anonymous
  • anonymous
is f(x)=xabs(x^2) an odd, even or neither function
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
even i belive
anonymous
  • anonymous
you just look at the expontes
anonymous
  • anonymous
it is online hw and i only have one attempt to get it right

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anonymous
  • anonymous
u have to be 100% sure
anonymous
  • anonymous
\[f(x)=x \left| x^2 \right|\]
anonymous
  • anonymous
is it odd, even or neither
anonymous
  • anonymous
So in theory, IF: \(f(x)=f(-x)\) THEN It is even. \(-f(x)=f(-x)\) THEN it is odd.
anonymous
  • 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
  • anonymous
can u give me an answer?
anonymous
  • anonymous
please
anonymous
  • anonymous
i get confused with the absolute value
anonymous
  • anonymous
I don't get what hes saying what my teacher told me is you just look at the exponets
anonymous
  • 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
  • anonymous
I am \(200\%\) sure that \(f(x)=x|x^2|\) is odd
anonymous
  • anonymous
thanks
anonymous
  • anonymous
what do i do if it gives me two graphs
anonymous
  • anonymous
If what does...?
anonymous
  • anonymous
the hw
anonymous
  • 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
  • anonymous
don't know if the link works
anonymous
  • anonymous
@KeithAfasCalcLover
anonymous
  • anonymous
good luck t(-_-)t
anonymous
  • anonymous
i wil do it on my own i just want u to help me start
anonymous
  • anonymous
And what you want to do with the graph?
anonymous
  • anonymous
find out if they r odd, even or neither
anonymous
  • anonymous
@KeithAfasCalcLover
anonymous
  • 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
  • anonymous
so what do u think
anonymous
  • anonymous
lol why don't you think instead of just asking for answers
anonymous
  • anonymous
i think its red=even and blue=odd. what do u think
anonymous
  • anonymous
Well think about it, could you flip one side of the red over the y axis to match the other?
anonymous
  • anonymous
have u looked at the graph
anonymous
  • anonymous
Yes muzzamil, I have.
anonymous
  • anonymous
buddy i really need help on this
anonymous
  • anonymous
LOL KEITH WALK AWAY NOW
anonymous
  • 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
  • anonymous
i said i think red is even and blue is odd. atlesat i am trying
anonymous
  • anonymous
i asked if u thouht i was right
anonymous
  • anonymous
Well why do you think that?
anonymous
  • anonymous
and he is asking you a question and you haven't answered it and if you did it answers your question llol
anonymous
  • anonymous
kramer instead of wasting ur precious time maybe u should leave. i did what u told me to
anonymous
  • anonymous
trust me my time at the moment is not precious
anonymous
  • 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
  • anonymous
didn't you get taught the lesson in school?
anonymous
  • 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
  • anonymous
bro that isn't calculus that's pre calc
anonymous
  • anonymous
its th first chapter of the book
anonymous
  • anonymous
u dont do derivatives the first day
tkhunny
  • 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
  • anonymous
we already solved that. we r on the graph one. i posted a link @tkhunny
anonymous
  • 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
  • anonymous
because hes a liar lol
anonymous
  • anonymous
UMTYMP. university of minnesota talented youth math program
anonymous
  • anonymous
look it up. mathcep.umn.edu
anonymous
  • anonymous
@llkramer mathcep.umn.edu
tkhunny
  • tkhunny
I saw the solution. I just wanted to make sure we used the word "redundant".
anonymous
  • anonymous
@tkhunny this guy just wants answers no work don't waste your time
anonymous
  • anonymous
@llkramer did u look it up?
anonymous
  • anonymous
\[f(x)=x|x|^2\] \[f(2)=2\times |2|^2=\] \[f(-2)=-2\times |-2|^2=-8\]
anonymous
  • anonymous
we already solved this @satellite73 . we r now on the link i posted with the two graphs
anonymous
  • 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
  • anonymous
DONT GIVE HIM A ANSWER BRO
anonymous
  • anonymous
what a pretty picture what are you supposed to do with it?
anonymous
  • anonymous
AN**
anonymous
  • anonymous
r the graphs even, odd, or neither
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
i think red=even and blue=odd
anonymous
  • 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
  • anonymous
yes, you are right
anonymous
  • anonymous
LOL @satellite73 "What a pretty picture" haha that got a good laugh a out of me.
anonymous
  • anonymous
THANK YOU @satellite73 FINALLY SOMEBODY WHO COULD ACTUALLY TELL ME IF I WAS RIGHT OR WRONG.
anonymous
  • anonymous
@satellite73 That's EXACTLY what I thought when It was first shown to me
anonymous
  • anonymous
THEN Y COULDN''T U SAY THAT
anonymous
  • anonymous
U COULD HAVE SAVED US LIKE 30 MIN
anonymous
  • 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
  • anonymous
WELL THANKS FOR THE FIRST ONE ANYWAY
anonymous
  • anonymous
Is this some sort of an "ALL CAPS RAGE" or so?
anonymous
  • anonymous
idk. bye
anonymous
  • anonymous
Bye. GOOD LUCK

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