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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?

so the numerator and denominator are 2 different domains?

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

so it's all reals for the range?

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

because it goes on forever?

yes!

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.

So, how come range will be there. Not possible

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

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))

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

oh cool @ParthKohli

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

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...

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

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

hmmm... analyzing

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