Here, you will learn what is odd even functions and how to determine if given function is odd or even with example.

Let’s begin –

## Odd Even Function :

Let a function f(x) such that both x and -x are in its domain then

If f(-x) = f(x) then f is said to be an even function.

If f(-x) = -f(x) then f is said to be an odd function.

## Properties of Odd Even Functions :

(i) f(x) – f(-x) = 0 \(\implies\) f(x) is even & f(x) + f(-x) = 0 \(\implies\) f(x) is odd.

(ii) A function may be neither be odd nor even.

(iii) The only function which is defined on the entire number line & is even and odd at the same time is f(x) = 0.

(iv) Every constant function is even function.

(v) Inverse of an even function is not defined.

(vi) Every even function is symmetric about the y-axis & every odd function is symmetric about origin.

(vii) If a function is defined as f(a + x) = f(a – x) then this function is symmetric about the line x = a.

Example : Identify the given functions as odd, even or neither :

(i) f(x) = \(x\over {e^x – 1}\) + \(x\over 2\) + 1

(ii) f(x + y) = f(x) + f(y) for all x, y \(\in\) R

Solution : (i) f(x) = \(x\over {e^x – 1}\) + \(x\over 2\) + 1

Clearly domain of f(x) is R – {0}, We have

f(-x) = \(-x\over {e^{-x} – 1}\) – \(x\over 2\) + 1

= \(-e^{x}.x\over {1 – e^x}\) – \(x\over 2\) + 1

= x + \(x\over {e^x – 1}\) – \(x\over 2\) + 1 = \(x\over {e^x – 1}\) + \(x\over 2\) + 1 = f(x)

Hence, f(x) is an even.

(ii) f(x + y) = f(x) + f(y) for all x, y \(\in\) R

Replacing x, y by zero, we get f(0) = 2f(0) \(\implies\) f(0) = 0

Replacing y by -x, we get f(x) + f(-x) = f(0) = 0 \(\implies\) f(x) = -f(-x)

Hence, f(x) is an odd.

Hope you learnt how to determine if given function is odd or even, learn more concepts of function and practice more questions to get ahead in the competition. Good luck!