{"id":3381,"date":"2021-07-23T08:47:51","date_gmt":"2021-07-23T08:47:51","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3381"},"modified":"2021-11-24T16:42:40","modified_gmt":"2021-11-24T11:12:40","slug":"odd-even-function","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/odd-even-function\/","title":{"rendered":"How to Determine Odd Even Function"},"content":{"rendered":"
Here, you will learn what is odd even functions and how to determine if given function is odd or even with example.<\/p>\n
Let’s begin –<\/p>\n
Let a function f(x) such that both x and -x are in its domain then<\/p>\n
\nIf f(-x) = f(x) then f is said to be an even function.<\/p>\n
If f(-x) = -f(x) then f is said to be an odd function.<\/p>\n<\/blockquote>\n
Properties of Odd Even Functions :<\/h2>\n
(i) f(x) – f(-x) = 0 \\(\\implies\\) f(x) is even & f(x) + f(-x) = 0 \\(\\implies\\) f(x) is odd.<\/p>\n
(ii) A function may be neither be odd nor even.<\/p>\n
(iii) The only function which is defined on the entire number line & is even and odd at the same time is f(x) = 0.<\/p>\n
(iv) Every constant function is even function.<\/p>\n
(v) Inverse of an even function is not defined.<\/p>\n
(vi) Every even function is symmetric about the y-axis & every odd function is symmetric about origin.<\/p>\n
(vii) If a function is defined as f(a + x) = f(a – x) then this function is symmetric about the line x = a.<\/p>\n\n\n
Example : <\/span>Identify the given functions as odd, even or neither :
\n(i) f(x) = \\(x\\over {e^x – 1}\\) + \\(x\\over 2\\) + 1
\n(ii) f(x + y) = f(x) + f(y) for all x, y \\(\\in\\) R<\/p>\nSolution : <\/span>(i) f(x) = \\(x\\over {e^x – 1}\\) + \\(x\\over 2\\) + 1
\nClearly domain of f(x) is R – {0}, We have
\nf(-x) = \\(-x\\over {e^{-x} – 1}\\) – \\(x\\over 2\\) + 1
\n= \\(-e^{x}.x\\over {1 – e^x}\\) – \\(x\\over 2\\) + 1
\n= x + \\(x\\over {e^x – 1}\\) – \\(x\\over 2\\) + 1 = \\(x\\over {e^x – 1}\\) + \\(x\\over 2\\) + 1 = f(x)
\nHence, f(x) is an even.
\n(ii) f(x + y) = f(x) + f(y) for all x, y \\(\\in\\) R
\nReplacing x, y by zero, we get f(0) = 2f(0) \\(\\implies\\) f(0) = 0
\nReplacing y by -x, we get f(x) + f(-x) = f(0) = 0 \\(\\implies\\) f(x) = -f(-x)
\nHence, f(x) is an odd.
\n<\/p>\n\n\nHope 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!<\/p>\n\n\n