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Find the range of \(\Huge \frac {3x^2-2x-1}{x^2+x+2}\) if x is real.

Mathematics
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Is calculus available as a tool for this?
I dun think it need calculus.....
No, but it would be quite straightforward if you used it.

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Other answers:

but this is a 4 mark question and i need the steps to get the mark...... if use calculus i dun think i can get full mark for this question.... (at least 1 mark 1 step)
okay, well that sounds like there's a particular way you're supposed to do it. What class is it?
Polynomial/
I'm not sure what you mean by that. Is it an algebra class or something? And does it ask you to complete a series of 4 steps?
somtething like: http://www.stpmwiki.com/index.php/Polynomials_Part2
and http://www.stpmwiki.com/index.php/Polynomials_Part1
whoa whoa whoa. Are you looking for a range or a remainder?
i duno which part can be use to find the range :( but i know the syllabus is just in that 2 webpage
I say again, just to clarify... you're looking for the range of the function? Which means all of the possible values it could take? Or you're looking for the remainder of the fraction once you've divided?
find the range for the fraction possible
lets say, \(\frac {3x^2-2x-1}{x^2+x+2} = k \) => \(3x^2-2x-1 = k(x^2+x+2)\) \((3-k)x^2 + (-2-k)x + (-1-2k) = 0\)
Thank you veryyyy much!!
since, x is real, \(b^2 - 4ac >= 0 \) => \((2+k)^2 - 4(k-3)(1+2k) >= 0\) \((k-4)(k+4/7) >= 0\)
np :)
I like that! I've never done it that way before. @cwtan I'm somewhat surprised that this is a viable technique for you in this instance given what I read in your syllabus, but I suppose if it works it works.
yea this is fallback option when calculus is not allowed.... mukushla taught me this few few days back... :)

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