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anonymous

  • one year ago

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = x/x+6 [1, 12] If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. (Enter your answers as a comma-separated list. If it does not satisfy the hypotheses, enter DNE). Ive tried to do it multiple ways but i cant get the right answer. Please help!

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  1. jim_thompson5910
    • one year ago
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    the function is \[\Large f(x) = \frac{x}{x+6}\] right?

  2. anonymous
    • one year ago
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    yes

  3. jim_thompson5910
    • one year ago
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    which x value makes the denominator zero?

  4. anonymous
    • one year ago
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    -6

  5. jim_thompson5910
    • one year ago
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    is -6 in the interval [1,12] ?

  6. anonymous
    • one year ago
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    no

  7. jim_thompson5910
    • one year ago
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    so the function is continuous on [1,12] that allows us to use the MVT properly here

  8. jim_thompson5910
    • one year ago
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    if there was a discontinuity on [1,12] then we couldn't use the MVT

  9. anonymous
    • one year ago
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    okay, I understand that part, the equation part is whats given me a lot of trouble, i dont have examples like this one

  10. jim_thompson5910
    • one year ago
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    Mean Value Theorem If f(x) is continuous on [a,b] and differentiable on (a,b), then there exists at least one value of c such that \[\Large f \ '(c) = \frac{f(b) - f(a)}{b-a}\]

  11. jim_thompson5910
    • one year ago
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    what you have to do is find the secant slope through (1, f(1)) and (12,f(12)) then find f ' (x) set f ' (x) equal to the secant slope and solve for x

  12. anonymous
    • one year ago
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    so i would substitute 1 for a and 12 for b? that gives me 1?

  13. jim_thompson5910
    • one year ago
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    you said `that gives me 1?` what do you mean exactly? that's the secant slope you got?

  14. anonymous
    • one year ago
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    yeah if i substitute the a for 1 and 12 for b, that equals to 11/11 so 1? i dont think im doing this right

  15. jim_thompson5910
    • one year ago
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    that's not the correct secant slope

  16. jim_thompson5910
    • one year ago
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    \[\Large f(x) = \frac{x}{x+6}\] \[\Large f(1) = \frac{1}{1+6}\] \[\Large f(1) = \frac{1}{7}\]

  17. jim_thompson5910
    • one year ago
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    \[\Large f(x) = \frac{x}{x+6}\] \[\Large f(12) = \frac{12}{12+6}\] \[\Large f(12) = \frac{12}{18}\] \[\Large f(12) = \frac{2}{3}\]

  18. anonymous
    • one year ago
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    and then i would do the same for f(12)?

  19. jim_thompson5910
    • one year ago
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    yes

  20. anonymous
    • one year ago
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    okay I got that part!

  21. jim_thompson5910
    • one year ago
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    you should have this next \[\Large f \ '(c) = \frac{f(b) - f(a)}{b-a}\] \[\Large f \ '(c) = \frac{f(12) - f(1)}{12-1}\] \[\Large f \ '(c) = \frac{\frac{2}{3} - \frac{1}{7}}{12-1}\] \[\Large f \ '(c) = \frac{\frac{11}{21}}{11}\] \[\Large f \ '(c) = \frac{1}{21}\]

  22. jim_thompson5910
    • one year ago
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    now you need f ' (x)

  23. anonymous
    • one year ago
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    so then i would substitute 1/21 for x in this 6/(6+x)^2 right?

  24. jim_thompson5910
    • one year ago
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    no

  25. jim_thompson5910
    • one year ago
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    you set them equal to one another and solve for x

  26. anonymous
    • one year ago
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    I set what equal to each other?

  27. jim_thompson5910
    • one year ago
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    1/21 and that f ' (x) you got

  28. anonymous
    • one year ago
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    i end up with 6=1/21x^2+4/7x+12/7

  29. anonymous
    • one year ago
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    what would i do w the other x?

  30. jim_thompson5910
    • one year ago
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    \[\Large f \ '(x) = \frac{6}{(x+6)^2}\] \[\Large \frac{1}{21} = \frac{6}{(x+6)^2}\] \[\Large 1*(x+6)^2 = 21*6\] \[\Large (x+6)^2 = 126\] keep going to solve for x

  31. anonymous
    • one year ago
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    got -6+3root14 and -6-3root 14?

  32. jim_thompson5910
    • one year ago
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    those are the correct solutions to the equation you now need to check if they lie in the interval [1,12]

  33. anonymous
    • one year ago
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    okay so it would only be -6+3root14

  34. jim_thompson5910
    • one year ago
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    yes since \(\large -6+3\sqrt{14} \approx 5.22\) which is in the interval [1,12]

  35. jim_thompson5910
    • one year ago
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    the other value is -17.22 which is not in that interval

  36. anonymous
    • one year ago
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    thank you so much!!

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