# Thread: Even, Odd or Neither Functions

1. ## Even, Odd or Neither Functions

I don't quiet understand the concept of even, odd and neither functions.

Test to see if each function below is even, odd or neither.

What makes a function NEITHER?

I somewhat grasp the odd and even idea (somewhat) but what actually makes it neither?

(1) f(x) = 2x + 5

(2) y = x^3 - 7

(3) h(x) = -5x^5 + 3x^4

2. Even and odd function are functions which satisfy some symmetry relation.

Geometrically, the graph of an even function is symmetric with respect to the y-axis. Therefore, let $f(x)$ be a real-valued function with a real variable. Then f is even if the following equation holds for all x in the domain of f.

$f(x) = f(-x)$.

Geometrically, the graph of an odd function has rotational symmetry with respect to the origin; i.e., the graph remains unchanged after rotation of 180 degrees about the origin. Therefore, Therefore, let $f(x)$ be a real-valued function with a real variable. Then f is even if the following equation holds for all x in the domain of f.

$-f(x) = f(-x)$.

A function is neither even nor odd if it fails to hold for the above equations.

$f(x) = 2x+5$,
$f(-x) = -2x+5$, therefore $f(x) \neq f(-x)$ and $-f(x) \neq f(-x)$ so $f(x)$ is neither even nor odd.

$h(x) = -5x^{5}+3x^4$,
$h(-x) = 5x^{5}+3x^4$, therefore $h(x) \neq h(-x)$ and $-h(x) \neq h(-x)$ so $h(x)$ is also neither even nor odd.

3. Hello Magentarita,

4. Hello, magentarita!

I don't quite understand the concept of even, odd and neither functions.

I somewhat grasp the odd and even idea (somewhat).
. Somewhat?
How would you feel if your cardiologist said, "I somewhat understand how to do your quadruple bypass
... I mean, I got the general idea, y'know ... but I don't quite understand the concept ..."

But what actually makes it neither?

Test to see if each function below is even odd, or neither.

$(1)\;\;f(x) \:=\:2x + 5 \qquad(2)\;\;y \:=\:x^3 - 7 \qquad (3)\;\;h(x) \:=\: -5x^5 + 3x^4$

For functions comprised of polynomials (only), there's a simpe rule . . .

If all the exponents are even, it is an even function.

If all the exponents are odd, it is an odd function.

[If the exponents are "mixed", it's a Neither.]

Note: Recall that a constant has x-to-the-zero-power.
. . . . That is: . $6 \:=\:6x^0$ . . . and 0 is even.

Now, let's look at the problems . . .

$(1)\;\;f(x) \:=\:2x+5 \:=\:2x^{\color{red}1} + 5x^{\color{red}0}$ . . . odd and even: Neither

$(2)\;\;y \:=\:x^3-7 \:=\:x^{\color{red}3}-7x^{\color{red}0}$ . . . odd and even: Neither

$(3)\;\;h(x) \:=\:-5x^{\color{red}5} + 3x^{\color{red}4}$ . . . odd and even: Neither

5. With polynomials or rational functions, as all of these are, there is an easy short cut. If all powers of the variable are even, the function is even (constant terms are 0 power of the variable). If all powers of the variable are odd, the function is odd. If there are both even and odd powers then the function is neither.

That is because, if n is even, $(-x)^n= x^n$, if n is odd, $(-x)^n= -x^n$. In fact, that's where the terms "even" and "odd" function came from.

6. ## finally...

Originally Posted by Pn0yS0ld13r
Even and odd function are functions which satisfy some symmetry relation.

Geometrically, the graph of an even function is symmetric with respect to the y-axis. Therefore, let $f(x)$ be a real-valued function with a real variable. Then f is even if the following equation holds for all x in the domain of f.

$f(x) = f(-x)$.

Geometrically, the graph of an odd function has rotational symmetry with respect to the origin; i.e., the graph remains unchanged after rotation of 180 degrees about the origin. Therefore, Therefore, let $f(x)$ be a real-valued function with a real variable. Then f is even if the following equation holds for all x in the domain of f.

$-f(x) = f(-x)$.

A function is neither even nor odd if it fails to hold for the above equations.

$f(x) = 2x+5$,
$f(-x) = -2x+5$, therefore $f(x) \neq f(-x)$ and $-f(x) \neq f(-x)$ so $f(x)$ is neither even nor odd.

$h(x) = -5x^{5}+3x^4$,
$h(-x) = 5x^{5}+3x^4$, therefore $h(x) \neq h(-x)$ and $-h(x) \neq h(-x)$ so $h(x)$ is also neither even nor odd.
I finally understand. Thanks!