## 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.

1. 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?

2. BloomLocke367

Sorry for making it so wordy XD

3. Nnesha

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

4. DanJS

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

5. BloomLocke367

yes @DanJS. That's what I learned.

6. DanJS

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

7. BloomLocke367

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

8. BloomLocke367

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

9. DanJS

yep, that makes sense

10. 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

11. BloomLocke367

Now you're confusing me :O

12. DanJS

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

13. BloomLocke367

14. DanJS

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

15. 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.

16. DanJS

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

17. DanJS

that is basically all you need to do

18. 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?

19. DanJS

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

20. BloomLocke367

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

21. DanJS

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

22. 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

23. DanJS

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

24. DanJS

you get it all good now?

25. BloomLocke367

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

26. 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

27. BloomLocke367

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

28. BloomLocke367

andddd he left.

29. BloomLocke367

@DanJS

30. BloomLocke367

@Nnesha do you know that one?

31. BloomLocke367

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

32. BloomLocke367

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

33. Nnesha

$$\huge\color{green}{\checkmark}$$

34. 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!

35. mathmate

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

36. BloomLocke367

What? What do you mean mm?

37. mathmate

and a constant is even.

38. Nnesha

did*:XD

39. mathmate

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

40. BloomLocke367

I really don't understand what you mean mm XD

41. mathmate

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

42. mathmate

Check out each term individually.

43. mathmate

Check if "each" term is even or odd.

44. mathmate

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

45. 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.

46. BloomLocke367

but your method seems easier XD Thank you.

47. 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.

48. BloomLocke367

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

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

50. BloomLocke367

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

51. BloomLocke367

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

52. mathmate

exactly!

53. mathmate

Exactly again!

54. BloomLocke367

Awesome, thank you!

55. mathmate

Great! You're welcome! :)

56. BloomLocke367

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

57. BloomLocke367

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

58. mathmate

Apply the rules and see what you get! :)

59. 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.

60. mathmate

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

61. BloomLocke367

what? I'm so confused

62. BloomLocke367

so it's odd?

63. BloomLocke367

because -h(x)

64. 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.

65. mathmate

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

66. BloomLocke367

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

67. BloomLocke367

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

68. mathmate

|dw:1441229730362:dw|

69. BloomLocke367

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

70. mathmate

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

71. BloomLocke367

I'm pretty sure I understand completely now.

72. mathmate

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