If $f\left( x \right)$ is an even function and $g\left( x \right)$ is an odd function, how do you prove that $h\left( x \right) = f\left( x \right) \times g\left( x \right)$ is odd?
Answer
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Hint: In this question, one even and one odd function is given, and we want to prove that the multiplication of those functions is odd.
Even functions: Let f be a real-valued function of a real variable. Then f is even if the following equation holds for all x such that x and –x in the domain of f.
$f\left( x \right) = f\left( { - x} \right)$
Odd function: Let f be a real-valued function of a real variable. Then f is odd if the following equation holds for all x such that x and –x are in the domain of f.
$ - f\left( x \right) = f\left( { - x} \right)$
Complete step-by-step answer:
In this question, $f\left( x \right)$ is an even function and $g\left( x \right)$ is an odd function.
Here, $f\left( x \right)$ is an even function.
Therefore, we can write:
$ \Rightarrow f\left( { - x} \right) = f\left( x \right)$
Also given that $g\left( x \right)$ is an odd function.
Therefore, we can write:
$ \Rightarrow g\left( { - x} \right) = - g\left( x \right)$
Now, we want to prove that function $h\left( { - x} \right)$ is an odd function.
$ \Rightarrow h\left( x \right) = f\left( x \right) \times g\left( x \right)$...(1)
Now,
$ \Rightarrow h\left( { - x} \right) = f\left( { - x} \right) \times g\left( { - x} \right)$
Let us put the values of $f\left( { - x} \right)$ and $g\left( { - x} \right)$.
That is equal to,
$ \Rightarrow h\left( { - x} \right) = f\left( x \right) \times - g\left( x \right)$
The multiplication with negative number -1 gives the negative sign before the answer.
Therefore,
$ \Rightarrow h\left( { - x} \right) = - \left( {f\left( x \right) \times g\left( x \right)} \right)$
Let us put the value of $f\left( x \right) \times g\left( x \right)$ from the equation (1).
$ \Rightarrow h\left( { - x} \right) = - h\left( x \right)$
Hence, we can say that $h\left( x \right) = f\left( x \right) \times g\left( x \right)$ is an odd function.
Note:
The even function can also be written as $f\left( x \right) - f\left( { - x} \right) = 0$, and the graph of an even function is symmetric with respect to the y-axis that means the graph remains unchanged after reflection about the y-axis. The odd function can be written as $f\left( x \right) + f\left( { - x} \right) = 0$, the graph of an odd function has rotational symmetry with respect to the origin that means the graph remains unchanged after rotation of 180 degrees about the origin.
Even functions: Let f be a real-valued function of a real variable. Then f is even if the following equation holds for all x such that x and –x in the domain of f.
$f\left( x \right) = f\left( { - x} \right)$
Odd function: Let f be a real-valued function of a real variable. Then f is odd if the following equation holds for all x such that x and –x are in the domain of f.
$ - f\left( x \right) = f\left( { - x} \right)$
Complete step-by-step answer:
In this question, $f\left( x \right)$ is an even function and $g\left( x \right)$ is an odd function.
Here, $f\left( x \right)$ is an even function.
Therefore, we can write:
$ \Rightarrow f\left( { - x} \right) = f\left( x \right)$
Also given that $g\left( x \right)$ is an odd function.
Therefore, we can write:
$ \Rightarrow g\left( { - x} \right) = - g\left( x \right)$
Now, we want to prove that function $h\left( { - x} \right)$ is an odd function.
$ \Rightarrow h\left( x \right) = f\left( x \right) \times g\left( x \right)$...(1)
Now,
$ \Rightarrow h\left( { - x} \right) = f\left( { - x} \right) \times g\left( { - x} \right)$
Let us put the values of $f\left( { - x} \right)$ and $g\left( { - x} \right)$.
That is equal to,
$ \Rightarrow h\left( { - x} \right) = f\left( x \right) \times - g\left( x \right)$
The multiplication with negative number -1 gives the negative sign before the answer.
Therefore,
$ \Rightarrow h\left( { - x} \right) = - \left( {f\left( x \right) \times g\left( x \right)} \right)$
Let us put the value of $f\left( x \right) \times g\left( x \right)$ from the equation (1).
$ \Rightarrow h\left( { - x} \right) = - h\left( x \right)$
Hence, we can say that $h\left( x \right) = f\left( x \right) \times g\left( x \right)$ is an odd function.
Note:
The even function can also be written as $f\left( x \right) - f\left( { - x} \right) = 0$, and the graph of an even function is symmetric with respect to the y-axis that means the graph remains unchanged after reflection about the y-axis. The odd function can be written as $f\left( x \right) + f\left( { - x} \right) = 0$, the graph of an odd function has rotational symmetry with respect to the origin that means the graph remains unchanged after rotation of 180 degrees about the origin.
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