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anonymous

  • 2 years ago

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

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  1. anonymous
    • 2 years ago
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    \[f(x) = \frac{ x^2+x }{ x-1 }\]

  2. anonymous
    • 2 years ago
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    function is only discontinuous at 1, which is not in the interval theorem holds

  3. anonymous
    • 2 years ago
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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

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

  5. anonymous
    • 2 years ago
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    oh here

  6. anonymous
    • 2 years ago
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    you need \(f(4)\) first

  7. anonymous
    • 2 years ago
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    [5/2 , 4] f(c) = 6

  8. anonymous
    • 2 years ago
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    the statement \(f(c)=6\) makes no sense

  9. anonymous
    • 2 years ago
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    well, it's given

  10. anonymous
    • 2 years ago
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    mvt says under the suitable conditions there is a \(c\in (a, b)\) where \[f'(c)=\frac{f(b)-f(a)}{b-a}\]

  11. anonymous
    • 2 years ago
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    oooh abort sorry wrong i am wrong

  12. anonymous
    • 2 years ago
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    oh I didn't learn that yet but okay.. what should I do?

  13. anonymous
    • 2 years ago
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    it thought it said mean value theorem, it says intermediate value theorem we have to start over sorry

  14. anonymous
    • 2 years ago
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    it's fine

  15. anonymous
    • 2 years ago
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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 \(f(4)\)

  16. anonymous
    • 2 years ago
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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

  17. anonymous
    • 2 years ago
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    since the function is continuous on that interval and cannot skip values

  18. anonymous
    • 2 years ago
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    yes

  19. anonymous
    • 2 years ago
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    sorry for the interruption but

  20. anonymous
    • 2 years ago
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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?

  21. anonymous
    • 2 years ago
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    you job is to solve \[f(x)=6\] but that is for the second part of the question

  22. anonymous
    • 2 years ago
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    there are two parts 1) verify that the intermediate value theorem applies to the indicated interval

  23. anonymous
    • 2 years ago
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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

  24. anonymous
    • 2 years ago
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    i think \(f(\frac{5}{2})=5\tfrac{5}{6}\) and \(f(4)=\frac{20}{3}\)

  25. anonymous
    • 2 years ago
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    how can 4 be larger than 6?

  26. anonymous
    • 2 years ago
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    k lets go slow

  27. anonymous
    • 2 years ago
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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

  28. anonymous
    • 2 years ago
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    in other words \[f(\frac{5}{2})<6<f(4)\]

  29. anonymous
    • 2 years ago
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    okay

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

  31. anonymous
    • 2 years ago
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    now solving it is just a matter of solving \[\frac{x^2+x}{x-1}=6\]

  32. anonymous
    • 2 years ago
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    you would start with \[x^2+x=6(x-1)\] and solve the resulting quadratic equation

  33. anonymous
    • 2 years ago
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    ooh okay okay I finally get it :)

  34. anonymous
    • 2 years ago
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    Thank you so much for your kind explanation! :)

  35. anonymous
    • 2 years ago
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    yw

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