Question

The function $f\left( x \right) = \left| {px - q} \right| + r\left| x \right|$, $x \in \left( { - \infty ,\infty } \right)$ where $p > 0$, $q > 0$, $r > 0$ and it has a minimum value. Find the condition so the function has a minimum value.
A. $p \ne q$
B. $q \ne r$
C. $p \ne r$$
D. $p = q = r$

Answer 1

Hint: We rewrite the given function in the piecewise manner. After that, we will draw a diagram for the conditions $p = r$, $p < r$, and $p > r$. From the diagram, we will find the point where the function is minimum.

Complete step by step solution:
Given function is $f\left( x \right) = \left| {px - q} \right| + r\left| x \right|$, $x \in \left( { - \infty ,\infty } \right)$.
We know that, the absolute function$f\left( x \right) = \left| x \right|$ can be written as$f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{ - x,}&{x < 0}\\{x,}&{x \ge 0}\end{array}} \right.$.
The function $f\left( x \right)$ is a sum of two absolute functions.
Assume that, $f\left( x \right) = \left| {px - q} \right| + r\left| x \right| = g\left( x \right) + h\left( x \right)$
The piecewise from of $g\left( x \right) = \left| {px - q} \right|$ is $g\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{ - \left( {px - p} \right),}&{px - p < 0}\\{px - p,}&{px - p \ge 0}\end{array}} \right.$
The piecewise form of $h\left( x \right) = r\left| x \right|$ is $h\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{ - rx}&{x < 0}\\{rx}&{x \ge 0}\end{array}} \right.$
The $g\left( x \right)$function can be written is $g\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{ - \left( {px - p} \right)}&{x < 0}\\{ - \left( {px - p} \right)}&{px - p < 0}\\{px - p}&{px - p \ge 0}\end{array}} \right.$.
The piecewise form of the $h\left( x \right)$is $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{ - \left( {px - q} \right) - rx}&{x < 0}\\{ - \left( {px - q} \right) + rx}&{0 < x < \dfrac{q}{p}}\\{px - q + rx}&{x > 0}\end{array}} \right.$
$ = \left\{ {\begin{array}{*{20}{c}}{ - \left( {p + r} \right)x + q}&{x < 0}\\{\left( {p - r} \right)x + q}&{0 < x < \dfrac{q}{p}}\\{\left( {p + r} \right)x - q}&{x > 0}\end{array}} \right.$
Given that, $p > 0$, $q > 0$, $r > 0$.
So, the slope of $ - \left( {p + r} \right)x + q$ must be negative.
The slope of $\left( {p + r} \right)x - q$must be positive.
But we cannot predict the slope of $\left( {p - r} \right)x + q$.
Case I:
When $p > r$, then the slope of $\left( {p - r} \right)x + q$ is positive.
 So, the graph will be

Image: Graph of function when r>p
The function $f\left( x \right)$ has a minimum point at $x = 0$.

Case II:
When $p = r$, then the slope of $\left( {p - r} \right)x + q$ is zero.
 So, the graph will be

Image: Graph of function when r=p
The $f\left( x \right)$ has an infinite number of minimum points.

Case III:
When $p < r$, then the slope of $\left( {p - r} \right)x + q$ is negative.
 So, the graph will be

Image: Graph of function when rThe function has a minimum point at $x = \dfrac{q}{p}$.
Thus, the function has a minimum point when$p < r$ and $p > r$. This implies the function has a minimum point $p \ne r$.

Option ‘C’ is correct

Note: The maximum value is the highest point or value that a function can achieve in a graph; it is also referred to as the global maxima while the other peaks is referred to as local maxima. The same concept applies to the minimum value; the lowest value of a function for a graph is known as the global minima, while the other lower valleys are known as local minima.