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onegirl

  • 2 years ago

Check the hypotheses of Rolle’s Theorem and the Mean Value Theorem and find a value of c that makes the appropriate conclusion true. Illustrate the conclusion with a graph. f(x) = sin x , [0,pi/2]

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  1. onegirl
    • 2 years ago
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    @Mertsj can u help?

  2. onegirl
    • 2 years ago
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    @experimentX how would i do this one?

  3. experimentX
    • 2 years ago
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    let f(x) = sin(x) then f'(x) = ?

  4. onegirl
    • 2 years ago
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    okay

  5. onegirl
    • 2 years ago
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    cos(x)

  6. experimentX
    • 2 years ago
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    when is cos(x) zero?

  7. onegirl
    • 2 years ago
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    zero? you said find the derivative of sinx right?

  8. experimentX
    • 2 years ago
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    yes ... at what point is cos(x) equal to zero?

  9. onegirl
    • 2 years ago
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    there is not point where it is equal to zero

  10. experimentX
    • 2 years ago
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    why not?

  11. onegirl
    • 2 years ago
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    when i graph it...it doesn't

  12. experimentX
    • 2 years ago
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    you got the wrong graph then.

  13. onegirl
    • 2 years ago
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    |dw:1363197320465:dw| thats how it looks liek ohh okay

  14. experimentX
    • 2 years ago
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    |dw:1363197334937:dw|

  15. onegirl
    • 2 years ago
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    ok

  16. experimentX
    • 2 years ago
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    |dw:1363197397226:dw|

  17. onegirl
    • 2 years ago
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    yea it is

  18. onegirl
    • 2 years ago
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    okay i got it

  19. experimentX
    • 2 years ago
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    so ... what's the point?

  20. onegirl
    • 2 years ago
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    (0,1) ?

  21. experimentX
    • 2 years ago
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    no ...

  22. experimentX
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=plot+cos%28x%29+from+-2pi+to+2pi

  23. onegirl
    • 2 years ago
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    so its more than 1?

  24. onegirl
    • 2 years ago
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    0,1.5 ?

  25. experimentX
    • 2 years ago
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    NOOOOOOOOOOOOOOO http://www.wolframalpha.com/input/?i=solve+cos%28x%29+%3D+0+for+x+between+0+and+pi

  26. onegirl
    • 2 years ago
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    so its 1.57080?

  27. experimentX
    • 2 years ago
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    |dw:1363198158808:dw|

  28. onegirl
    • 2 years ago
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    ok

  29. experimentX
    • 2 years ago
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    so pick any two points above pi/2 and below pi/2

  30. onegirl
    • 2 years ago
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    okay? do you do this different than the las tone tho? i'm not gonna use f(a) - f(b)/ b - a ?

  31. experimentX
    • 2 years ago
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    |dw:1363198343133:dw|

  32. onegirl
    • 2 years ago
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    ohh ok

  33. experimentX
    • 2 years ago
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    |dw:1363198401986:dw|

  34. onegirl
    • 2 years ago
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    ok

  35. experimentX
    • 2 years ago
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    |dw:1363198546707:dw| do you get that?

  36. onegirl
    • 2 years ago
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    yes i do

  37. experimentX
    • 2 years ago
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    Okay i gotta go ...

  38. onegirl
    • 2 years ago
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    ok thx

  39. experimentX
    • 2 years ago
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    |dw:1363198676612:dw|

  40. onegirl
    • 2 years ago
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    okay

  41. onegirl
    • 2 years ago
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    So i filled in those a b c's and here i got....f(pi/2) = f(3pi/4) - f(pi/4) over 3pi/4 - pi/4

  42. onegirl
    • 2 years ago
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    f(3pi/4) = 3pi^2 + 1 = ?

  43. onegirl
    • 2 years ago
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    can u tell me if i plugged it right @experimentX ?

  44. experimentX
    • 2 years ago
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    yes that's correct.

  45. experimentX
    • 2 years ago
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    \[ f(3 \pi /4) = \sin(3\pi/4) = 1/ \sqrt 2 \] Find the value of f(pi/4) from your calculator or wolfram.

  46. onegirl
    • 2 years ago
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    okay

  47. onegirl
    • 2 years ago
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    I got 1

  48. experimentX
    • 2 years ago
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    huh .. do it again, what is sin(pi/4) ... plug into wolfram.

  49. onegirl
    • 2 years ago
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    hold on

  50. onegirl
    • 2 years ago
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    1/sqrt(2)

  51. onegirl
    • 2 years ago
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    i got 1 over sqrt of 2

  52. onegirl
    • 2 years ago
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    what do i do next ?

  53. experimentX
    • 2 years ago
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    yes ... plug into your Rolle's condition.

  54. onegirl
    • 2 years ago
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    the formula right?

  55. onegirl
    • 2 years ago
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    so f(1/sqrt(2) - f(1/sqrt(2)/ 1/sqrt(2) - 1/sqrt(2) ?

  56. experimentX
    • 2 years ago
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    no ...

  57. experimentX
    • 2 years ago
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    |dw:1363384214639:dw|

  58. onegirl
    • 2 years ago
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    ohh ok

  59. onegirl
    • 2 years ago
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    how do i solve that?

  60. onegirl
    • 2 years ago
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    so it will be 3/3 right?

  61. experimentX
    • 2 years ago
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    |dw:1363384415083:dw|

  62. onegirl
    • 2 years ago
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    okay

  63. onegirl
    • 2 years ago
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    so those will be the numerator?

  64. experimentX
    • 2 years ago
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    yes ... you just found out the value. Do you know what is functional notation?

  65. onegirl
    • 2 years ago
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    no

  66. experimentX
    • 2 years ago
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    yeah ... that is the problem. you haven't advanced to the point of doing this. let f(x) = cos(x), what is value of f(0)

  67. onegirl
    • 2 years ago
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    f(x) = 1

  68. experimentX
    • 2 years ago
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    yep ... if f(x) = sin(x) ,,, then what is f(pi/4) ??

  69. onegirl
    • 2 years ago
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    1 over sqrt(2)

  70. experimentX
    • 2 years ago
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    yes ... what is f(3pi/4) ??

  71. onegirl
    • 2 years ago
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    its the same 1 over sqrt(2)

  72. experimentX
    • 2 years ago
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    yes ... plug in those values in |dw:1363385976116:dw|

  73. onegirl
    • 2 years ago
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    plug them in the denominator?

  74. experimentX
    • 2 years ago
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    |dw:1363386098600:dw|

  75. onegirl
    • 2 years ago
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    ohh okay hold on

  76. onegirl
    • 2 years ago
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    f(1/sqrt(2) - f(1/sqrt(2) over 3pi/4 - pi/4

  77. experimentX
    • 2 years ago
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    No no ... what are you dong?? why are you putting 1/sqrt(2) inside function?

  78. onegirl
    • 2 years ago
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    you said to plug them in?

  79. experimentX
    • 2 years ago
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    yes .. |dw:1363386528821:dw|

  80. onegirl
    • 2 years ago
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    okay

  81. onegirl
    • 2 years ago
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    i'll ask my teacher on this one thanks for your help.

  82. experimentX
    • 2 years ago
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    |dw:1363387122187:dw|

  83. onegirl
    • 2 years ago
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    so i plug those in pi?

  84. experimentX
    • 2 years ago
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    no ... you just calculated those values. plug them in.

  85. onegirl
    • 2 years ago
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    I just did and you told me it was incorrect..i'm confused

  86. experimentX
    • 2 years ago
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    yes ... this is where you got confused.

  87. onegirl
    • 2 years ago
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    ok

  88. experimentX
    • 2 years ago
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    |dw:1363387985440:dw|

  89. onegirl
    • 2 years ago
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    ohh ok

  90. experimentX
    • 2 years ago
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    what is the value of|dw:1363388256685:dw|

  91. onegirl
    • 2 years ago
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    0?

  92. experimentX
    • 2 years ago
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    yes ... so |dw:1363388362361:dw|

  93. onegirl
    • 2 years ago
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    the derivative of it?

  94. experimentX
    • 2 years ago
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    yes ... is there a point where derivative is zero?

  95. onegirl
    • 2 years ago
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    yea

  96. experimentX
    • 2 years ago
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    what is that point?

  97. onegirl
    • 2 years ago
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    umm at 0, 0?

  98. experimentX
    • 2 years ago
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    No ..

  99. onegirl
    • 2 years ago
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    idk

  100. experimentX
    • 2 years ago
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    we talked about that in the above posts. check again ... the point is pi/2

  101. onegirl
    • 2 years ago
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    ohh yes okay

  102. onegirl
    • 2 years ago
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    so it reached 0 at pi/2

  103. experimentX
    • 2 years ago
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    no ... the derivative is zero at pi/2

  104. onegirl
    • 2 years ago
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    ok

  105. experimentX
    • 2 years ago
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    gotta go ... you gotta work harder.

  106. experimentX
    • 2 years ago
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    good luck!!

  107. onegirl
    • 2 years ago
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    ok well can u tell me what i should do next before you go

  108. experimentX
    • 2 years ago
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    you wanted to verify Rolle's theorem, which you did.

  109. onegirl
    • 2 years ago
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    yea its continuos and differentiable i know that

  110. experimentX
    • 2 years ago
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    what do you want to do?

  111. onegirl
    • 2 years ago
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    find a value for c

  112. onegirl
    • 2 years ago
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    so this could be true a < c < b

  113. experimentX
    • 2 years ago
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    of course it is ..

  114. experimentX
    • 2 years ago
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    |dw:1363389508215:dw|

  115. onegirl
    • 2 years ago
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    ok so c is pi/2 ?

  116. experimentX
    • 2 years ago
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    |dw:1363389556563:dw|

  117. onegirl
    • 2 years ago
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    ohhh ok

  118. onegirl
    • 2 years ago
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    so c is 2pi/4

  119. experimentX
    • 2 years ago
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    yes

  120. onegirl
    • 2 years ago
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    ok

  121. experimentX
    • 2 years ago
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    any probs??

  122. onegirl
    • 2 years ago
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    no thx

  123. experimentX
    • 2 years ago
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    see you around ... gotta go offline :)

  124. onegirl
    • 2 years ago
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    ok

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