Question

The value of the derivative of $\left| {x - 1} \right| + \left| {x - 3} \right|$ at $x = 2$ is:
(a) 2
(b) 1
(c) 0
(d) $- 2$

Answer 1

Here, we need to find the value of the derivative of the function at the given value. First, we will find the critical values of the modulus function and find the values of the function at different intervals. Then, we will differentiate the value of the function at $x = 2$ to find the required value.

Complete step by step solution:
Let $f\left( x \right) = \left| {x - 1} \right| + \left| {x - 3} \right|$.
We can observe that the given function is the sum of two modulus function.
Thus, we will first find the critical points of the function, where the value of the function changes.
Equating $x - 1$ and $x - 3$ to 0, we get
$\Rightarrow x - 1 = 0$ and
Simplifying the expressions, we get
$\Rightarrow x = 1,3$
Thus, the critical points of the function are $x = 1$ and $x = 3$.
Case 1: When $x$ is less than 1.
When $x$ is less than 1, the value of the modulus function $\left| {x - 1} \right|$ can be written as $- \left( {x - 1} \right)$.
When $x$ is less than 1, the value of the modulus function $\left| {x - 3} \right|$ can be written as $- \left( {x - 3} \right)$.
Therefore, the function $f\left( x \right) = \left| {x - 1} \right| + \left| {x - 3} \right|$ becomes
$\begin{array}{l} \Rightarrow f\left( x \right) = - \left( {x - 1} \right) - \left( {x - 3} \right)\\\ \Rightarrow f\left( x \right) = - 1\left( {x - 1} \right) - 1\left( {x - 3} \right)\end{array}$
Multiplying the terms using the distributive law of multiplication, we get
$\Rightarrow f\left( x \right) = - x + 1 - x + 3$
Adding and subtracting the like terms in the expression, we get
$\Rightarrow f\left( x \right) = - 2x + 4$
Thus, $f\left( x \right) = - 2x + 4$ when $x$ is less than 1.
Case 2: When $x$ is between 1 and 3.
When $x$ is between 1 and 3, the value of the modulus function $\left| {x - 1} \right|$ can be written as $x - 1$.
When $x$ is between 1 and 3, the value of the modulus function $\left| {x - 3} \right|$ can be written as $- \left( {x - 3} \right)$.
Therefore, the function $f\left( x \right) = \left| {x - 1} \right| + \left| {x - 3} \right|$ becomes
$\begin{array}{l} \Rightarrow f\left( x \right) = x - 1 - \left( {x - 3} \right)\\\ \Rightarrow f\left( x \right) = x - 1 - 1\left( {x - 3} \right)\end{array}$
Multiplying the terms using the distributive law of multiplication, we get
$\Rightarrow f\left( x \right) = x - 1 - x + 3$
Adding and subtracting the like terms in the expression, we get
$\Rightarrow f\left( x \right) = 2$
Thus, $f\left( x \right) = 2$ when $x$ is between 1 and 3.
Case 3: When $x$ is more than 3.
When $x$ is more than 3, the value of the modulus function $\left| {x - 1} \right|$ can be written as $x - 1$.
When $x$ is more than 3, the value of the modulus function $\left| {x - 3} \right|$ can be written as $x - 3$.
Therefore, the function $f\left( x \right) = \left| {x - 1} \right| + \left| {x - 3} \right|$ becomes
$\Rightarrow f\left( x \right) = x - 1 + x - 3$
Adding and subtracting the like terms in the expression, we get
$\Rightarrow f\left( x \right) = 2x - 4$
Thus, $f\left( x \right) = 2x - 4$ when $x$ is more than 3.
Therefore, we get
f\left( x \right) = \left\\{ {\begin{array}{*{20}{c}}{ - 2x + 4}\\\2\\\\{2x - 4}\end{array}{\rm{ }}\begin{array}{*{20}{c}};\\\;\\\;\end{array}{\rm{ }}\begin{array}{*{20}{c}}{x < 1}\\\\{1 < x < 3}\\\\{x > 3}\end{array}} \right.
We need to find the derivative of the function at $x = 2$.
At $x = 2$, the value of the function is 2.
Thus, we get
$\Rightarrow f\left( x \right) = 2$
Differentiating both sides with respect to $x$, we get
$\Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left( 2 \right)}}{{dx}}$
The derivative of a constant is always 0.
Thus, we get
$\Rightarrow \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} = 0$
Substituting $f\left( x \right) = \left| {x - 1} \right| + \left| {x - 3} \right|$ in the equation, we get
$\Rightarrow \dfrac{{d\left[ {\left| {x - 1} \right| + \left| {x - 3} \right|} \right]}}{{dx}} = 0$
Therefore, we get the derivative of the function $\left| {x - 1} \right| + \left| {x - 3} \right|$ at $x = 2$ as 0.

Thus, the correct option is option (c).

Note:
We used the distributive law of multiplication in the solution. The distributive law of multiplication states that $a\left( {b + c} \right) = a \cdot b + a \cdot c$.
We used the term ‘modulus function’ in the solution. A function of the form $\left| x \right|$ is called a modulus function. The value of the function is $\left| x \right| = - x$ at $x < 0$, and $\left| x \right| = x$ at $x \ge 0$.