$ for $x\le 2,-2\le x<3$ or in $(-\infty ,3)$so $f\left(x\right)$ is decreasing function in $(-\infty ,3)$
$\frac{dy}{dx}=+ive$ for $3\le x<5,x\ge 5$ or in $(3,\infty )$ so that $f\left(x\right)$ is an increasing function in $(3,\infty )$.
Hence $x=3$ is a point of minimum value of $f\left(x\right)$ as the function changes from decreasing to increasing.
$\therefore f\left(3\right)=0+5+2=7$ is minimum value of $f\left(x\right)$.
Ans: A