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David_Ho

Give the number of values of c that verify the mean value theorem for f(x) = sin(x) on the interval [-1,5].

  • 2 years ago
  • 2 years ago

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  1. naf
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    |dw:1317571240881:dw|

    • 2 years ago
  2. David_Ho
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    Hiya! So far I have cos(x) = (sin(5) - sin(-1))/4 I really have no clue where to go with that.

    • 2 years ago
  3. naf
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    |dw:1317571301234:dw|

    • 2 years ago
  4. elecengineer
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    Find a value for the RHS

    • 2 years ago
  5. elecengineer
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    then you can figure out how many solns for x are in the interval

    • 2 years ago
  6. JamesJ
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    What is C here? (And btw, the derivative isn't a feature of the MVT.)

    • 2 years ago
  7. naf
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    |dw:1317571400618:dw|

    • 2 years ago
  8. naf
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    |dw:1317571475430:dw|

    • 2 years ago
  9. JamesJ
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    I think I see. Look at f(-1) and f(5). Then the MVP says for every y such that f(-1) < y < f(5) there is a c such that f(c) = y. Is that what you're asking about?

    • 2 years ago
  10. elecengineer
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    c is a value of x

    • 2 years ago
  11. David_Ho
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    Well, what I've recently learned is that MVT is (f(a) - f(b))/b -a = f' (c)

    • 2 years ago
  12. David_Ho
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    I had this problem earlier where I had to find the c value.

    • 2 years ago
  13. JamesJ
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    Ok, that's another MVT, sometimes called Rolle's theorem. But be that as it may, we'll use the one you're talking about. Here a = -1, b = 5. What is the LHS that expression?

    • 2 years ago
  14. David_Ho
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    What is LHS? sorry

    • 2 years ago
  15. elecengineer
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    Rolle's theorm is a specific case of MTV

    • 2 years ago
  16. David_Ho
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    oh ok, it would be the derivative of f(x) or cos(x)

    • 2 years ago
  17. JamesJ
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    the left hand side LHS

    • 2 years ago
  18. David_Ho
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    cos(x) = (sin(5) - sin(-1))/4

    • 2 years ago
  19. David_Ho
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    replace x for c

    • 2 years ago
  20. JamesJ
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    Right. so solutions to this equation are teh values c you're looking for. Remember also that c must lie between -1 and 5 So how many such solutions are there?

    • 2 years ago
  21. JamesJ
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    Actually no -- check your equation carefully . It is NOT divided by 4.

    • 2 years ago
  22. naf
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    |dw:1317572230820:dw|

    • 2 years ago
  23. David_Ho
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    ohh, divide by 6.

    • 2 years ago
  24. David_Ho
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    There would only be one value, right? Since there are only constants on the right side the value of c can only be one value. Is this logic sound?

    • 2 years ago
  25. JamesJ
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    No, it is not sound, because the function cos x isn't 1:1; it has periodicity; it goes up and down with period 2pi, which is almost exactly equal to the length of the interval [-1.5]. So there MIGHT be one c, but there could easily be two values of c. So you are going to have to calculate explicitly and check.

    • 2 years ago
  26. David_Ho
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    oohh, alright. I see what you mean now. I have to find all of the points on cos x that have a tangent line = to (sin(5)-sin(-1))/6.

    • 2 years ago
  27. JamesJ
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    No you need to find x such that cos x = (sin(5)-sin(-1))/6

    • 2 years ago
  28. JamesJ
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    cos x is the formula for the slope of the tangents to sin x. That's what we're looking for.

    • 2 years ago
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