Question
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) is an even function and g(x) is an odd function, how do you prove that h(x)=f(x)×g(x) is odd?
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)
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)
Complete step-by-step answer:
In this question, f(x) is an even function and g(x) is an odd function.
Here, f(x) is an even function.
Therefore, we can write:
⇒f(−x)=f(x)
Also given that g(x) is an odd function.
Therefore, we can write:
⇒g(−x)=−g(x)
Now, we want to prove that function h(−x) is an odd function.
⇒h(x)=f(x)×g(x)...(1)
Now,
⇒h(−x)=f(−x)×g(−x)
Let us put the values of f(−x) and g(−x).
That is equal to,
⇒h(−x)=f(x)×−g(x)
The multiplication with negative number -1 gives the negative sign before the answer.
Therefore,
⇒h(−x)=−(f(x)×g(x))
Let us put the value of f(x)×g(x) from the equation (1).
⇒h(−x)=−h(x)
Hence, we can say that h(x)=f(x)×g(x) is an odd function.
Note:
The even function can also be written as f(x)−f(−x)=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)=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.