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anonymous

  • one year ago

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

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  1. arindameducationusc
    • one year ago
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    Well for domain, the roots of the denominator should not be zero.

  2. anonymous
    • one year ago
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    Is the domain all real numbers except for -1, and 4?

  3. anonymous
    • one year ago
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    and is the range 1?

  4. anonymous
    • one year ago
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    so the numerator and denominator are 2 different domains?

  5. UsukiDoll
    • one year ago
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    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.

  6. UsukiDoll
    • one year ago
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    \[(x^2-3x-4) \rightarrow (x+1)(x-4)\] 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__

  7. UsukiDoll
    • one year ago
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    that doesn't sound right at all ^

  8. UsukiDoll
    • one year ago
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    @Michele_Laino

  9. anonymous
    • one year ago
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    Thanks!

  10. arindameducationusc
    • one year ago
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    Yes Yes, I did a calculation mistake, @Abbs_ is right

  11. anonymous
    • one year ago
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    It's okay. :)

  12. UsukiDoll
    • one year ago
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    range is going to be a little bit harder though...

  13. anonymous
    • one year ago
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    Is it all real numbers except y=1?

  14. UsukiDoll
    • one year ago
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    I gotta be honest. I have to graph that function first.. and desmos hates me and my fractions

  15. Michele_Laino
    • one year ago
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    I agree with the answer of @UsukiDoll

  16. UsukiDoll
    • one year ago
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    thanks, HEY YOUR QH symbol is back @Michele_Laino !

  17. Michele_Laino
    • one year ago
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    yes! I see nevertheless not in mathematics subsections, thanks! @UsukiDoll

  18. UsukiDoll
    • one year ago
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  19. UsukiDoll
    • one year ago
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    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

  20. Michele_Laino
    • one year ago
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    at points where the function is undefined, generally that function behaves as an infinity, and that is what is happening into your graph

  21. UsukiDoll
    • one year ago
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    so it's all reals for the range?

  22. UsukiDoll
    • one year ago
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    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.

  23. Michele_Laino
    • one year ago
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    the range, as in this case, is always a real set, nevertheless is not a limited set

  24. UsukiDoll
    • one year ago
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    because it goes on forever?

  25. Michele_Laino
    • one year ago
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    yes!

  26. UsukiDoll
    • one year ago
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    |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.

  27. Michele_Laino
    • one year ago
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    towards down or towards up

  28. arindameducationusc
    • one year ago
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    yes, range should be -1 to 1

  29. arindameducationusc
    • one year ago
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    y-axis represents Range

  30. Michele_Laino
    • one year ago
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    your function is a circular function, so its domain is the subsequent set: \[\Large [0,2\pi )\]

  31. UsukiDoll
    • one year ago
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    *I'm not going for trig... just regular domain and range*

  32. Michele_Laino
    • one year ago
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    in order to establish the range, we need to know the exact algebraic shape of your function

  33. arindameducationusc
    • one year ago
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    One second... A circle is not a function... right?

  34. arindameducationusc
    • one year ago
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    As intersects two times along x-axis.

  35. UsukiDoll
    • one year ago
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    it's not a function.. fails the vertical line test

  36. Empty
    • one year ago
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    Depends on what coordinate system you're using.

  37. UsukiDoll
    • one year ago
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    not 1 to 1 either.. fails horizontal line test.

  38. Michele_Laino
    • one year ago
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    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

  39. arindameducationusc
    • one year ago
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    So, how come range will be there. Not possible

  40. Michele_Laino
    • one year ago
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    there is not meaning in speaking about range, since we have not a function

  41. Empty
    • one year ago
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    $$r(\theta) = 1$$ is a function for the domain \([0,2 \pi) \) that is a circle. Vertical line test is for boxes not circles :P

  42. Michele_Laino
    • one year ago
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    here is why we hav not a function: |dw:1437991311333:dw|

  43. UsukiDoll
    • one year ago
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    actually the vertical line test can be used for circles. and yeah I know that.

  44. ParthKohli
    • one year ago
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    Guys, I have a neat way to find the range...

  45. arindameducationusc
    • one year ago
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    ok, try finding range of inverse of sin(sqrt(x^2+x+1))

  46. ParthKohli
    • one year ago
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    \[y = \frac{x^2 + x - 2}{x^2 - 3x - 4}\]\[\Rightarrow yx^2 - 3y x - 4y = x^2 + x - 2\]\[\Rightarrow (y-1)x^2 -(3y+1)x - (4y+2) = 0\]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 \(y \in \mathbb R\).

  47. ParthKohli
    • one year ago
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    Oh, and \(3/5, 1 \) won't be included.

  48. UsukiDoll
    • one year ago
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    oh cool @ParthKohli

  49. UsukiDoll
    • one year ago
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    @ParthKohli one question how did you get \[(3y+1)^2 + 4(4y-2)(y-1) \ge 0 \]

  50. Michele_Laino
    • one year ago
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    since we have: \[\Large {x^2} + x + 1 = {\left( {\sin y} \right)^2}\] the range is the subsequent set: \[\Large \begin{gathered} range = \left\{ {x \in {\mathbf{R}}\;{\text{such that:}}\;0 \leqslant {x^2} + x + 1 \leqslant 1} \right\} = \hfill \\ \hfill \\ = \left\{ {x \in {\mathbf{R}}\;{\text{such that:}}\; - 1 \leqslant x \leqslant 0} \right\} \hfill \\ \end{gathered} \]

  51. ParthKohli
    • one year ago
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    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.\[\rm discriminant \ge 0 \tag{for real solutions}\]

  52. Michele_Laino
    • one year ago
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    whereas the domain, is: \[\Large domain = \left\{ {\forall y \in {\mathbf{R}}} \right\}\]

  53. ParthKohli
    • one year ago
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    ...yes.

  54. UsukiDoll
    • one year ago
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    OH THAT IS COOL! I AM SO USING THAT! Thanks @ParthKohli

  55. ParthKohli
    • one year ago
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    :)

  56. arindameducationusc
    • one year ago
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    Thank you @Michele_Laino so in terms of pi? But in my book the answer is [pi/3,pi/2]

  57. Michele_Laino
    • one year ago
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    I'm pondering...

  58. arindameducationusc
    • one year ago
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    range | sin^(-1)(sqrt(x^2+x+1))

  59. UsukiDoll
    • one year ago
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    in degree mode [60 degrees , 90 degrees] hmmm...

  60. arindameducationusc
    • one year ago
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    I searched wolfram... answer is coming.. but if any of you guys can give an explanation it would be awesome

  61. Michele_Laino
    • one year ago
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    sorry my answer above is an error, here is the right answer: \[\Large {x^2} + x + 1 = {\left( {\arcsin y} \right)^2}\]

  62. Michele_Laino
    • one year ago
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    range is such that: \[\Large 0 \leqslant \sqrt {{x^2} + x + 1} \]

  63. Michele_Laino
    • one year ago
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    we have this: \[\Large {x^2} + x + 1 \geqslant \frac{3}{4}\]

  64. Michele_Laino
    • one year ago
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    we can try to solve this condition: \[\Large \frac{{\sqrt 3 }}{2} \leqslant \sqrt {{x^2} + x + 1} \leqslant 2\pi \]

  65. arindameducationusc
    • one year ago
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    hmmm... analyzing

  66. arindameducationusc
    • one year ago
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    nope, I am not getting. I will think more...

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