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anonymous
 one year ago
Whats the domain and range for
f(x)= x^2+x2/x^23x4
anonymous
 one year ago
Whats the domain and range for f(x)= x^2+x2/x^23x4

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arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1Well for domain, the roots of the denominator should not be zero.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is the domain all real numbers except for 1, and 4?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the numerator and denominator are 2 different domains?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the domain is in the xaxis and the range is in the yaxis. Since we are given a fraction as a function there is going to be restrictions in the domain.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2\[(x^23x4) \rightarrow (x+1)(x4)\] so the domain should be all real numbers except when x = 1 and x = 4. which is what I saw earlier. so yes you're correct on that one @Abbs__

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2that doesn't sound right at all ^

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1Yes Yes, I did a calculation mistake, @Abbs_ is right

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2range is going to be a little bit harder though...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is it all real numbers except y=1?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2I gotta be honest. I have to graph that function first.. and desmos hates me and my fractions

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I agree with the answer of @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2thanks, HEY YOUR QH symbol is back @Michele_Laino !

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I see nevertheless not in mathematics subsections, thanks! @UsukiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2the line has to cover the yvalues ... if the line doesn't cover the yvalues then we have a restriction. This is where I fumble a lot. It could be all reals except y = 1... not sure. What do you think @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1at points where the function is undefined, generally that function behaves as an infinity, and that is what is happening into your graph

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2so it's all reals for the range?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2it's like every time I do these problems, I understand the domain no problem. I see the gaps. It's the range that's frustrating.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the range, as in this case, is always a real set, nevertheless is not a limited set

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2because it goes on forever?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2dw:1437990839458:dw ok.. so consider this drawing of a circle .. would the domain be from 1 to 1 and then range be 1 to 1 as well.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1towards down or towards up

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1yes, range should be 1 to 1

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1yaxis represents Range

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1your function is a circular function, so its domain is the subsequent set: \[\Large [0,2\pi )\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2*I'm not going for trig... just regular domain and range*

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in order to establish the range, we need to know the exact algebraic shape of your function

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1One second... A circle is not a function... right?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1As intersects two times along xaxis.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2it's not a function.. fails the vertical line test

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Depends on what coordinate system you're using.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2not 1 to 1 either.. fails horizontal line test.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1another possible case is that of your graph is related to this equation: \[\Large {x^2} + {y^2} = 1\] nevertheless as we can easily check, that is not a function

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1So, how come range will be there. Not possible

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1there is not meaning in speaking about range, since we have not a function

Empty
 one year ago
Best ResponseYou've already chosen the best response.0$$r(\theta) = 1$$ is a function for the domain \([0,2 \pi) \) that is a circle. Vertical line test is for boxes not circles :P

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is why we hav not a function: dw:1437991311333:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2actually the vertical line test can be used for circles. and yeah I know that.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Guys, I have a neat way to find the range...

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1ok, try finding range of inverse of sin(sqrt(x^2+x+1))

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3\[y = \frac{x^2 + x  2}{x^2  3x  4}\]\[\Rightarrow yx^2  3y x  4y = x^2 + x  2\]\[\Rightarrow (y1)x^2 (3y+1)x  (4y+2) = 0\]We want to check the values of \(y\) for which this equation has a real root, so\[D \ge 0\]\[(3y+1)^2 + 4(4y2)(y1) \ge 0\]\[ \Rightarrow 9y^2 + 6y + 1 + 4(4y^2  2y  4y + 2) \ge 0\]\[\Rightarrow 25y^2 18y + 9 \ge 0 \]This inequality is always true, so \(y \in \mathbb R\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3Oh, and \(3/5, 1 \) won't be included.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2@ParthKohli one question how did you get \[(3y+1)^2 + 4(4y2)(y1) \ge 0 \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since we have: \[\Large {x^2} + x + 1 = {\left( {\sin y} \right)^2}\] the range is the subsequent set: \[\Large \begin{gathered} range = \left\{ {x \in {\mathbf{R}}\;{\text{such that:}}\;0 \leqslant {x^2} + x + 1 \leqslant 1} \right\} = \hfill \\ \hfill \\ = \left\{ {x \in {\mathbf{R}}\;{\text{such that:}}\;  1 \leqslant x \leqslant 0} \right\} \hfill \\ \end{gathered} \]

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.3OK, let me try that out again... we want to check the values of \(y\) where the quadratic equation\[(y1)x^2  (3y + 1) x  (4y  2) = 0 \]has real solutions.\[\rm discriminant \ge 0 \tag{for real solutions}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1whereas the domain, is: \[\Large domain = \left\{ {\forall y \in {\mathbf{R}}} \right\}\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2OH THAT IS COOL! I AM SO USING THAT! Thanks @ParthKohli

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1Thank you @Michele_Laino so in terms of pi? But in my book the answer is [pi/3,pi/2]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'm pondering...

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1range  sin^(1)(sqrt(x^2+x+1))

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.2in degree mode [60 degrees , 90 degrees] hmmm...

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1I searched wolfram... answer is coming.. but if any of you guys can give an explanation it would be awesome

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry my answer above is an error, here is the right answer: \[\Large {x^2} + x + 1 = {\left( {\arcsin y} \right)^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1range is such that: \[\Large 0 \leqslant \sqrt {{x^2} + x + 1} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have this: \[\Large {x^2} + x + 1 \geqslant \frac{3}{4}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can try to solve this condition: \[\Large \frac{{\sqrt 3 }}{2} \leqslant \sqrt {{x^2} + x + 1} \leqslant 2\pi \]

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1hmmm... analyzing

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.1nope, I am not getting. I will think more...
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