Question

The derivative of \[f\left( x \right) = 3\left| {2 + x} \right|\] at the point \[{x_0} = - 3\] is
A. \[3\]
B. \[ - 3\]
C. \[0\]
D. None of these

Answer 1

Hint: In this question, we need to find the derivative of \[f\left( x \right) = 3\left| {2 + x} \right|\] at point \[{x_0} = - 3\]. For this, we need to use the concept of limit and modulus. We will use the following formula of limit to find the end result.

Formula used: The following formula is used to solve this example.
We can say that the derivative of function \[f\left( x \right)\] at \[x = c\] is defined as the limit of the slope of the secant line from \[x = c\] to \[x = c + h\] as \[h\] tends to \[0\].
Mathematically, it is expressed as
\[f'\left( c \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {c + h} \right) - f\left( c \right)}}{h}\]

Complete step-by-step answer:
We know that \[f\left( x \right) = 3\left| {2 + x} \right|\]
So, consider \[f'\left( c \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {c + h} \right) - f\left( c \right)}}{h}\]
But \[c = - 3\]
Thus, we get
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {3\left| {2 + \left( { - 3 + h} \right)} \right|} \right] - \left[ {3\left| {2 + \left( { - 3} \right)} \right|} \right]}}{h}\]
By simplifying, we get
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {3\left| {2 - 3 + h} \right|} \right] - \left[ {3\left| {2 + \left( { - 3} \right)} \right|} \right]}}{h}\]
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {3\left| { - 1 + h} \right|} \right] - \left[ {3\left| { - 1} \right|} \right]}}{h}\]
We know that \[\left| x \right| = \left\{ {\begin{array}{*{20}{c}}
  {x;{\text{ x}} \geqslant {\text{0}}} \\
  { - x;{\text{ x}} < {\text{0 }}}
\end{array}{\text{ }}} \right.\]
So, \[\left| { - 1 + h} \right| = - \left( { - 1 + h} \right) = \left( {1 - h} \right)\] and \[\left| { - 1} \right| = - \left( { - 1} \right) = 1\]
Thus, we get
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left[ {3\left( {1 - h} \right)} \right] - \left[ {3\left( 1 \right)} \right]}}{h}\]
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3h - 3}}{h}\]
By simplifying further, we get
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ - 3h}}{h}\]
\[f'\left( { - 3} \right) = \mathop {\lim }\limits_{h \to 0} \left( { - 3} \right)\]
So, we get
\[f'\left( { - 3} \right) = - 3\]
Therefore, the correct option is (B).

Additional information: A limit is defined in mathematics as the value at which a function approaches the outcome for the given set of input values. Limits are being used to specify integrals, derivatives, as well as continuity in calculus and mathematical analysis. In simple terms, we can say that a limit describes the value that a function approaches as its inputs get nearer and nearer up to a specific number. Always remember that the function's limit exists between any two consecutive integers.

Note: Here, students generally make mistakes in determining the modulus. If they got the wrong modulus then the final result will be wrong. Also, the definition of limit plays a significant role here. It is necessary to write the function in terms of \[f\left( {c + h} \right)\].