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ChrisV

Solve. f(x)=2x^2-1/x ; x=sqrt2/2

  • 2 years ago
  • 2 years ago

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  1. ChrisV
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    \[2x^2-1/\sqrt{x} ; x=\sqrt{2/2}\]

    • 2 years ago
  2. Ishaan94
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    1 -sqrt2

    • 2 years ago
  3. Ishaan94
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    that makes x=1

    • 2 years ago
  4. ChrisV
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    oops\[x=\sqrt{2}/2\]

    • 2 years ago
  5. Ishaan94
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    Oh okay, then I think I must be right.

    • 2 years ago
  6. ChrisV
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    well the book says the answer is \[2\sqrt{2}\]

    • 2 years ago
  7. ChrisV
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    but i dont get that

    • 2 years ago
  8. Diyadiya
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    \[2( \frac{1}{\sqrt{2}})^2- \frac{1}{1/\sqrt{2}}=1-\sqrt{2}\]

    • 2 years ago
  9. ChrisV
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    well would help if i gave you guys the right equation

    • 2 years ago
  10. ChrisV
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    \[2x^2-1/x\]

    • 2 years ago
  11. Diyadiya
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    Yes still the answer is 1- sqqrt2

    • 2 years ago
  12. ChrisV
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    \[(2(\sqrt{2}/2)^2-1)/\sqrt{2}/2\]

    • 2 years ago
  13. ChrisV
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    thats what i need solved

    • 2 years ago
  14. ChrisV
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    \[(4(\sqrt{2}/2)-2)/\sqrt{2}\]

    • 2 years ago
  15. ChrisV
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    oops squared

    • 2 years ago
  16. ChrisV
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    \[4(2/4)-2/\sqrt{2}\]

    • 2 years ago
  17. ChrisV
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    which comes out to 0 for me

    • 2 years ago
  18. ChrisV
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    so i assume i did something wrong

    • 2 years ago
  19. Diyadiya
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    oh ok \[\frac{(2x)^2-1}{x}\]\[\frac{(2 \times \frac{1}{\sqrt{2}})^2-1}{ \frac{1}{\sqrt{2}}}\]\[(4 \times 1/2 \ -1)\sqrt{2}\]\[=\sqrt{2}\]

    • 2 years ago
  20. ChrisV
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    its only the x squared

    • 2 years ago
  21. ChrisV
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    not 2x

    • 2 years ago
  22. Diyadiya
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    \[\frac{2(x)^2-1}{x}\]is this the equation?

    • 2 years ago
  23. ChrisV
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    yes

    • 2 years ago
  24. Diyadiya
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    It can't be because it gives zero unless zero is the answer

    • 2 years ago
  25. ChrisV
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    thats what i was thinking

    • 2 years ago
  26. ChrisV
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    but the relative minimum given by the book is

    • 2 years ago
  27. ChrisV
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    \[(\sqrt{2}/2, 2\sqrt{2})\]

    • 2 years ago
  28. ChrisV
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    which means the f(sqrt2/2) somehow is not 0

    • 2 years ago
  29. Diyadiya
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    Wait \[x=\frac{\sqrt{2}}{2} \ or \ \frac{2}{\sqrt{2}}\]

    • 2 years ago
  30. ChrisV
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    was 1/sqrt2 but rationalized is sqrt2/2

    • 2 years ago
  31. Diyadiya
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    i'm getting the same answer sorry! once again check your question

    • 2 years ago
  32. ChrisV
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    ok ill go through the entire problem with you

    • 2 years ago
  33. Diyadiya
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    ok

    • 2 years ago
  34. Diyadiya
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    wait

    • 2 years ago
  35. ChrisV
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    i have to find the critical numbers of f(x)=2x+1/x using the first derivative test

    • 2 years ago
  36. Diyadiya
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    Ok then its different

    • 2 years ago
  37. ChrisV
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    so first critical number is 0 because f(x) is undefinedat 0

    • 2 years ago
  38. ChrisV
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    then i find the derivative

    • 2 years ago
  39. Diyadiya
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    Wait

    • 2 years ago
  40. Diyadiya
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    i dont know much about this ..relative minimum

    • 2 years ago
  41. ChrisV
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    oh im sure i got this part right

    • 2 years ago
  42. ChrisV
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    2x/=1/x=(2x^2+1)/x

    • 2 years ago
  43. ChrisV
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    first derivative

    • 2 years ago
  44. ChrisV
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    (x(4x)-(2x^2+1))/X^2

    • 2 years ago
  45. ChrisV
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    (4x^2-2x*2-1)/x^2

    • 2 years ago
  46. ChrisV
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    (2x^2-1)/x^2

    • 2 years ago
  47. ChrisV
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    thats the first derivative

    • 2 years ago
  48. ChrisV
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    so 2x-1=0

    • 2 years ago
  49. ChrisV
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    2x^2-1=0 oops

    • 2 years ago
  50. ChrisV
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    x^2=1/2

    • 2 years ago
  51. ChrisV
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    x=+/-1/sqrt2

    • 2 years ago
  52. ChrisV
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    so critical numbers = 0,-sqrt2/2, and sqrt2/2

    • 2 years ago
  53. ChrisV
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    0 is a vertical asymtope so its not a min or max

    • 2 years ago
  54. ChrisV
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    but by the first derivative test

    • 2 years ago
  55. ChrisV
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    i can see the interval ( -infinity, -sqrt2/2) is increasing which tells me -sqrt2/2 is the max

    • 2 years ago
  56. Diyadiya
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    i don't understand this ..sorry But derivative of 2x+1/x=(2x^2-1)/x^2 right?

    • 2 years ago
  57. ChrisV
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    the interval (sqrt2/2, infinty) is decreasing which means minimum

    • 2 years ago
  58. ChrisV
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    yes it is

    • 2 years ago
  59. ChrisV
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    but since crit numbers = 0 or undefined

    • 2 years ago
  60. ChrisV
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    you set the numerator to 0 to find crit numbers

    • 2 years ago
  61. ChrisV
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    and since x=0 would make the denominator 0 that would be undefined

    • 2 years ago
  62. ChrisV
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    hince the 3 critical numbers

    • 2 years ago
  63. ChrisV
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    so this far i am correct

    • 2 years ago
  64. ChrisV
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    now i need the y values for my max and min

    • 2 years ago
  65. ChrisV
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    which means i plug my crit numbers into f(x)

    • 2 years ago
  66. ChrisV
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    so i have to solve for f(sqrt2/2) and f(-sqrt2/2)

    • 2 years ago
  67. ChrisV
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    f(x)= 2x+1/x

    • 2 years ago
  68. Diyadiya
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    Ok so\[2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\]

    • 2 years ago
  69. ChrisV
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    which equals (2x^2-1)/x

    • 2 years ago
  70. ChrisV
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    well common denominator would be x

    • 2 years ago
  71. ChrisV
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    so 2x(x)/x +1/x

    • 2 years ago
  72. ChrisV
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    (2x^2+1)/x

    • 2 years ago
  73. ChrisV
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    now plug in \[x=\sqrt{2}/2; x=-\sqrt{2}/2\]

    • 2 years ago
  74. ChrisV
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    and the answers in the book i verified via cramster and calcchat.com

    • 2 years ago
  75. Diyadiya
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    But in your question f(x) was \[2x+ \frac{1}{x}\]thats how you found derivative to be \[\frac{2x^2-1}{x^2}\] i'm sorry if i'm wrong i have no idea about this i'm just trying

    • 2 years ago
  76. ChrisV
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    no those two are equal

    • 2 years ago
  77. ChrisV
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    that just combining via common denominator

    • 2 years ago
  78. ChrisV
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    oh wait that is the derivative

    • 2 years ago
  79. ChrisV
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    but we plug the crit numbers back into f(x) to find y

    • 2 years ago
  80. ChrisV
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    not the first derivative

    • 2 years ago
  81. Diyadiya
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    Yes when we plug into f(x)\[2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\]

    • 2 years ago
  82. Callisto
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    I'm so confused... Is f(x) = (2x^2 -1)/x or f(x) = (2x^2 +1)/x or f(x) = 2x^2 + (1/x) or f(x) = 2x^2 - (1/x)?

    • 2 years ago
  83. ChrisV
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    and f(x) = (2x^2+1)/x

    • 2 years ago
  84. ChrisV
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    or 2x+1/x

    • 2 years ago
  85. ChrisV
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    both of those are the same

    • 2 years ago
  86. Diyadiya
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    ok so when we plug in 1/sqrt2 to f(x) we get 2sqrt2

    • 2 years ago
  87. ChrisV
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    ok

    • 2 years ago
  88. ChrisV
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    2(1/sqrt2)+1/1/sqrt2

    • 2 years ago
  89. ChrisV
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    2/2sqrt +sqrt2

    • 2 years ago
  90. ChrisV
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    2sqrt2/2+sqrt2

    • 2 years ago
  91. ChrisV
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    =2sqrt2

    • 2 years ago
  92. ChrisV
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    thats right thanks

    • 2 years ago
  93. Diyadiya
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    welcome :)

    • 2 years ago
  94. ChrisV
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    and now the negative

    • 2 years ago
  95. ChrisV
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    must equal -2sqrt2

    • 2 years ago
  96. Callisto
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    It's so messy here ... I dunno if this is sth you want 'cause i don't even know the question

    • 2 years ago
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  97. Callisto
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    (Nah.. seems problem's solved)

    • 2 years ago
  98. ChrisV
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    so

    • 2 years ago
  99. Diyadiya
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    yup -2sqrt2

    • 2 years ago
  100. ChrisV
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    2(-1/sqrt2) +1/-1/sqrt2

    • 2 years ago
  101. Diyadiya
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    @Callisto has given you the entire solution :)

    • 2 years ago
  102. ChrisV
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    -2/sqrt2 -sqrt2

    • 2 years ago
  103. ChrisV
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    2sqrt2/2 - sqrt2

    • 2 years ago
  104. ChrisV
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    -2sqrt2

    • 2 years ago
  105. ChrisV
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    yea just making sure i understand

    • 2 years ago
  106. Diyadiya
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    yeah :)

    • 2 years ago
  107. ChrisV
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    i have a test in a couple of days

    • 2 years ago
  108. ChrisV
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    i have a test in a couple of days

    • 2 years ago
  109. ChrisV
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    not the kind of person to have people do my homework for me :)

    • 2 years ago
  110. ChrisV
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    you guys were a great help thanks

    • 2 years ago
  111. Diyadiya
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    glad to help !

    • 2 years ago
  112. Callisto
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    lol that 'solution' serves as a reference only.. I don' t even know if it is correct, BTW, Good luck with your test :)

    • 2 years ago
  113. Diyadiya
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    Yeah Goodluck!

    • 2 years ago
  114. ChrisV
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    no problem, i am sure what i got now is correct. thanks guys

    • 2 years ago
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