## Trending News

# Are these functions even, odd, or neither?

h(x) = x / x2 - 1

h(x) = -x3 / 3x2 - 9

"x2" is x squared and "x3" is x to the power of 3.

### 6 Answers

- 1 decade agoFavorite Answer
h(-x) = (-x)/(x2+1) , therefore the first is neither.

h(-x) = (-(-x3))/(3(-x2) - 9) = x3/3x2 - 9 therefore this is also neither.

A function is even when f(-x)=f(x), or (x,y) goes to (-x,y). It symmetrical to the y axis.

A function is odd when f(-x) = - f(x), or (x,y) goes to (-x,-y). It is symmetrical to the origin.

I hope that helps you!!

Please take into account that I'm a precalculus student, not a teacher, so I might have made a mistake :D. I don't THINK I did....

Source(s): My math class! - Anonymous1 decade ago
I'm going to assume that x^2 - 1 and 3x^2 - 9 are both denominators.

A function is even if f(x) = f(-x). A function is odd if f(-x) = -f(x). It's neither if neither works.

h(x) = x/(x^2 - 1)

h(-x) = -x/((-x)^2 - 1) = -x/(x^2 - 1)

so h(-x) = -h(x) and it's odd

h(x) = -x^3/(3x^2 - 9)

h(-x) = -(-x)^3/(3(-x)^2 - 9)

= x^3/(3x^2 - 9)

So h(-x) = - h(x) and it's odd

If I've misread the question - if those weren't both denominators - then my answers will be wrong, but the procedure will still work. In the future, parentheses will make it a WHOLE lot easier to figure out what you're asking.

- 1 decade ago
The way to figure out the type of symmetrical function it to put negatives in front of all Xs and then trying to rearrange the equation so it is exactly like before (sort of like inverses).

1]

h(x) = x/(x^2-1)

=-x/(-x^2-1)

=-x/(x^2-1)

*From here, we can see that the numerator of X is negative, there is no way to change it, so the function is NEITHER.

EDIT : Put down Odd instead of neither :p

2] h(x) = -x^3 / (3x^2 - 9)

=-(-x^3)/(3(-x^2) - 9)

= x^3/ (3x^2-9)

*Since the numerator is now positive, the function is no longer like before in any way, thus, it is NEITHER.

- Anonymous4 years ago
in case you have a graphing calculator, it fairly is amazingly elementary to tell by using finding on the graph the equation makes. There are 2 curved lines that are symmetric to the very center (the foundation). A line on one area of the graph might desire to be the precise opposite on the different area of the graph. and that they are.

- How do you think about the answers? You can sign in to vote the answer.
- 1 decade ago
Here's a great source to answer your questions... http://en.wikipedia.org/wiki/Even_and_odd_function... Good luck!!!