is f(x)=xabs(x^2) an odd, even or neither function

- anonymous

is f(x)=xabs(x^2) an odd, even or neither function

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- schrodinger

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- anonymous

even i belive

- anonymous

you just look at the expontes

- anonymous

it is online hw and i only have one attempt to get it right

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## More answers

- anonymous

u have to be 100% sure

- anonymous

\[f(x)=x \left| x^2 \right|\]

- anonymous

is it odd, even or neither

- anonymous

So in theory, IF:
\(f(x)=f(-x)\) THEN It is even.
\(-f(x)=f(-x)\) THEN it is odd.

- anonymous

So if we know that \(f(x)=x|x^2|\), Lets check to see if the identities are correct!
Try computing f(-x) and -f(x) and see what those return and if it fits any of the above equations.

- anonymous

can u give me an answer?

- anonymous

please

- anonymous

i get confused with the absolute value

- anonymous

I don't get what hes saying what my teacher told me is you just look at the exponets

- anonymous

Oh yeah...good point. Try this:
\[f(x)=x|x^2|\]
\[f(-x)=(-x)|(-x)^2|=-x|x^2|\]
\[-f(x)=-(x)|x^2|=-x|x^2|\]
So since \(f(-x)=-f(x)\) The function is odd.

- anonymous

I am \(200\%\) sure that \(f(x)=x|x^2|\) is odd

- anonymous

thanks

- anonymous

what do i do if it gives me two graphs

- anonymous

If what does...?

- anonymous

the hw

- anonymous

it gives me http://roy.math.umn.edu/webwork2_course_files/umtymp-calculus1-f13/tmp/gif/muzzammil-Q-raza-524-setReview_and_Rate_of_Changeprob4image1.png

- anonymous

don't know if the link works

- anonymous

@KeithAfasCalcLover

- anonymous

good luck t(-_-)t

- anonymous

i wil do it on my own i just want u to help me start

- anonymous

And what you want to do with the graph?

- anonymous

find out if they r odd, even or neither

- anonymous

@KeithAfasCalcLover

- anonymous

Visually you can identify a function as odd even or neither by imagining this:
If the right or left side of \(f(x)\) can be rotated about the y-axis to match the other side, it is even.
If the If the right or left side of \(f(x)\) can be rotated about the line of \(y=x\) or \(y=-x\) to match the other side, it is odd.
Otherwise it is neither.

- anonymous

so what do u think

- anonymous

lol why don't you think instead of just asking for answers

- anonymous

i think its red=even and blue=odd. what do u think

- anonymous

Well think about it, could you flip one side of the red over the y axis to match the other?

- anonymous

have u looked at the graph

- anonymous

Yes muzzamil, I have.

- anonymous

buddy i really need help on this

- anonymous

LOL KEITH WALK AWAY NOW

- anonymous

Have you?
Take a look. Imagine you grab one side and fold it over the y-axis. Does it fit on the other side?

- anonymous

i said i think red is even and blue is odd. atlesat i am trying

- anonymous

i asked if u thouht i was right

- anonymous

Well why do you think that?

- anonymous

and he is asking you a question and you haven't answered it and if you did it answers your question llol

- anonymous

kramer instead of wasting ur precious time maybe u should leave. i did what u told me to

- anonymous

trust me my time at the moment is not precious

- anonymous

i keep asking ur opinion cuz my a$$ is on the line. my parents get pissed if i mess up on my hw. its an asian thing

- anonymous

didn't you get taught the lesson in school?

- anonymous

i also want to watch football this weekend and i need this done. i am a ninth grader taking calculus at the university of minnesota. the professors suck there

- anonymous

bro that isn't calculus that's pre calc

- anonymous

its th first chapter of the book

- anonymous

u dont do derivatives the first day

- tkhunny

x is odd.
x^2 is even
abs(x^2) is redundant because x^2 is always positive.
Therefore, f(x) = x^3 and it is definitely odd.

- anonymous

we already solved that. we r on the graph one. i posted a link @tkhunny

- anonymous

I would have to agree that it is not calculus but it may be review of pre calc in the beginning.
How youre accepted in Minnesota University at grade 9 is beyond me.

- anonymous

because hes a liar lol

- anonymous

UMTYMP. university of minnesota talented youth math program

- anonymous

look it up. mathcep.umn.edu

- anonymous

@llkramer mathcep.umn.edu

- tkhunny

I saw the solution. I just wanted to make sure we used the word "redundant".

- anonymous

@tkhunny this guy just wants answers no work don't waste your time

- anonymous

@llkramer did u look it up?

- anonymous

\[f(x)=x|x|^2\]
\[f(2)=2\times |2|^2=\]
\[f(-2)=-2\times |-2|^2=-8\]

- anonymous

we already solved this @satellite73 . we r now on the link i posted with the two graphs

- anonymous

http://roy.math.umn.edu/webwork2_course_files/umtymp-calculus1-f13/tmp/gif/muzzammil-Q-raza-524-setReview_and_Rate_of_Changeprob4image1.png

- anonymous

DONT GIVE HIM A ANSWER BRO

- anonymous

what a pretty picture
what are you supposed to do with it?

- anonymous

AN**

- anonymous

r the graphs even, odd, or neither

- anonymous

@satellite73

- anonymous

i think red=even and blue=odd

- anonymous

the one that is symmetric with respect to the \(y\) axis is even
the one that is symmetric with respect to the origin \((0,0)\) is odd

- anonymous

yes, you are right

- anonymous

LOL @satellite73 "What a pretty picture" haha that got a good laugh a
out of me.

- anonymous

THANK YOU @satellite73 FINALLY SOMEBODY WHO COULD ACTUALLY TELL ME IF I WAS RIGHT OR WRONG.

- anonymous

@satellite73 That's EXACTLY what I thought when It was first shown to me

- anonymous

THEN Y COULDN''T U SAY THAT

- anonymous

U COULD HAVE SAVED US LIKE 30 MIN

- anonymous

Though math is about right and wrong, It is even more about the WHY. "yes" and "no" wont get you very far without the WHY.

- anonymous

WELL THANKS FOR THE FIRST ONE ANYWAY

- anonymous

Is this some sort of an "ALL CAPS RAGE" or so?

- anonymous

idk. bye

- anonymous

Bye. GOOD LUCK

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