Question

Let $f\left(x\right)=|x-1|+|x-2|+|x-3|+|x-4|xϵR$ . Then

• A. $x=3$ is the point of local minima
• B. $x=2$ is the point of local minima
• C. $x=1$ is the point of local minima
• D. $x=4$ is the point of local minima

Answer 1

Answer: (A), (B)

$x=2$ is the point of local minima
$x=3$ is the point of local minima

$f\left(x\right)=|x-1|+|x-2|+|x-3|+|x-4|xϵR$

Let us plot $f\left(x\right)$ for that we will need different cases.

Case $1,x<1$

$f\left(x\right)=-(x-1)-(x-2)-(x-3)-(x-4)$

$f\left(x\right)=-4x+10$

Case $2,x>1$ and $x<2$

$f\left(x\right)=(x-1)-(x-2)-(x-3)-(x-4)$

$f\left(x\right)=-2x+8$

Case $3,x>2$ and $x<3$

$f\left(x\right)=(x-1)+(x-2)-(x-3)-(x-4)$

$=4$

Case $4,x>3$ and $x<4$

$f\left(x\right)=(x-1)+(x-2)+(x-3)-(x-4)$

$f\left(x\right)=2x-2$

Case $5,x>4$

$f\left(x\right)=(x-1)+(x-2)+(x-3)+(x-4)$

$=4x-10$

(Image)

Clearly, there are two points of minima $x=2$ and $x=3$