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BloomLocke367
 one year ago
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
 one year ago
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.

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BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0For \(g(x)=x^3\) 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
 one year ago
Best ResponseYou've already chosen the best response.0Sorry for making it so wordy XD

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1you would multiply by 1 to check if it's odd or neither

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Even Function is symmetric about Y axis F(x) = F(x)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0yes @DanJS. That's what I learned.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0the odd is reflected over the origin, like the line y=x , or y = x^3

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0and for odd you have f(x)=f(x)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay. I'm still confused on what she did in class today XD

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1for exxample if f(X)=x^3x 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^3x

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0Now you're confusing me :O

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Even functions are symmetric about the yaxis, Odd functions are symmetric about both x and yaxis, that's it

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0odd are symmetric about both? so about the origin...or...?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0right, about the origin, same as reflected over both axis

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0okay 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
 one year ago
Best ResponseYou've already chosen the best response.0Just switch x for x and see if you get either the original f(x)  or the negative  f(x) or neither

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0that is basically all you need to do

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0yea you got it...but it is not parabola, y^2  x^2 = k hyperbola, upper half +root

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0oops XD I only saw the upper half. My bad. Thanks.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0it is only the upper half, not + or  root, parabola is y = x^2 no square on the y.. eh nevermind

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0And 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
 one year ago
Best ResponseYou've already chosen the best response.0yeah , if the only x term is squared, it will be even prolly

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0you get it all good now?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0Yes, I THINK so. I'll tag you if I have any more questions.

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0just 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
 one year ago
Best ResponseYou've already chosen the best response.0WAIT. I think this is neither. \(f(x)=x^2+0.03x+5\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0@Nnesha do you know that one?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0http://prntscr.com/8bwwtw there's the graph.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0it's symmetrical over the yaxis but when you plug in x it doesn't work out so I'm confused.

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1\(\huge\color{green}{\checkmark}\)

Nnesha
 one year ago
Best ResponseYou've already chosen the best response.1well 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 \[1(x^20.03x+5)=x^2+0.035\] 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
 one year ago
Best ResponseYou've already chosen the best response.2Some simple rules, Sum of odd functions is odd. Sum of even functions is even. Sum of odd and even is neither!

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0What? What do you mean mm?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2and a constant is even.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2f(x)=x^2+x is a sum of even and odd functions, so f(x) is neither odd nor even!

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I really don't understand what you mean mm XD

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2If you are given a function which is the sum of many terms,

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Check out each term individually.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Check if "each" term is even or odd.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2For example, f(x)=x^4+3x^2 is even because x^4 and 3x^2 are both even.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0ohhhh. 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
 one year ago
Best ResponseYou've already chosen the best response.0but your method seems easier XD Thank you.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2No, 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
 one year ago
Best ResponseYou've already chosen the best response.0yes. and I know, I have to prove these problems algebraically and graphically anyways. But it's good to know for the future.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Good! Are you all clear now, for example, is y=x+3 even or odd?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0so \(g(x)=2x^33x\) is odd, right? because after you plug in x you get \(g(x)=2x^3+3x\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0and that's neither... right? The one you showed?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0Awesome, thank you!

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Great! You're welcome! :)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0woe woe woe. I have one more. This one is like the example she gave us.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0\(\large h(x)=\frac{1}{x}\)

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Apply the rules and see what you get! :)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0you get \(h(x)=\large\frac{1}{x}\) but you can take \(h(x)=\large\frac{1}{1(x)}\) which would make it even.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2given h(x)=1/x h(x)=1/(x)=1/x=h(x) so...

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0what? I'm so confused

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2After 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
 one year ago
Best ResponseYou've already chosen the best response.2yes, it is odd. It's easier to visualize it graphically.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0so it's odd then. because it would equal \(\frac{1}{x}\)?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0and this is the graph. http://prntscr.com/8bx7i1

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2dw:1441229730362:dw

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0yes. okay. thank you. I'm going to open a new question for asymptotes.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2yes, it is odd because h(x) =  h(x) (where h(x)=1/x)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.0I'm pretty sure I understand completely now.

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Good! In math, the more exercise we do, the better we understand.
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