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Question: If \(f\left( x \right)\) is an even function and \(g\left( x \right)\) is an odd function, how do yo...

If f(x)f\left( x \right) is an even function and g(x)g\left( x \right) is an odd function, how do you prove that h(x)=f(x)×g(x)h\left( x \right) = f\left( x \right) \times g\left( x \right) is odd?

Explanation

Solution

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(x)=f(x)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(x)=f(x)- f\left( x \right) = f\left( { - x} \right)

Complete step-by-step answer:
In this question, f(x)f\left( x \right) is an even function and g(x)g\left( x \right) is an odd function.
Here, f(x)f\left( x \right) is an even function.
Therefore, we can write:
f(x)=f(x)\Rightarrow f\left( { - x} \right) = f\left( x \right)
Also given that g(x)g\left( x \right) is an odd function.
Therefore, we can write:
g(x)=g(x)\Rightarrow g\left( { - x} \right) = - g\left( x \right)
Now, we want to prove that function h(x)h\left( { - x} \right) is an odd function.
h(x)=f(x)×g(x)\Rightarrow h\left( x \right) = f\left( x \right) \times g\left( x \right)...(1)
Now,
h(x)=f(x)×g(x)\Rightarrow h\left( { - x} \right) = f\left( { - x} \right) \times g\left( { - x} \right)
Let us put the values of f(x)f\left( { - x} \right) and g(x)g\left( { - x} \right).
That is equal to,
h(x)=f(x)×g(x)\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,
h(x)=(f(x)×g(x))\Rightarrow h\left( { - x} \right) = - \left( {f\left( x \right) \times g\left( x \right)} \right)
Let us put the value of f(x)×g(x)f\left( x \right) \times g\left( x \right) from the equation (1).
h(x)=h(x)\Rightarrow h\left( { - x} \right) = - h\left( x \right)
Hence, we can say that h(x)=f(x)×g(x)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(x)f(x)=0f\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(x)+f(x)=0f\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.