Question

The range of the function $f\left(x\right)=|x-1|+|x-8|$ is

• A. $[7,\infty )$
• B. $R-\{1,8\}$
• C. $[0,\infty )$
• D. $[8,\infty )$

Answer 1

The correct option is A $[7,\infty )$

$f\left(x\right)=|x-1|+|x-8|$
Clearly, domain is $R$

$f\left(x\right)=\{\begin{array}{cc}1-x+8-x,& x<1\\ x-1+8-x,& 1\le x<8\\ x-1+x-8,& x\ge 8\end{array}$

$f\left(x\right)=\{\begin{array}{cc}9-2x,& x<1\\ 7,& 1\le x<8\\ 2x-9,& x\ge 8\end{array}$


Clearly, from the graph, we can conclude that range of $f$ is $[7,\infty )$

Alternate:
If $f\left(x\right)=|x-a_1|+|x-a_2|+|x-a_3|+\cdots +|x-a_n|,$ then the minimum value of $f$ occurs at the median of the critical points of $f$,
i.e., the minimum value of $f$ occurs at $\text{median}\{a_1,a_2,a_3,\dots ,a_n\}$

For $f\left(x\right)=|x-1|+|x-8|$,
critical points are $1$ and $8$.
Since, there are two (even number of) critical points, the minimum value of $f$ occurs at any $x\in [1,8]$