Whats the domain and range for f(x)= x^2+x-2/x^2-3x-4

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Whats the domain and range for f(x)= x^2+x-2/x^2-3x-4

Mathematics
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Well for domain, the roots of the denominator should not be zero.
Is the domain all real numbers except for -1, and 4?
and is the range 1?

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

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