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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- arindameducationusc

Well for domain, the roots of the denominator should not be zero.

- anonymous

Is the domain all real numbers except for -1, and 4?

- anonymous

and is the range 1?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

so the numerator and denominator are 2 different domains?

- 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

\[(x^2-3x-4) \rightarrow (x+1)(x-4)\]
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

that doesn't sound right at all ^

- UsukiDoll

@Michele_Laino

- anonymous

Thanks!

- arindameducationusc

Yes Yes, I did a calculation mistake, @Abbs_ is right

- anonymous

It's okay. :)

- UsukiDoll

range is going to be a little bit harder though...

- anonymous

Is it all real numbers except y=1?

- UsukiDoll

I gotta be honest. I have to graph that function first.. and desmos hates me and my fractions

- Michele_Laino

I agree with the answer of @UsukiDoll

- UsukiDoll

thanks, HEY YOUR QH symbol is back @Michele_Laino !

- Michele_Laino

yes! I see nevertheless not in mathematics subsections, thanks! @UsukiDoll

- UsukiDoll

##### 1 Attachment

- 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

at points where the function is undefined, generally that function behaves as an infinity, and that is what is happening into your graph

- UsukiDoll

so it's all reals for the range?

- 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

the range, as in this case, is always a real set, nevertheless is not a limited set

- UsukiDoll

because it goes on forever?

- Michele_Laino

yes!

- 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

towards down or towards up

- arindameducationusc

yes, range should be -1 to 1

- arindameducationusc

y-axis represents Range

- Michele_Laino

your function is a circular function, so its domain is the subsequent set:
\[\Large [0,2\pi )\]

- UsukiDoll

*I'm not going for trig... just regular domain and range*

- Michele_Laino

in order to establish the range, we need to know the exact algebraic shape of your function

- arindameducationusc

One second... A circle is not a function... right?

- arindameducationusc

As intersects two times along x-axis.

- UsukiDoll

it's not a function.. fails the vertical line test

- Empty

Depends on what coordinate system you're using.

- UsukiDoll

not 1 to 1 either.. fails horizontal line test.

- 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

So, how come range will be there. Not possible

- Michele_Laino

there is not meaning in speaking about range, since we have not a function

- 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

here is why we hav not a function:
|dw:1437991311333:dw|

- UsukiDoll

actually the vertical line test can be used for circles.
and yeah I know that.

- ParthKohli

Guys, I have a neat way to find the range...

- arindameducationusc

ok, try finding range of inverse of sin(sqrt(x^2+x+1))

- 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

Oh, and \(3/5, 1 \) won't be included.

- UsukiDoll

oh cool @ParthKohli

- UsukiDoll

@ParthKohli one question
how did you get
\[(3y+1)^2 + 4(4y-2)(y-1) \ge 0 \]

- 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

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

whereas the domain, is:
\[\Large domain = \left\{ {\forall y \in {\mathbf{R}}} \right\}\]

- ParthKohli

...yes.

- UsukiDoll

OH THAT IS COOL! I AM SO USING THAT! Thanks @ParthKohli

- ParthKohli

:)

- arindameducationusc

Thank you @Michele_Laino
so in terms of pi?
But in my book the answer is [pi/3,pi/2]

- Michele_Laino

I'm pondering...

- arindameducationusc

range | sin^(-1)(sqrt(x^2+x+1))

- UsukiDoll

in degree mode
[60 degrees , 90 degrees]
hmmm...

- arindameducationusc

I searched wolfram... answer is coming.. but if any of you guys can give an explanation it would be awesome

- 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

range is such that:
\[\Large 0 \leqslant \sqrt {{x^2} + x + 1} \]

- Michele_Laino

we have this:
\[\Large {x^2} + x + 1 \geqslant \frac{3}{4}\]

- Michele_Laino

we can try to solve this condition:
\[\Large \frac{{\sqrt 3 }}{2} \leqslant \sqrt {{x^2} + x + 1} \leqslant 2\pi \]

- arindameducationusc

hmmm... analyzing

- arindameducationusc

nope, I am not getting. I will think more...

Looking for something else?

Not the answer you are looking for? Search for more explanations.