Question

If f and g are both even functions, is the product \(fg\) even? If f and g are both odd functions, is \(fg\) odd? What if f is even and g is odd? Justify your answers.

Short answer

• Yes, if f and g both are even functions,\(fg\)is an even function.
• Yes, if f and g both are odd functions,\(fg\)is an even function.
• If f is even g is odd function\(fg\)is odd an odd function.

Step-by-step solution

Even and odd function

Function f is an even function if,\(f\left( x \right) = f\left( { - x} \right)\).

Function f is an odd function if,\(f\left( x \right) = - f\left( { - x} \right)\).

Finding the function \(fg\)is even or odd.

Let the multiplication of two functions is written as:\(h\left( x \right) = f\left( x \right)g\left( x \right)\).

1. If f and g both are even functions.

If f is even, then the function is written as\(f\left( x \right) = f\left( { - x} \right)\).

If g is even, then the function is written as\(g\left( x \right) = g\left( { - x} \right)\).

Then, multiplication of these functions is written as:

\(\begin{array}{l}h\left( { - x} \right) = f\left( { - x} \right)g\left( { - x} \right)\\h\left( { - x} \right) = f\left( x \right)g\left( x \right)\\h\left( { - x} \right) = h\left( x \right)\end{array}\)

Therefore, h is an even function, hence,\(fg\)is an even function.

1. If f and g both are odd

If f is odd, then the function is written as\(f\left( x \right) = - f\left( { - x} \right)\).

If g is odd, then the function is written as\(g\left( x \right) = - g\left( { - x} \right)\).

Then, multiplication of these functions is written as:

\(\begin{array}{l}h\left( { - x} \right) = f\left( { - x} \right)g\left( { - x} \right)\\h\left( { - x} \right) = \left( { - f\left( x \right)} \right)\left( { - g\left( x \right)} \right)\\h\left( { - x} \right) = h\left( x \right)\end{array}\)

Therefore, h is an even function, hence,\(fg\)is an even function.

1. If f is even and g is odd

If f is even, then the function is written as\(f\left( x \right) = f\left( { - x} \right)\).

If g is odd, then the function is written as\(g\left( x \right) = - g\left( { - x} \right)\).

Then, multiplication of these functions is written as:

\(\begin{array}{l}h\left( { - x} \right) = f\left( { - x} \right)g\left( { - x} \right)\\h\left( { - x} \right) = f\left( x \right)\left( { - g\left( x \right)} \right)\\h\left( { - x} \right) = - h\left( x \right)\end{array}\)

Therefore, h is odd, hence,\(fg\) is an odd function.