Here, you will learn what is odd even functions and how to determine if given function is odd or even with example.
Letโs begin โ
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) โ 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 โ x) then this function is symmetric about the line x = a.
Example : Identify the given functions as odd, even or neither :
(i) f(x) = \(x\over {e^x โ 1}\) + \(x\over 2\) + 1
(ii) f(x + y) = f(x) + f(y) for all x, y \(\in\) R
Solution : (i) f(x) = \(x\over {e^x โ 1}\) + \(x\over 2\) + 1
Clearly domain of f(x) is R โ {0}, We have
f(-x) = \(-x\over {e^{-x} โ 1}\) โ \(x\over 2\) + 1
= \(-e^{x}.x\over {1 โ e^x}\) โ \(x\over 2\) + 1
= x + \(x\over {e^x โ 1}\) โ \(x\over 2\) + 1 = \(x\over {e^x โ 1}\) + \(x\over 2\) + 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 functionย and practice more questions to get ahead in the competition. Good luck!
