I need help determining whether a function is even or odd. I know HOW to...for the most part, I just need a little bit of clarification.

- BloomLocke367

- jamiebookeater

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- BloomLocke367

For \(g(x)=x^3\) you have to plug in -x and see if the function is the same or completely the opposite. I know if it's the same that the function is even, and if it's completely the opposite it's odd. BUT my teacher did a few examples on the board where she took out -1, so the function would be even when you'd think it would be odd. When would you take out a -1?

- BloomLocke367

Sorry for making it so wordy XD

- Nnesha

you would multiply by -1 to check if it's odd or neither

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## More answers

- DanJS

Even Function is symmetric about Y axis
F(x) = F(-x)

- BloomLocke367

yes @DanJS. That's what I learned.

- DanJS

the odd is reflected over the origin, like the line y=x , or y = x^3

- BloomLocke367

and for odd you have f(-x)=-f(x)

- BloomLocke367

ohh okay. I'm still confused on what she did in class today XD

- DanJS

yep, that makes sense

- Nnesha

for exxample
if
f(X)=x^3-x
i would substitute x for -x
(-x)^2 -(-x)
-x^3 +x <-- not the original equation so it's not even
now i would multiply by -1
(-x^3 -x) times -1 =
x^3-x

- BloomLocke367

Now you're confusing me :O

- DanJS

Even functions are symmetric about the y-axis, Odd functions are symmetric about both x and y-axis, that's it

- BloomLocke367

odd are symmetric about both? so about the origin...or...?

- DanJS

right, about the origin, same as reflected over both axis

- BloomLocke367

okay thanks XD so \(f(x)=\sqrt{x^2+2}\) is even, right? because when you plug in -x the equation stays the same and when you graph it it's a porabola.

- DanJS

Just switch x for -x and see if you get either the original f(x) - or the negative - f(x) or neither

- DanJS

that is basically all you need to do

- BloomLocke367

yes, I know. you get \(f(-x)=\sqrt{(-x)^2+2}\) which simplifies to \(f(-x)=\sqrt{x^2+2}\) which is the same as the original, right?

- DanJS

yea you got it...but it is not parabola, y^2 - x^2 = k hyperbola, upper half +root

- BloomLocke367

oops XD I only saw the upper half. My bad. Thanks.

- DanJS

it is only the upper half, not + or - root, parabola is y = x^2 no square on the y.. eh nevermind

- BloomLocke367

And I also got even for \(\large g(x)=\frac{3}{1+x^2}\)
and yes, I should have realized. I did study hyperbolas and conics last year in Alg 2

- DanJS

yeah , if the only x term is squared, it will be even prolly

- DanJS

you get it all good now?

- BloomLocke367

Yes, I THINK so. I'll tag you if I have any more questions.

- DanJS

just change to -x
if you get f(x) back ---even
if you get -f(x) back ---odd
---------------------------
most functions are neither even or odd

- BloomLocke367

WAIT. I think this is neither. \(f(x)=-x^2+0.03x+5\)

- BloomLocke367

andddd he left.

- BloomLocke367

- BloomLocke367

@Nnesha do you know that one?

- BloomLocke367

http://prntscr.com/8bwwtw there's the graph.

- BloomLocke367

it's symmetrical over the y-axis but when you plug in -x it doesn't work out so I'm confused.

- Nnesha

\(\huge\color{green}{\checkmark}\)

- Nnesha

well i don't know abt graph
but when u replace x with -x
you will get -x^2-.03x+5 right ? so it's not even
now i would multiply by -1 \[-1(-x^2-0.03x+5)=x^2+0.03-5\]
i didn't get the original equation -x^2+0.03x+5
that's how its neither
and ~~this is how we di it!

- mathmate

Some simple rules,
Sum of odd functions is odd.
Sum of even functions is even.
Sum of odd and even is neither!

- BloomLocke367

What? What do you mean mm?

- mathmate

and a constant is even.

- Nnesha

did*:XD

- mathmate

f(x)=x^2+x
is a sum of even and odd functions, so f(x) is neither odd nor even!

- BloomLocke367

I really don't understand what you mean mm XD

- mathmate

If you are given a function which is the sum of many terms,

- mathmate

Check out each term individually.

- mathmate

Check if "each" term is even or odd.

- mathmate

For example, f(x)=x^4+3x^2 is even because x^4 and 3x^2 are both even.

- BloomLocke367

ohhhh. That's not what I learned but okay. XD I just plug -x where every x is and solve to see if the equation matches the original or not.

- BloomLocke367

but your method seems easier XD Thank you.

- mathmate

No, this does not replace what you learned. This makes your life easier when you have a function consisting of multiple terms. I assume you already know that x^2, x^4... are even, and x, x^3.... are odd.

- BloomLocke367

yes. and I know, I have to prove these problems algebraically and graphically anyways. But it's good to know for the future.

- mathmate

Good! Are you all clear now, for example, is
y=x+3 even or odd?

- BloomLocke367

so \(g(x)=2x^3-3x\) is odd, right? because after you plug in -x you get \(g(-x)=-2x^3+3x\)

- BloomLocke367

and that's neither... right? The one you showed?

- mathmate

exactly!

- mathmate

Exactly again!

- BloomLocke367

Awesome, thank you!

- mathmate

Great! You're welcome! :)

- BloomLocke367

woe woe woe. I have one more. This one is like the example she gave us.

- BloomLocke367

\(\large h(x)=\frac{1}{x}\)

- mathmate

Apply the rules and see what you get! :)

- BloomLocke367

you get \(h(-x)=\large\frac{1}{-x}\) but you can take \(h(-x)=\large\frac{1}{-1(x)}\) which would make it even.

- mathmate

given h(x)=1/x
h(-x)=1/(-x)=-1/x=-h(x)
so...

- BloomLocke367

what? I'm so confused

- BloomLocke367

so it's odd?

- BloomLocke367

because -h(x)

- mathmate

After you have put in the -x, you want to rearrange the expression in terms of h(x) so you can decide if h(x) is even or odd.
You have these rules:
if h(-x)=-h(x), then h(x) is odd
if h(-x)=h(x), then h(x) is even.

- mathmate

yes, it is odd.
It's easier to visualize it graphically.

- BloomLocke367

so it's odd then. because it would equal \(-\frac{1}{x}\)?

- BloomLocke367

and this is the graph. http://prntscr.com/8bx7i1

- mathmate

|dw:1441229730362:dw|

- BloomLocke367

yes. okay. thank you. I'm going to open a new question for asymptotes.

- mathmate

yes, it is odd because h(-x) = - h(x) (where h(x)=1/x)

- BloomLocke367

I'm pretty sure I understand completely now.

- mathmate

Good!
In math, the more exercise we do, the better we understand.

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