Question

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.

Answer 1

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?

Answer 2

Sorry for making it so wordy XD

Answer 3

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

Answer 4

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

Answer 5

yes @DanJS. That's what I learned.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

yep, that makes sense

Answer 10

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

Answer 11

Now you're confusing me :O

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

that is basically all you need to do

Answer 18

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?

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

you get it all good now?

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

andddd he left.

Answer 29

@DanJS

Answer 30

@Nnesha do you know that one?

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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!

Answer 35

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

Answer 36

What? What do you mean mm?

Answer 37

and a constant is even.

Answer 38

did*:XD

Answer 39

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

Answer 40

I really don't understand what you mean mm XD

Answer 41

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

Answer 42

Check out each term individually.

Answer 43

Check if "each" term is even or odd.

Answer 44

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

Answer 45

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.

Answer 46

but your method seems easier XD Thank you.

Answer 47

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.

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

exactly!

Answer 53

Exactly again!

Answer 54

Awesome, thank you!

Answer 55

Great! You're welcome! :)

Answer 56

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

Answer 57

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

Answer 58

Apply the rules and see what you get! :)

Answer 59

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

Answer 60

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

Answer 61

what? I'm so confused

Answer 62

so it's odd?

Answer 63

because -h(x)

Answer 64

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.

Answer 65

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

Answer 66

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

Answer 67

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