
How do you determine if \[f\left( x \right) = \left| {{x^2} + x} \right|\] is an even or odd function?
Answer
524.4k+ views
Hint: A function can be defined as even or odd if it satisfies an even or odd definition. In the given function, we need to determine algebraically whether a function is even or odd. Hence, to determine if \[f\left( x \right) = \left| {{x^2} + x} \right|\] is an even or odd function, we need to consider the following:
If \[f\left( x \right) = f\left( { - x} \right)\], then \[f\left( x \right)\] is even: Even functions have symmetry about the y-axis.
If \[f\left( { - x} \right) = - f\left( x \right)\], then \[f\left( x \right)\] is odd: Odd functions have symmetry about the origin.
Complete step by step answer:
Given,
\[f\left( x \right) = \left| {{x^2} + x} \right|\]
Let us determine for even function:
We know that for even: \[f\left( x \right) = f\left( { - x} \right)\], hence:
\[f\left( { - x} \right) = {\left( { - x} \right)^2} - \left( { - x} \right)\]
Simplifying the terms, we get:
\[ \Rightarrow f\left( { - x} \right) = {x^2} + x \ne f\left( x \right)\]
Since, \[f\left( x \right) \ne f\left( { - x} \right)\], then the given function \[f\left( x \right)\] is not even.
Now, let us determine for odd function:
We know that for odd: \[f\left( { - x} \right) = - f\left( x \right)\], hence:
\[ - f\left( x \right) = - \left( {{x^2} - x} \right)\]
Simplifying the terms, we get:
\[ \Rightarrow - f\left( x \right) = - {x^2} + x \ne f\left( { - x} \right)\]
Since, \[f\left( { - x} \right) \ne - f\left( x \right)\], then the given function \[f\left( x \right)\] is not odd.
Thus, the given function \[f\left( x \right) = \left| {{x^2} + x} \right|\] is neither odd nor even.
Note: The key point to note is that, we must know the conditions, if \[f\left( x \right) = f\left( { - x} \right)\], then \[f\left( x \right)\] is even and if\[f\left( { - x} \right) = - f\left( x \right)\], then \[f\left( x \right)\] is odd. A function can be neither even nor odd if it does not exhibit either symmetry. Also, the only function that is both even and odd is the constant function. A function ‘f’ is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if \[f\left( { - x} \right) = f\left( x \right)\] for all x in the domain of f. A function f is odd if the graph of f is symmetric with respect to the origin.
If \[f\left( x \right) = f\left( { - x} \right)\], then \[f\left( x \right)\] is even: Even functions have symmetry about the y-axis.
If \[f\left( { - x} \right) = - f\left( x \right)\], then \[f\left( x \right)\] is odd: Odd functions have symmetry about the origin.
Complete step by step answer:
Given,
\[f\left( x \right) = \left| {{x^2} + x} \right|\]
Let us determine for even function:
We know that for even: \[f\left( x \right) = f\left( { - x} \right)\], hence:
\[f\left( { - x} \right) = {\left( { - x} \right)^2} - \left( { - x} \right)\]
Simplifying the terms, we get:
\[ \Rightarrow f\left( { - x} \right) = {x^2} + x \ne f\left( x \right)\]
Since, \[f\left( x \right) \ne f\left( { - x} \right)\], then the given function \[f\left( x \right)\] is not even.
Now, let us determine for odd function:
We know that for odd: \[f\left( { - x} \right) = - f\left( x \right)\], hence:
\[ - f\left( x \right) = - \left( {{x^2} - x} \right)\]
Simplifying the terms, we get:
\[ \Rightarrow - f\left( x \right) = - {x^2} + x \ne f\left( { - x} \right)\]
Since, \[f\left( { - x} \right) \ne - f\left( x \right)\], then the given function \[f\left( x \right)\] is not odd.
Thus, the given function \[f\left( x \right) = \left| {{x^2} + x} \right|\] is neither odd nor even.
Note: The key point to note is that, we must know the conditions, if \[f\left( x \right) = f\left( { - x} \right)\], then \[f\left( x \right)\] is even and if\[f\left( { - x} \right) = - f\left( x \right)\], then \[f\left( x \right)\] is odd. A function can be neither even nor odd if it does not exhibit either symmetry. Also, the only function that is both even and odd is the constant function. A function ‘f’ is even if the graph of f is symmetric with respect to the y-axis. Algebraically, f is even if and only if \[f\left( { - x} \right) = f\left( x \right)\] for all x in the domain of f. A function f is odd if the graph of f is symmetric with respect to the origin.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 8 Maths: Engaging Questions & Answers for Success

Class 8 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

