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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.

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  1. BloomLocke367
    • one year ago
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    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?

  2. BloomLocke367
    • one year ago
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    Sorry for making it so wordy XD

  3. Nnesha
    • one year ago
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    you would multiply by -1 to check if it's odd or neither

  4. DanJS
    • one year ago
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    Even Function is symmetric about Y axis F(x) = F(-x)

  5. BloomLocke367
    • one year ago
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    yes @DanJS. That's what I learned.

  6. DanJS
    • one year ago
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    the odd is reflected over the origin, like the line y=x , or y = x^3

  7. BloomLocke367
    • one year ago
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    and for odd you have f(-x)=-f(x)

  8. BloomLocke367
    • one year ago
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    ohh okay. I'm still confused on what she did in class today XD

  9. DanJS
    • one year ago
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    yep, that makes sense

  10. Nnesha
    • one year ago
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    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

  11. BloomLocke367
    • one year ago
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    Now you're confusing me :O

  12. DanJS
    • one year ago
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    Even functions are symmetric about the y-axis, Odd functions are symmetric about both x and y-axis, that's it

  13. BloomLocke367
    • one year ago
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    odd are symmetric about both? so about the origin...or...?

  14. DanJS
    • one year ago
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    right, about the origin, same as reflected over both axis

  15. BloomLocke367
    • one year ago
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    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.

  16. DanJS
    • one year ago
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    Just switch x for -x and see if you get either the original f(x) - or the negative - f(x) or neither

  17. DanJS
    • one year ago
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    that is basically all you need to do

  18. BloomLocke367
    • one year ago
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    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?

  19. DanJS
    • one year ago
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    yea you got it...but it is not parabola, y^2 - x^2 = k hyperbola, upper half +root

  20. BloomLocke367
    • one year ago
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    oops XD I only saw the upper half. My bad. Thanks.

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

  22. BloomLocke367
    • one year ago
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    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

  23. DanJS
    • one year ago
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    yeah , if the only x term is squared, it will be even prolly

  24. DanJS
    • one year ago
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    you get it all good now?

  25. BloomLocke367
    • one year ago
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    Yes, I THINK so. I'll tag you if I have any more questions.

  26. DanJS
    • one year ago
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    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

  27. BloomLocke367
    • one year ago
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    WAIT. I think this is neither. \(f(x)=-x^2+0.03x+5\)

  28. BloomLocke367
    • one year ago
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    andddd he left.

  29. BloomLocke367
    • one year ago
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    @DanJS

  30. BloomLocke367
    • one year ago
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    @Nnesha do you know that one?

  31. BloomLocke367
    • one year ago
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    http://prntscr.com/8bwwtw there's the graph.

  32. BloomLocke367
    • one year ago
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    it's symmetrical over the y-axis but when you plug in -x it doesn't work out so I'm confused.

  33. Nnesha
    • one year ago
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    \(\huge\color{green}{\checkmark}\)

  34. Nnesha
    • one year ago
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    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!

  35. mathmate
    • one year ago
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    Some simple rules, Sum of odd functions is odd. Sum of even functions is even. Sum of odd and even is neither!

  36. BloomLocke367
    • one year ago
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    What? What do you mean mm?

  37. mathmate
    • one year ago
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    and a constant is even.

  38. Nnesha
    • one year ago
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    did*:XD

  39. mathmate
    • one year ago
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    f(x)=x^2+x is a sum of even and odd functions, so f(x) is neither odd nor even!

  40. BloomLocke367
    • one year ago
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    I really don't understand what you mean mm XD

  41. mathmate
    • one year ago
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    If you are given a function which is the sum of many terms,

  42. mathmate
    • one year ago
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    Check out each term individually.

  43. mathmate
    • one year ago
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    Check if "each" term is even or odd.

  44. mathmate
    • one year ago
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    For example, f(x)=x^4+3x^2 is even because x^4 and 3x^2 are both even.

  45. BloomLocke367
    • one year ago
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    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.

  46. BloomLocke367
    • one year ago
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    but your method seems easier XD Thank you.

  47. mathmate
    • one year ago
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    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.

  48. BloomLocke367
    • one year ago
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    yes. and I know, I have to prove these problems algebraically and graphically anyways. But it's good to know for the future.

  49. mathmate
    • one year ago
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    Good! Are you all clear now, for example, is y=x+3 even or odd?

  50. BloomLocke367
    • one year ago
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    so \(g(x)=2x^3-3x\) is odd, right? because after you plug in -x you get \(g(-x)=-2x^3+3x\)

  51. BloomLocke367
    • one year ago
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    and that's neither... right? The one you showed?

  52. mathmate
    • one year ago
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    exactly!

  53. mathmate
    • one year ago
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    Exactly again!

  54. BloomLocke367
    • one year ago
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    Awesome, thank you!

  55. mathmate
    • one year ago
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    Great! You're welcome! :)

  56. BloomLocke367
    • one year ago
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    woe woe woe. I have one more. This one is like the example she gave us.

  57. BloomLocke367
    • one year ago
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    \(\large h(x)=\frac{1}{x}\)

  58. mathmate
    • one year ago
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    Apply the rules and see what you get! :)

  59. BloomLocke367
    • one year ago
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    you get \(h(-x)=\large\frac{1}{-x}\) but you can take \(h(-x)=\large\frac{1}{-1(x)}\) which would make it even.

  60. mathmate
    • one year ago
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    given h(x)=1/x h(-x)=1/(-x)=-1/x=-h(x) so...

  61. BloomLocke367
    • one year ago
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    what? I'm so confused

  62. BloomLocke367
    • one year ago
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    so it's odd?

  63. BloomLocke367
    • one year ago
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    because -h(x)

  64. mathmate
    • one year ago
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    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.

  65. mathmate
    • one year ago
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    yes, it is odd. It's easier to visualize it graphically.

  66. BloomLocke367
    • one year ago
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    so it's odd then. because it would equal \(-\frac{1}{x}\)?

  67. BloomLocke367
    • one year ago
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    and this is the graph. http://prntscr.com/8bx7i1

  68. mathmate
    • one year ago
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    |dw:1441229730362:dw|

  69. BloomLocke367
    • one year ago
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    yes. okay. thank you. I'm going to open a new question for asymptotes.

  70. mathmate
    • one year ago
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    yes, it is odd because h(-x) = - h(x) (where h(x)=1/x)

  71. BloomLocke367
    • one year ago
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    I'm pretty sure I understand completely now.

  72. mathmate
    • one year ago
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    Good! In math, the more exercise we do, the better we understand.

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