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musiklover317 Group Title

pleasee help verify that the intermediate value theorem applies to the indicated interval and find the value of c guaranteed by the theorem. f(x) = (x^2+x)/(x-1), [5/2,4], f(c) = 6

  • one year ago
  • one year ago

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  1. musiklover317 Group Title
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    \[f(x) = \frac{ x^2+x }{ x-1 }\]

    • one year ago
  2. satellite73 Group Title
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    function is only discontinuous at 1, which is not in the interval theorem holds

    • one year ago
  3. musiklover317 Group Title
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    hmm... for the other questions, I plugged in c to x and set that number (in this case, 6) equal to the equation and solved it. but for this one, I can't

    • one year ago
  4. satellite73 Group Title
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    \[f(x) = \frac{ x^2+x }{ x-1 }\] you need some numbers to get the second part

    • one year ago
  5. musiklover317 Group Title
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    oh here

    • one year ago
  6. satellite73 Group Title
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    you need \(f(4)\) first

    • one year ago
  7. musiklover317 Group Title
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    [5/2 , 4] f(c) = 6

    • one year ago
  8. satellite73 Group Title
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    the statement \(f(c)=6\) makes no sense

    • one year ago
  9. musiklover317 Group Title
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    well, it's given

    • one year ago
  10. satellite73 Group Title
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    mvt says under the suitable conditions there is a \(c\in (a, b)\) where \[f'(c)=\frac{f(b)-f(a)}{b-a}\]

    • one year ago
  11. satellite73 Group Title
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    oooh abort sorry wrong i am wrong

    • one year ago
  12. musiklover317 Group Title
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    oh I didn't learn that yet but okay.. what should I do?

    • one year ago
  13. satellite73 Group Title
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    it thought it said mean value theorem, it says intermediate value theorem we have to start over sorry

    • one year ago
  14. musiklover317 Group Title
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    it's fine

    • one year ago
  15. satellite73 Group Title
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    they way you show there must be a number in \([\frac{5}{2},4]\) with \[f(c)=6\] is to compute \(f(\frac{5}{2})\) and \(f(4)\)

    • one year ago
  16. satellite73 Group Title
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    one number will be larger than 6, one number will be smaller than 6, so by the intermediate value theorem the function must be 6 somewhere in that interval

    • one year ago
  17. satellite73 Group Title
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    since the function is continuous on that interval and cannot skip values

    • one year ago
  18. musiklover317 Group Title
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    yes

    • one year ago
  19. musiklover317 Group Title
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    sorry for the interruption but

    • one year ago
  20. musiklover317 Group Title
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    well, in class, I learned that c has to be replaced with x and have to set that equation to 6. when I get the x values, the values between 5/2 and 4 is the answer is this right?

    • one year ago
  21. satellite73 Group Title
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    you job is to solve \[f(x)=6\] but that is for the second part of the question

    • one year ago
  22. satellite73 Group Title
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    there are two parts 1) verify that the intermediate value theorem applies to the indicated interval

    • one year ago
  23. satellite73 Group Title
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    the reason that you know \[f(x)=6\] for some \(x\) in the interval is because \(f(4)\) is larger than 6, and \(f(\frac{5}{2})\) is smaller than 6

    • one year ago
  24. satellite73 Group Title
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    i think \(f(\frac{5}{2})=5\tfrac{5}{6}\) and \(f(4)=\frac{20}{3}\)

    • one year ago
  25. musiklover317 Group Title
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    how can 4 be larger than 6?

    • one year ago
  26. satellite73 Group Title
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    k lets go slow

    • one year ago
  27. satellite73 Group Title
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    it is not 4 that is larger than 6, it is \(f(4)=\frac{20}{3}\) that is larger than 6

    • one year ago
  28. satellite73 Group Title
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    in other words \[f(\frac{5}{2})<6<f(4)\]

    • one year ago
  29. musiklover317 Group Title
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    okay

    • one year ago
  30. satellite73 Group Title
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    so since 6 lies between the two function values, there must be some \[c\in (\frac{5}{2},4)\] where \(f(c)=6\)

    • one year ago
  31. satellite73 Group Title
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    now solving it is just a matter of solving \[\frac{x^2+x}{x-1}=6\]

    • one year ago
  32. satellite73 Group Title
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    you would start with \[x^2+x=6(x-1)\] and solve the resulting quadratic equation

    • one year ago
  33. musiklover317 Group Title
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    ooh okay okay I finally get it :)

    • one year ago
  34. musiklover317 Group Title
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    Thank you so much for your kind explanation! :)

    • one year ago
  35. satellite73 Group Title
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    yw

    • one year ago
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