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ChrisV

  • 2 years ago

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

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

  2. Ishaan94
    • 2 years ago
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    1 -sqrt2

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

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

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

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

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

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

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

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

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

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

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

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

  15. ChrisV
    • 2 years ago
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    oops squared

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

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

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

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

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

  21. ChrisV
    • 2 years ago
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    not 2x

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

  23. ChrisV
    • 2 years ago
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    yes

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

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

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

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

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

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

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

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

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

  33. Diyadiya
    • 2 years ago
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    ok

  34. Diyadiya
    • 2 years ago
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    wait

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

  36. Diyadiya
    • 2 years ago
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    Ok then its different

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

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

  39. Diyadiya
    • 2 years ago
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    Wait

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

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

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

  43. ChrisV
    • 2 years ago
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    first derivative

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

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

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

  47. ChrisV
    • 2 years ago
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    thats the first derivative

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

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

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

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

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

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

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

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

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

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

  58. ChrisV
    • 2 years ago
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    yes it is

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  75. Diyadiya
    • 2 years ago
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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

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

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

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

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

  80. ChrisV
    • 2 years ago
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    not the first derivative

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

  82. Callisto
    • 2 years ago
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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)?

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

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

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

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

  87. ChrisV
    • 2 years ago
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    ok

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

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

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

  91. ChrisV
    • 2 years ago
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    =2sqrt2

  92. ChrisV
    • 2 years ago
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    thats right thanks

  93. Diyadiya
    • 2 years ago
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    welcome :)

  94. ChrisV
    • 2 years ago
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    and now the negative

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

  96. Callisto
    • 2 years ago
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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

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

  98. ChrisV
    • 2 years ago
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    so

  99. Diyadiya
    • 2 years ago
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    yup -2sqrt2

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

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

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

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

  104. ChrisV
    • 2 years ago
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    -2sqrt2

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

  106. Diyadiya
    • 2 years ago
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    yeah :)

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

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

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

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

  111. Diyadiya
    • 2 years ago
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    glad to help !

  112. Callisto
    • 2 years ago
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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 :)

  113. Diyadiya
    • 2 years ago
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    Yeah Goodluck!

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

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