Answer
348.6k+ views
Hint: Here, we are given a function and we need to find \[{f'}(x)\] 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 \[{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\] . 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,
\[f(x) = |{x^3}|\]
Thus, to open the mode sign, we will have the following values of x as below,
\[f(x) = - {x^3},x < 0\] and \[f(x) = {x^3},x \geqslant 0\]
Since our given function is continuous, because
Left hand limit (LHL) = Right hand limit (RHL) at x = 0.
\[{f'}(x) = 3{x^2}\] at \[x \geqslant 0\] and \[{f'}(x) = - 3{x^2}\] at x<0
We will find the derivative through the given formula,
\[{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
First, for \[x \geqslant 0\], then we will have,
\[\therefore {f'}(0) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(0 + h) - f(0)}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^3} - 0}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^3}}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} {h^2}\]
\[ = 0\]
Next, for x<0, then we will have,
\[\therefore {f'}(0) = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(0 - h) - f(0)}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{ - {h^3} - 0}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{ - {h^3}}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} ( - {h^2})\]
\[ = 0\]
Since, LHL = RHL. This means that the limit exists.
Thus the function is derivable.
Hence, for the given function \[f(x) = |{x^3}|\] at x = 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.
Complete step-by-step answer:
Given that,
\[f(x) = |{x^3}|\]
Thus, to open the mode sign, we will have the following values of x as below,
\[f(x) = - {x^3},x < 0\] and \[f(x) = {x^3},x \geqslant 0\]
Since our given function is continuous, because
Left hand limit (LHL) = Right hand limit (RHL) at x = 0.
\[{f'}(x) = 3{x^2}\] at \[x \geqslant 0\] and \[{f'}(x) = - 3{x^2}\] at x<0
We will find the derivative through the given formula,
\[{f'}(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
First, for \[x \geqslant 0\], then we will have,
\[\therefore {f'}(0) = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{f(0 + h) - f(0)}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^3} - 0}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{{h^3}}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} {h^2}\]
\[ = 0\]
Next, for x<0, then we will have,
\[\therefore {f'}(0) = \mathop {\lim }\limits_{h \to {0^ - }} \dfrac{{f(0 - h) - f(0)}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{ - {h^3} - 0}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} \dfrac{{ - {h^3}}}{h}\]
\[ = \mathop {\lim }\limits_{h \to {0^ + }} ( - {h^2})\]
\[ = 0\]
Since, LHL = RHL. This means that the limit exists.
Thus the function is derivable.
Hence, for the given function \[f(x) = |{x^3}|\] at x = 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.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)