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cwtan

  • 2 years ago

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

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  1. Jemurray3
    • 2 years ago
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    Is calculus available as a tool for this?

  2. cwtan
    • 2 years ago
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    I dun think it need calculus.....

  3. Jemurray3
    • 2 years ago
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    No, but it would be quite straightforward if you used it.

  4. cwtan
    • 2 years ago
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    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)

  5. Jemurray3
    • 2 years ago
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    okay, well that sounds like there's a particular way you're supposed to do it. What class is it?

  6. cwtan
    • 2 years ago
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    Polynomial/

  7. Jemurray3
    • 2 years ago
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    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?

  8. cwtan
    • 2 years ago
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    somtething like: http://www.stpmwiki.com/index.php/Polynomials_Part2

  9. cwtan
    • 2 years ago
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    and http://www.stpmwiki.com/index.php/Polynomials_Part1

  10. Jemurray3
    • 2 years ago
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    whoa whoa whoa. Are you looking for a range or a remainder?

  11. cwtan
    • 2 years ago
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    i duno which part can be use to find the range :( but i know the syllabus is just in that 2 webpage

  12. Jemurray3
    • 2 years ago
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    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?

  13. cwtan
    • 2 years ago
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    find the range for the fraction possible

  14. ganeshie8
    • 2 years ago
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    lets say, \(\frac {3x^2-2x-1}{x^2+x+2} = k \) => \(3x^2-2x-1 = k(x^2+x+2)\) \((3-k)x^2 + (-2-k)x + (-1-2k) = 0\)

  15. cwtan
    • 2 years ago
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    Thank you veryyyy much!!

  16. ganeshie8
    • 2 years ago
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    since, x is real, \(b^2 - 4ac >= 0 \) => \((2+k)^2 - 4(k-3)(1+2k) >= 0\) \((k-4)(k+4/7) >= 0\)

  17. ganeshie8
    • 2 years ago
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    np :)

  18. Jemurray3
    • 2 years ago
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    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.

  19. ganeshie8
    • 2 years ago
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    yea this is fallback option when calculus is not allowed.... mukushla taught me this few few days back... :)

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