
How do you determine if \[f\left( x \right) = \left| {{x^2} + x} \right|\] is an even or odd function?
Answer
547.8k+ 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.
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