anonymous
  • anonymous
Whats the domain and range for f(x)= x^2+x-2/x^2-3x-4
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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arindameducationusc
  • arindameducationusc
Well for domain, the roots of the denominator should not be zero.
anonymous
  • anonymous
Is the domain all real numbers except for -1, and 4?
anonymous
  • anonymous
and is the range 1?

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More answers

anonymous
  • anonymous
so the numerator and denominator are 2 different domains?
UsukiDoll
  • UsukiDoll
the domain is in the x-axis and the range is in the y-axis. Since we are given a fraction as a function there is going to be restrictions in the domain.
UsukiDoll
  • UsukiDoll
\[(x^2-3x-4) \rightarrow (x+1)(x-4)\] 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
  • UsukiDoll
that doesn't sound right at all ^
UsukiDoll
  • UsukiDoll
@Michele_Laino
anonymous
  • anonymous
Thanks!
arindameducationusc
  • arindameducationusc
Yes Yes, I did a calculation mistake, @Abbs_ is right
anonymous
  • anonymous
It's okay. :)
UsukiDoll
  • UsukiDoll
range is going to be a little bit harder though...
anonymous
  • anonymous
Is it all real numbers except y=1?
UsukiDoll
  • UsukiDoll
I gotta be honest. I have to graph that function first.. and desmos hates me and my fractions
Michele_Laino
  • Michele_Laino
I agree with the answer of @UsukiDoll
UsukiDoll
  • UsukiDoll
thanks, HEY YOUR QH symbol is back @Michele_Laino !
Michele_Laino
  • Michele_Laino
yes! I see nevertheless not in mathematics subsections, thanks! @UsukiDoll
UsukiDoll
  • UsukiDoll
UsukiDoll
  • UsukiDoll
the line has to cover the y-values ... if the line doesn't cover the y-values 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
  • Michele_Laino
at points where the function is undefined, generally that function behaves as an infinity, and that is what is happening into your graph
UsukiDoll
  • UsukiDoll
so it's all reals for the range?
UsukiDoll
  • UsukiDoll
it'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
  • Michele_Laino
the range, as in this case, is always a real set, nevertheless is not a limited set
UsukiDoll
  • UsukiDoll
because it goes on forever?
Michele_Laino
  • Michele_Laino
yes!
UsukiDoll
  • UsukiDoll
|dw: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
  • Michele_Laino
towards down or towards up
arindameducationusc
  • arindameducationusc
yes, range should be -1 to 1
arindameducationusc
  • arindameducationusc
y-axis represents Range
Michele_Laino
  • Michele_Laino
your function is a circular function, so its domain is the subsequent set: \[\Large [0,2\pi )\]
UsukiDoll
  • UsukiDoll
*I'm not going for trig... just regular domain and range*
Michele_Laino
  • Michele_Laino
in order to establish the range, we need to know the exact algebraic shape of your function
arindameducationusc
  • arindameducationusc
One second... A circle is not a function... right?
arindameducationusc
  • arindameducationusc
As intersects two times along x-axis.
UsukiDoll
  • UsukiDoll
it's not a function.. fails the vertical line test
Empty
  • Empty
Depends on what coordinate system you're using.
UsukiDoll
  • UsukiDoll
not 1 to 1 either.. fails horizontal line test.
Michele_Laino
  • Michele_Laino
another 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
  • arindameducationusc
So, how come range will be there. Not possible
Michele_Laino
  • Michele_Laino
there is not meaning in speaking about range, since we have not a function
Empty
  • Empty
$$r(\theta) = 1$$ is a function for the domain \([0,2 \pi) \) that is a circle. Vertical line test is for boxes not circles :P
Michele_Laino
  • Michele_Laino
here is why we hav not a function: |dw:1437991311333:dw|
UsukiDoll
  • UsukiDoll
actually the vertical line test can be used for circles. and yeah I know that.
ParthKohli
  • ParthKohli
Guys, I have a neat way to find the range...
arindameducationusc
  • arindameducationusc
ok, try finding range of inverse of sin(sqrt(x^2+x+1))
ParthKohli
  • ParthKohli
\[y = \frac{x^2 + x - 2}{x^2 - 3x - 4}\]\[\Rightarrow yx^2 - 3y x - 4y = x^2 + x - 2\]\[\Rightarrow (y-1)x^2 -(3y+1)x - (4y+2) = 0\]We want to check the values of \(y\) for which this equation has a real root, so\[D \ge 0\]\[(3y+1)^2 + 4(4y-2)(y-1) \ge 0\]\[ \Rightarrow 9y^2 + 6y + 1 + 4(4y^2 - 2y - 4y + 2) \ge 0\]\[\Rightarrow 25y^2 -18y + 9 \ge 0 \]This inequality is always true, so \(y \in \mathbb R\).
ParthKohli
  • ParthKohli
Oh, and \(3/5, 1 \) won't be included.
UsukiDoll
  • UsukiDoll
oh cool @ParthKohli
UsukiDoll
  • UsukiDoll
@ParthKohli one question how did you get \[(3y+1)^2 + 4(4y-2)(y-1) \ge 0 \]
Michele_Laino
  • Michele_Laino
since 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
  • ParthKohli
OK, let me try that out again... we want to check the values of \(y\) where the quadratic equation\[(y-1)x^2 - (3y + 1) x - (4y - 2) = 0 \]has real solutions.\[\rm discriminant \ge 0 \tag{for real solutions}\]
Michele_Laino
  • Michele_Laino
whereas the domain, is: \[\Large domain = \left\{ {\forall y \in {\mathbf{R}}} \right\}\]
ParthKohli
  • ParthKohli
...yes.
UsukiDoll
  • UsukiDoll
OH THAT IS COOL! I AM SO USING THAT! Thanks @ParthKohli
ParthKohli
  • ParthKohli
:)
arindameducationusc
  • arindameducationusc
Thank you @Michele_Laino so in terms of pi? But in my book the answer is [pi/3,pi/2]
Michele_Laino
  • Michele_Laino
I'm pondering...
arindameducationusc
  • arindameducationusc
range | sin^(-1)(sqrt(x^2+x+1))
UsukiDoll
  • UsukiDoll
in degree mode [60 degrees , 90 degrees] hmmm...
arindameducationusc
  • arindameducationusc
I searched wolfram... answer is coming.. but if any of you guys can give an explanation it would be awesome
Michele_Laino
  • Michele_Laino
sorry my answer above is an error, here is the right answer: \[\Large {x^2} + x + 1 = {\left( {\arcsin y} \right)^2}\]
Michele_Laino
  • Michele_Laino
range is such that: \[\Large 0 \leqslant \sqrt {{x^2} + x + 1} \]
Michele_Laino
  • Michele_Laino
we have this: \[\Large {x^2} + x + 1 \geqslant \frac{3}{4}\]
Michele_Laino
  • Michele_Laino
we can try to solve this condition: \[\Large \frac{{\sqrt 3 }}{2} \leqslant \sqrt {{x^2} + x + 1} \leqslant 2\pi \]
arindameducationusc
  • arindameducationusc
hmmm... analyzing
arindameducationusc
  • arindameducationusc
nope, I am not getting. I will think more...

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