BloomLocke367
  • BloomLocke367
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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BloomLocke367
  • 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
  • BloomLocke367
Sorry for making it so wordy XD
Nnesha
  • Nnesha
you would multiply by -1 to check if it's odd or neither

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DanJS
  • DanJS
Even Function is symmetric about Y axis F(x) = F(-x)
BloomLocke367
  • BloomLocke367
yes @DanJS. That's what I learned.
DanJS
  • DanJS
the odd is reflected over the origin, like the line y=x , or y = x^3
BloomLocke367
  • BloomLocke367
and for odd you have f(-x)=-f(x)
BloomLocke367
  • BloomLocke367
ohh okay. I'm still confused on what she did in class today XD
DanJS
  • DanJS
yep, that makes sense
Nnesha
  • 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
  • BloomLocke367
Now you're confusing me :O
DanJS
  • DanJS
Even functions are symmetric about the y-axis, Odd functions are symmetric about both x and y-axis, that's it
BloomLocke367
  • BloomLocke367
odd are symmetric about both? so about the origin...or...?
DanJS
  • DanJS
right, about the origin, same as reflected over both axis
BloomLocke367
  • 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
  • DanJS
Just switch x for -x and see if you get either the original f(x) - or the negative - f(x) or neither
DanJS
  • DanJS
that is basically all you need to do
BloomLocke367
  • 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
  • DanJS
yea you got it...but it is not parabola, y^2 - x^2 = k hyperbola, upper half +root
BloomLocke367
  • BloomLocke367
oops XD I only saw the upper half. My bad. Thanks.
DanJS
  • DanJS
it is only the upper half, not + or - root, parabola is y = x^2 no square on the y.. eh nevermind
BloomLocke367
  • 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
  • DanJS
yeah , if the only x term is squared, it will be even prolly
DanJS
  • DanJS
you get it all good now?
BloomLocke367
  • BloomLocke367
Yes, I THINK so. I'll tag you if I have any more questions.
DanJS
  • 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
  • BloomLocke367
WAIT. I think this is neither. \(f(x)=-x^2+0.03x+5\)
BloomLocke367
  • BloomLocke367
andddd he left.
BloomLocke367
  • BloomLocke367
@DanJS
BloomLocke367
  • BloomLocke367
@Nnesha do you know that one?
BloomLocke367
  • BloomLocke367
http://prntscr.com/8bwwtw there's the graph.
BloomLocke367
  • BloomLocke367
it's symmetrical over the y-axis but when you plug in -x it doesn't work out so I'm confused.
Nnesha
  • Nnesha
\(\huge\color{green}{\checkmark}\)
Nnesha
  • 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
  • mathmate
Some simple rules, Sum of odd functions is odd. Sum of even functions is even. Sum of odd and even is neither!
BloomLocke367
  • BloomLocke367
What? What do you mean mm?
mathmate
  • mathmate
and a constant is even.
Nnesha
  • Nnesha
did*:XD
mathmate
  • mathmate
f(x)=x^2+x is a sum of even and odd functions, so f(x) is neither odd nor even!
BloomLocke367
  • BloomLocke367
I really don't understand what you mean mm XD
mathmate
  • mathmate
If you are given a function which is the sum of many terms,
mathmate
  • mathmate
Check out each term individually.
mathmate
  • mathmate
Check if "each" term is even or odd.
mathmate
  • mathmate
For example, f(x)=x^4+3x^2 is even because x^4 and 3x^2 are both even.
BloomLocke367
  • 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
  • BloomLocke367
but your method seems easier XD Thank you.
mathmate
  • 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
  • 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
  • mathmate
Good! Are you all clear now, for example, is y=x+3 even or odd?
BloomLocke367
  • BloomLocke367
so \(g(x)=2x^3-3x\) is odd, right? because after you plug in -x you get \(g(-x)=-2x^3+3x\)
BloomLocke367
  • BloomLocke367
and that's neither... right? The one you showed?
mathmate
  • mathmate
exactly!
mathmate
  • mathmate
Exactly again!
BloomLocke367
  • BloomLocke367
Awesome, thank you!
mathmate
  • mathmate
Great! You're welcome! :)
BloomLocke367
  • BloomLocke367
woe woe woe. I have one more. This one is like the example she gave us.
BloomLocke367
  • BloomLocke367
\(\large h(x)=\frac{1}{x}\)
mathmate
  • mathmate
Apply the rules and see what you get! :)
BloomLocke367
  • 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
  • mathmate
given h(x)=1/x h(-x)=1/(-x)=-1/x=-h(x) so...
BloomLocke367
  • BloomLocke367
what? I'm so confused
BloomLocke367
  • BloomLocke367
so it's odd?
BloomLocke367
  • BloomLocke367
because -h(x)
mathmate
  • 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
  • mathmate
yes, it is odd. It's easier to visualize it graphically.
BloomLocke367
  • BloomLocke367
so it's odd then. because it would equal \(-\frac{1}{x}\)?
BloomLocke367
  • BloomLocke367
and this is the graph. http://prntscr.com/8bx7i1
mathmate
  • mathmate
|dw:1441229730362:dw|
BloomLocke367
  • BloomLocke367
yes. okay. thank you. I'm going to open a new question for asymptotes.
mathmate
  • mathmate
yes, it is odd because h(-x) = - h(x) (where h(x)=1/x)
BloomLocke367
  • BloomLocke367
I'm pretty sure I understand completely now.
mathmate
  • mathmate
Good! In math, the more exercise we do, the better we understand.

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