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Question: The derivative of \[f(x) = |{x^3}|\] at x = 0 is 1) 0 2) 1 3) -1 4) not defined...

The derivative of f(x)=∣x3∣f(x) = |{x^3}| at x = 0 is

  1. 0
  2. 1
  3. -1
  4. not defined
Explanation

Solution

Hint : Here, we are given a function and we need to find f′(x){f'}(x) at x = 0. First, we will find the value of the given function when the mode sign is removed. Then we will find the derivative from the given formula f′(x)=lim⁡h→0f(x+h)−f(x)h{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} . Using this formula, we will find the values of left hand limit (LHL) and right hand limit (RHL). If both LHL = RHL then the limit of the function exists and we will get the final output.

Complete step-by-step answer :
Given that,
f(x)=∣x3∣f(x) = |{x^3}|
Thus, to open the mode sign, we will have the following values of x as below,
f(x)=−x3,x<0f(x) = - {x^3},x < 0 and f(x)=x3,x⩾0f(x) = {x^3},x \geqslant 0
Since our given function is continuous, because
Left hand limit (LHL) = Right hand limit (RHL) at x = 0.
f′(x)=3x2{f'}(x) = 3{x^2} at x⩾0x \geqslant 0 and f′(x)=−3x2{f'}(x) = - 3{x^2} at x<0
We will find the derivative through the given formula,
f′(x)=lim⁡h→0f(x+h)−f(x)h{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
First, for x⩾0x \geqslant 0, then we will have,
∴f′(0)=lim⁡h→0+f(0+h)−f(0)h\therefore {f'}(0) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(0 + h) - f(0)}}{h}
=lim⁡h→0+h3−0h= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^3} - 0}}{h}
=lim⁡h→0+h3h= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^3}}}{h}
=lim⁡h→0+h2= \mathop {\lim }\limits_{h \to {0^ + }} {h^2}
=0= 0
Next, for x<0, then we will have,
∴f′(0)=lim⁡h→0−f(0−h)−f(0)h\therefore {f'}(0) = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(0 - h) - f(0)}}{h}
=lim⁡h→0+−h3−0h= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{ - {h^3} - 0}}{h}
=lim⁡h→0+−h3h= \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{ - {h^3}}}{h}
=lim⁡h→0+(−h2)= \mathop {\lim }\limits_{h \to {0^ + }} ( - {h^2})
=0= 0
Since, LHL = RHL. This means that the limit exists.
Thus the function is derivable.
Hence, for the given function f(x)=∣x3∣f(x) = |{x^3}| at x = 0, f′(0)=0{f'}(0) = 0.
So, the correct answer is “Option B”.

Note : A limit is defined as a value that a function approaches the output for the given input values. It is used in the analysis process, and it always concerns the behaviour of the function at a particular point. A derivative is defined as the instantaneous rate of change in function based on one of its variables. It is similar to finding the slope of a tangent to the function at a point. Integration is a method to find definite and indefinite integrals.