Ask your own question, for FREE!
Mathematics 9 Online
OpenStudy (anonymous):

please explain (: @hartnn

OpenStudy (loser66):

it is neither.

OpenStudy (anonymous):

please explain?

OpenStudy (loser66):

another way to find it out is drawing it out.

OpenStudy (anonymous):

is there any way to figure it out from like an equation?

OpenStudy (loser66):

Yes, there are. By plugging x by -x into the function Definition of even: f(x) = f(-x) odd : f(-x) =-f(x) Now, figure out all of them:

OpenStudy (loser66):

YOur f(x) = \(\sqrt {2x+1}\) right?

OpenStudy (loser66):

Is it f(x) = \(2\sqrt{2x+1}\) I need find 1) -f(x) 2) f(-x) and then compare.

OpenStudy (loser66):

1) -f(x) is easy, just plug minus sign in the front, I get -f(x) = -\(2\sqrt{2x+1}\) got this part??

OpenStudy (anonymous):

sorry my computer is acting up, it is this

OpenStudy (loser66):

ok, no problem, just find -f(x) f(-x) show me your -f(x)

OpenStudy (anonymous):

-f(x)= -x-5/x

OpenStudy (loser66):

where is ^3?

OpenStudy (anonymous):

-x^3-5/x

OpenStudy (loser66):

good, now replace all x in your original one by (-x) , what do you get? (it is f(-x))

OpenStudy (anonymous):

it is the same thing..

OpenStudy (loser66):

Don't be lazy, have to label them. As above you have \(\color{red}{-f(x)}= -x^3-\dfrac{5}{x}\)

OpenStudy (loser66):

Show me your work with the name of the function \(\color {blue}{f(-x)}\)

OpenStudy (anonymous):

im not! im confused!!

OpenStudy (loser66):

I know!! that's why I want you to do STEP BY STEP to NOT be confused anymore.

OpenStudy (anonymous):

didnt i already replace the x's in the first part? thats what im confused about

OpenStudy (loser66):

Those are different steps. for the first part , you just put the minus sign in the front of f(x) to get -f(x) For this part, you replace x by -x They are not the same.

OpenStudy (anonymous):

ohh!

OpenStudy (loser66):

For example: if f(x) = x^2 --> -f(x) = -x^2 but f(-x) = (-x)^2 = x^2 Trivially, they are not the same

OpenStudy (anonymous):

but youre multiplying the x by - 3 times which makes it still -x^2-5/x

OpenStudy (anonymous):

-x^3-5/x*

OpenStudy (loser66):

I told you!! write the whole thing!!! f(x) = x^3 + 5/x now put minus sign in the front to get -f(x) = -[ x^3 +5/x] = -x^3 -5/x and you are correct there. Now the second step. Replace x by (-x) into \(\Huge f(x) \) to get \(\Huge f(-x)\)

OpenStudy (loser66):

No question, just do. It will turn clear later.

OpenStudy (anonymous):

f(-x)= x+5/x

OpenStudy (loser66):

nope!!

OpenStudy (anonymous):

omg

OpenStudy (anonymous):

i think cause its over this that im not understanding, because im not that stupid

OpenStudy (loser66):

\(f(\color{red}{x})= \color{red}{x}^3+\dfrac{5}{\color{red}{x}}\) \(f(\color{red}{-x})= \color{red}{-x}^3+\dfrac{5}{\color{red}{-x}}= \color{red}{-x}^3-\dfrac{5}{\color{red}{x}}\)

OpenStudy (anonymous):

thats what i put!!!

OpenStudy (anonymous):

ok so moving on haha

OpenStudy (loser66):

Yes, now compare. you have \((\color{red}{-}f(x))= \color{red}{-x}^3-\dfrac{5}{\color{red}{x}}=f\color{red}{(-x)}\) right? so?? by definition, is it odd or even?

OpenStudy (anonymous):

even(:

OpenStudy (anonymous):

right?..

OpenStudy (loser66):

You guess??? type out the definition of odd and even , please

OpenStudy (anonymous):

wait so f(-x) and -f(x) was the same right?!

OpenStudy (loser66):

yes.

OpenStudy (anonymous):

lol im not guessing if f(-x) =- f(x) then the function is odd if f(-x) = f(x) then the function is even

OpenStudy (anonymous):

i said it was the same earlier!!!!! you said no!!!

OpenStudy (anonymous):

it is odd

OpenStudy (loser66):

Hihihi... you jumped from this question to other one. You wrote the functions without label them. You made question on a mess and you expected me understand what you are saying. hihihihi Yes, you are correct. My boss.!!

OpenStudy (anonymous):

such an easy question, took an hour lol

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!