Question

If $f\left(x\right)=\left|x\right|+|x-1|+|x-2|$, then

• A. $f\left(x\right)$ has minima at $x=1$
• B. $f\left(x\right)$ has maxima at $x=0$
• C. $f\left(x\right)$ has neither maxima not minima at $x=0$
• D. $f\left(x\right)$ has neither maxima nor minima at $x=2$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(x\right)$ has minima at $x=1$
C $f\left(x\right)$ has neither maxima not minima at $x=0$
D $f\left(x\right)$ has neither maxima nor minima at $x=2$

We have,

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

$=\{\begin{array}{c}\begin{array}{cc}-3x+3,& x<0\end{array}\\ \begin{array}{cc}-x+3,& 0\le x<1\end{array}\\ \begin{array}{cc}x+1,& 1\le x<2\end{array}\\ \begin{array}{cc}3x-3,& x\ge 2\end{array}\end{array}$

$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}\begin{array}{cc}-3& x<0\end{array}\\ \begin{array}{cc}doesnotexist& x=0\end{array}\\ \begin{array}{cc}-1& 0<x<1\end{array}\\ \begin{array}{cc}doesnotexist& x=1\end{array}\\ \begin{array}{cc}1& 1<x<2\end{array}\\ \begin{array}{cc}doesnotexist& x=2\end{array}\\ \begin{array}{cc}3& x>2\end{array}\end{array}$

Clearly, $f\left(x\right)$ has minima at $x=1$ and nether maxima nor minima at $x=0$ and $x=2.$