Question

If f and g are both even functions, is \(f + g\) even? If f and g

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

Short answer

• Yes, if f and g both are even functions, \(f + g\) is an even function.
• Yes, if f and g both are odd functions, \(f + g\) is an odd function.
• If f is even g is odd \(f + g\)is neither even nor odd

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 \(f + g\)is even or odd.

Let sum of two functions is\(h\left( x \right) = f\left( x \right) + g\left( x \right)\).

1. If f and g both are even functions.

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

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

Then, sum of these functions is written as:

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

Therefore, h is an even function, hence,\(f + g\)is an even function.

1. If f and g both are odd

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

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

Then, sum of these functions is written as:

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

Therefore, h is an odd function, hence,\(f + g\)is an odd function.

1. If f is even and g is odd

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

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

Then, sum of these functions is written as:

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

Therefore, h is neither even not odd, hence,\(f + g\) is neither even nor odd.