anonymous
  • anonymous
Why is the domain and range of x^2+y^2=1 this answer?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[D = {-1\le1 \le1}\]
anonymous
  • anonymous
maybe \[-1\leq x\leq 1\]
TuringTest
  • TuringTest
you mean\[-1\le x\le 1\]what if x=2 ? what would y be?

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anonymous
  • anonymous
\[R = -1 \le y \le 1\]
anonymous
  • anonymous
ohh I meant that
anonymous
  • anonymous
yes you are right, it is the unit circle
anonymous
  • anonymous
I get so confused with domains and ranges D:
anonymous
  • anonymous
like why is it -1 less than or equal to x less than or equal to 1?
TuringTest
  • TuringTest
what if \[|y|>1\]what would x be?
anonymous
  • anonymous
i don't know.
TuringTest
  • TuringTest
try it with x=2 what would y be?
anonymous
  • anonymous
what do you mean?
experimentX
  • experimentX
can't call it a function though ... it's a closed curve.
anonymous
  • anonymous
I just don't understand why it's greater than or less than.. o.o
TuringTest
  • TuringTest
@milliex51 can you solve the above for y ? an experimX is correct as usual, but that doesn't really matter for what we're talking about
TuringTest
  • TuringTest
solve\[x^2+y^2=1\]for \(y\) please :)
anonymous
  • anonymous
\[y = +/- \sqrt{-x}+1\]
anonymous
  • anonymous
-x^2
TuringTest
  • TuringTest
not quite...
anonymous
  • anonymous
really?
anonymous
  • anonymous
or do I close bracket the 1 too?
TuringTest
  • TuringTest
\[y=\pm\sqrt{1-x^2}\]
anonymous
  • anonymous
ohh..
TuringTest
  • TuringTest
\[x^2+y^2=1\]\[y^2=1-x^2\]\[y=\pm\sqrt{1-x^2}\]now what if x=2 ? what do you get for y ?
experimentX
  • experimentX
now if we redefine it x^2+y^2=1 by y = sqrt(1-x^2) & y = -sqrt(1-x^2) on a field of real numbers than ..
TuringTest
  • TuringTest
^echo, echo.... :P
experimentX
  • experimentX
TuringTest was thinking exactly same as me
anonymous
  • anonymous
isn't it the same from my answer? um, x=2 would be y=non real?
TuringTest
  • TuringTest
and if y is not real, then x=2 is not in the domain, right?
anonymous
  • anonymous
right.
TuringTest
  • TuringTest
what about any \[|x|>1\]? what will y be?
anonymous
  • anonymous
how do I solve that?
TuringTest
  • TuringTest
I asking you if you plug in any number greater than 1 in for x into the formula we have for y, what will y e?
TuringTest
  • TuringTest
*what will y be?
anonymous
  • anonymous
i don't know D:
TuringTest
  • TuringTest
ok, look at it this way...
TuringTest
  • TuringTest
what is under the radical in\[y=\pm\sqrt{1-x^2}\]cannot be negative, right? otherwise y would be imaginary. do we agree?
anonymous
  • anonymous
yes, yup.
TuringTest
  • TuringTest
so that mean that we have a requirement that\[1-x^2\ge0\]right?
anonymous
  • anonymous
yup. or else it will be undefined
anonymous
  • anonymous
or imaginary
TuringTest
  • TuringTest
so solve that requirement for x and you will get your domain
anonymous
  • anonymous
but how? ahhh
TuringTest
  • TuringTest
\[1-x^2\ge0\]\[1\ge x^2\]perhaps here is where you get a little confused...\[1\ge x\ge -1\]
anonymous
  • anonymous
ohh.. I transpose the negative x^2
TuringTest
  • TuringTest
remember that x^2 is always positive, so all that matters is that\[|x|\le1\]and y will be real
anonymous
  • anonymous
how about when I have to look at a graph?
anonymous
  • anonymous
Like:|dw:1333035682317:dw|
anonymous
  • anonymous
what signs am I going to use? o.o
TuringTest
  • TuringTest
|dw:1333035736039:dw|we are saying that x can take on any value in the black region, right?
anonymous
  • anonymous
yes
TuringTest
  • TuringTest
assuming it includes the endpoints, what is the most that x can be?
anonymous
  • anonymous
-3 and 4
TuringTest
  • TuringTest
good :) which of those is the maximum value of x ?
anonymous
  • anonymous
4
TuringTest
  • TuringTest
right, and this translates mathematically to the statement\[x\le4\]agreed?
anonymous
  • anonymous
why because 4 is greater than or equal to x?
anonymous
  • anonymous
can I read it like that?
TuringTest
  • TuringTest
"less-than or equal to"
anonymous
  • anonymous
why is it less than? when 4 is the maximum?
TuringTest
  • TuringTest
x is less-than or equal-to 4 is how you can read it... we agree that x cannot be more than four, so x must me "less-than or equal-to" 4 think about that until it makes sense to you
anonymous
  • anonymous
that's when I get confused. :(
anonymous
  • anonymous
i know that's how I'll read it
TuringTest
  • TuringTest
read what I wrote and think about it x cannot be more than 4 therefor x is less-than or equal to four does that make sense?
anonymous
  • anonymous
ohhhh
anonymous
  • anonymous
so x has to be less than or equal to 4 so that it won't exceed 4?
anonymous
  • anonymous
how about for 3?
anonymous
  • anonymous
*-3
TuringTest
  • TuringTest
exactly so what about the minimum value of x ? what is the least x can be? -3 right?
anonymous
  • anonymous
yes
TuringTest
  • TuringTest
so then we can say x cannot be less than -3 therefor x is greater-than or equal-to -3 same logic as the other one, but make sure you really understand what I wrote
TuringTest
  • TuringTest
brb in the meantime meditate on this idea: "x cannot be less than -3" implies that "x is greater-than or equal-to -3"
anonymous
  • anonymous
so say <-- -3 are all less than?
anonymous
  • anonymous
since they are smaller?
TuringTest
  • TuringTest
so are you saying to write\[x\le-3\]?
TuringTest
  • TuringTest
"x cannot be less than -3" implies that "x is greater-than or equal-to -3" is that what the above statement says?
anonymous
  • anonymous
no.
anonymous
  • anonymous
I mean yes
TuringTest
  • TuringTest
no\[x\le-3\]means "x less-than or equal-to -3" but we said that x \(cannot\) be less than -3, so this is not the statement we want we want to say "x greater-than or equal-to -3" how do we write that?

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