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mathmath333

  • one year ago

Find the minimum and maximum value of

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} \dfrac{x^2-x+1}{x^2+x+1},\ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}\)

  2. anonymous
    • one year ago
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    It's hard to explain...but I will see if I can show the work.

  3. anonymous
    • one year ago
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    are you talking local or absolute maxes and mins?

  4. mathmath333
    • one year ago
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    i would also prefer a way without calculus ( idk calculus)

  5. anonymous
    • one year ago
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    Do you need to show your work for this problem?

  6. mathmath333
    • one year ago
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    not necessary i should understand the idea

  7. anonymous
    • one year ago
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    Based on what I know algebraically , I got a minimum of (1, 1/3) and a maximum of (-1,3).

  8. mathmath333
    • one year ago
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    u calculated the co-ordinates where as it should be the numerical value

  9. anonymous
    • one year ago
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    Ok. Let me try a different way.

  10. anonymous
    • one year ago
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    Ok. I got the minimum is 1/3, and the maximum is -3.

  11. anonymous
    • one year ago
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    oops +3. not neg

  12. mathmath333
    • one year ago
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    how did u calculate

  13. anonymous
    • one year ago
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    \[\frac{x^2 - x + 1 }{ x^2 + x + 1 } = 3, when (x) = -1\] I got this after simplifying to figure out what x = for the numerator

  14. anonymous
    • one year ago
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    Same for the denominator, except it is when x=1

  15. anonymous
    • one year ago
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    It's not the way I was taught, but my aunt is a math teacher, so she taught me her way of solving these pesky sorts

  16. xapproachesinfinity
    • one year ago
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    eh i like your questions lol

  17. xapproachesinfinity
    • one year ago
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    make me struggle :)

  18. mathmath333
    • one year ago
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    i have tons of question but no time

  19. xapproachesinfinity
    • one year ago
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    i cheated haha min=1/3<y<3 let's how we get to this

  20. mathmath333
    • one year ago
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    one way seems to put random \(x=\pm 0, \pm 1,\pm 2, \pm 3\) but doesnt work always http://www.wolframalpha.com/input/?i=table%5Bf%28x%29%3D%5Cdfrac%7Bx%5E2-x%2B1%7D%7Bx%5E2%2Bx%2B1%7D%2C%7Bx%2C-10%2C10%7D%5D

  21. Empty
    • one year ago
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    Honestly just graphing this is the best way to do this lol.

  22. Empty
    • one year ago
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    That's why Descartes invented graphing in the first place, to make stuff like this easier. And then we got computers, lol.

  23. xapproachesinfinity
    • one year ago
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    i suggest that we study \[y=1+\frac{2x}{x^2+x+1} ~~x\in \mathbb{R}\]

  24. xapproachesinfinity
    • one year ago
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    minus not plus mistake haha

  25. xapproachesinfinity
    • one year ago
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    \[y=1-\frac{2x}{x^2+x+1}\]

  26. xapproachesinfinity
    • one year ago
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    hmm need to find upper and lower bound m<y<M some how i think that finding the range of x for 2x/x^2+x+1 might be our way!!!

  27. xapproachesinfinity
    • one year ago
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    I'm just throwing stuff as i go lol

  28. xapproachesinfinity
    • one year ago
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    x^2+x+1=0 has no real solution yes? so this parabola is >0

  29. Empty
    • one year ago
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    We can factor it and for just simplicity of writing I'll make this: \[\omega = e^{i 2 \pi / 3}\] \[\frac{x^2-x+1}{x^2+x+1}=\frac{(x+\omega)(x+\omega^2)}{(x-\omega)(x-\omega^2)}\]

  30. xapproachesinfinity
    • one year ago
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    y1=2x/x^2+x+1 can we find the range for this going by what i just did?

  31. anonymous
    • one year ago
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    let \(x=\tan z\) with \(z \in (-\frac{\pi}{2}, \frac{\pi}{2})\) you will get\[f(\tan z)=\frac{\tan^2 z -\tan z+1}{\tan^2 z +\tan z+1}=\frac{4}{2+\sin (2z)}-1\]

  32. xapproachesinfinity
    • one year ago
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    hmm interesting @mukushla

  33. xapproachesinfinity
    • one year ago
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    i feel that my way can be a lot easier haha though don't you think

  34. xapproachesinfinity
    • one year ago
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    just need to find the range for 2x/(x^2+x+1)?

  35. anonymous
    • one year ago
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    yeah, making a perfect square in denom, that's a lot better than mine

  36. xapproachesinfinity
    • one year ago
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    need pen and paper!

  37. mathmath333
    • one year ago
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    i like that tan substitution looks something creative

  38. anonymous
    • one year ago
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    let me continue with yours\[f(x)=1-\frac{2x}{x^2+x+1}=1-\frac{2}{x+\frac{1}{x}+1}\]actually there is no need to make perfect square

  39. anonymous
    • one year ago
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    we just need to note that\[\left|x+\frac{1}{x} \right| \ge 2 \]

  40. xapproachesinfinity
    • one year ago
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    hmm seems you a different approach

  41. xapproachesinfinity
    • one year ago
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    i was trying to use parabola and eyeball the range

  42. xapproachesinfinity
    • one year ago
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    how is |x+1/x|>=2

  43. xapproachesinfinity
    • one year ago
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    first time i see that :)

  44. anonymous
    • one year ago
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    AM-GM inequality\[x+\frac{1}{x} \ge 2\sqrt{x.\frac{1}{x}}\]

  45. anonymous
    • one year ago
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    if \(x<0\) it will be\[x+\frac{1}{x} \le -2\]

  46. xapproachesinfinity
    • one year ago
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    hmm i see

  47. anonymous
    • one year ago
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    equality occurs when x=-1 and x=1 for two cases

  48. xapproachesinfinity
    • one year ago
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    so for x=-1 we have x+1/x<-2==> x+1/x+1<-1 then y<3

  49. xapproachesinfinity
    • one year ago
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    for x=1 by the same reasoning 1/3<y

  50. xapproachesinfinity
    • one year ago
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    excellent @mukushla

  51. Empty
    • one year ago
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    That's a nice trick with that AM-GM inequality.

  52. xapproachesinfinity
    • one year ago
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    beautiful indeed :)

  53. anonymous
    • one year ago
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    @mathmath333 ty, actually it is a nice trick to use trig sub for finding range, you may try this one\[f(x)=\sqrt{x-2}+\sqrt{4-x}\]

  54. mathmath333
    • one year ago
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    thats incorrect

  55. xapproachesinfinity
    • one year ago
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    hmm if one realize they can use trig sub the problem becomes really easy since sin and cos are easily bounded

  56. anonymous
    • one year ago
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    that's right

  57. mathmath333
    • one year ago
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    the range of x here is \(2<x<4\)

  58. anonymous
    • one year ago
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    when you wanna do a trig sub, find the domain, it will give you the clue for the form of sub

  59. xapproachesinfinity
    • one year ago
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    well domain above is R how can that help

  60. anonymous
    • one year ago
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    tan covers R

  61. xapproachesinfinity
    • one year ago
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    but tan has pi/2 singularity?

  62. xapproachesinfinity
    • one year ago
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    does not matter ?

  63. anonymous
    • one year ago
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    no we will treat bounds in limit, so it doesn't matter

  64. anonymous
    • one year ago
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    now, what about the second function?

  65. anonymous
    • one year ago
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    note that\[2 \le x \le4\]is a restricted domain

  66. xapproachesinfinity
    • one year ago
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    yeah just about to say that math did it already

  67. xapproachesinfinity
    • one year ago
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    so what sub is that telling us to use?

  68. xapproachesinfinity
    • one year ago
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    sin or cos ?

  69. xapproachesinfinity
    • one year ago
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    hold on don't tell me yet

  70. anonymous
    • one year ago
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    both of then can be

  71. anonymous
    • one year ago
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    you know why?

  72. xapproachesinfinity
    • one year ago
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    hmm no really because they are both have same bounds -1 and 1

  73. anonymous
    • one year ago
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    yes and because\[ -1\le x-3 \le 1\]

  74. xapproachesinfinity
    • one year ago
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    oh i see

  75. Empty
    • one year ago
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    They're both the same function just shifted argument by 90 degrees. \(\cos x + 3\) will work just as well as \(\sin x + 3\)

  76. anonymous
    • one year ago
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    quite right

  77. xapproachesinfinity
    • one year ago
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    so in that we put x-3=sinz yes?

  78. xapproachesinfinity
    • one year ago
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    damn i feel weak in trig lol :)

  79. anonymous
    • one year ago
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    oh yeah, \(3+\cos z\) will be better maybe

  80. xapproachesinfinity
    • one year ago
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    hmm you said they all work fine No? okay lets go with x=3+cosz

  81. anonymous
    • one year ago
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    well derive with cos will be easier

  82. xapproachesinfinity
    • one year ago
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    so \[y=\sqrt{\cos z+1}+\sqrt{1-\cos z}\]

  83. xapproachesinfinity
    • one year ago
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    so \[0\leq y \leq 2\sqrt{2}\]

  84. anonymous
    • one year ago
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    oh no you can't just plug in there values, your trig function is not alone, you'll need more simplification

  85. xapproachesinfinity
    • one year ago
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    if we simplify we just get radical 2?

  86. xapproachesinfinity
    • one year ago
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    if i didn't make an error that is

  87. anonymous
    • one year ago
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    \[f=\sqrt{2} \left( \left|\sin \frac{z}{2} \right|+\left|\cos \frac{z}{2} \right| \right)\]

  88. xapproachesinfinity
    • one year ago
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    cosz=2cos^2x/2-1 oh i see i threw the square when i did my simplification haha

  89. xapproachesinfinity
    • one year ago
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    i didn't i mean*

  90. anonymous
    • one year ago
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    ok, good discussion, tnx guys

  91. xapproachesinfinity
    • one year ago
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    well isn't that the same result above? 0<y<2root2

  92. xapproachesinfinity
    • one year ago
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    thanks to you learned bunch of stuff here :)

  93. xapproachesinfinity
    • one year ago
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    this one i never remember \[\frac{a+b}{2}\leq \sqrt{ab}\]

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