cwtan
  • cwtan
Find the range of \(\Huge \frac {3x^2-2x-1}{x^2+x+2}\) if x is real.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Is calculus available as a tool for this?
cwtan
  • cwtan
I dun think it need calculus.....
anonymous
  • anonymous
No, but it would be quite straightforward if you used it.

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cwtan
  • cwtan
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)
anonymous
  • anonymous
okay, well that sounds like there's a particular way you're supposed to do it. What class is it?
cwtan
  • cwtan
Polynomial/
anonymous
  • anonymous
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?
cwtan
  • cwtan
somtething like: http://www.stpmwiki.com/index.php/Polynomials_Part2
cwtan
  • cwtan
and http://www.stpmwiki.com/index.php/Polynomials_Part1
anonymous
  • anonymous
whoa whoa whoa. Are you looking for a range or a remainder?
cwtan
  • cwtan
i duno which part can be use to find the range :( but i know the syllabus is just in that 2 webpage
anonymous
  • anonymous
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?
cwtan
  • cwtan
find the range for the fraction possible
ganeshie8
  • ganeshie8
lets say, \(\frac {3x^2-2x-1}{x^2+x+2} = k \) => \(3x^2-2x-1 = k(x^2+x+2)\) \((3-k)x^2 + (-2-k)x + (-1-2k) = 0\)
cwtan
  • cwtan
Thank you veryyyy much!!
ganeshie8
  • ganeshie8
since, x is real, \(b^2 - 4ac >= 0 \) => \((2+k)^2 - 4(k-3)(1+2k) >= 0\) \((k-4)(k+4/7) >= 0\)
ganeshie8
  • ganeshie8
np :)
anonymous
  • anonymous
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.
ganeshie8
  • ganeshie8
yea this is fallback option when calculus is not allowed.... mukushla taught me this few few days back... :)

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