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anonymous

  • 5 years ago

Suppose that we know that f(x) is a continuous and differentiable function on [2,14]. Also, suppose we know that f(2)=9 and f'(x) is greater than or equal to -12. What is the smallest possible value of f(14)?

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  1. amistre64
    • 5 years ago
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    would this be a straight line from 2,14 running at a slope of 12?

  2. anonymous
    • 5 years ago
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    (I'm supposed to use the mean value theorem, or at least that's what it says on the worksheet.)

  3. anonymous
    • 5 years ago
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    (I'm supposed to use the mean value theorem, or at least that's what it says on the worksheet.)

  4. anonymous
    • 5 years ago
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    @amistre64 - I have no idea! I'm complete lost on this one. Don't even know where to start.

  5. amistre64
    • 5 years ago
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    f'(2) =>12....6x? maybe

  6. anonymous
    • 5 years ago
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    Well, if f'(x) = -12 throughout the interval then it would be a straight line from (2,9) with a slope of -12. So that would be the minimum possible value given what we know.

  7. anonymous
    • 5 years ago
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    Sorry, @polpak - how do we know that?

  8. amistre64
    • 5 years ago
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    ahh,....(-)12 ... that was hiding in the corner

  9. anonymous
    • 5 years ago
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    We are given that f'(x) is greater than or equal to -12.

  10. anonymous
    • 5 years ago
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    So if it maintained the smallest possible derivative over the whole interval we would have a line with slope of -12.

  11. myininaya
    • 5 years ago
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    i got the smallest possible value is -135 for f(14)

  12. amistre64
    • 5 years ago
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    thats what I thought :)

  13. amistre64
    • 5 years ago
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    but with a 12 lol

  14. anonymous
    • 5 years ago
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    9-12*12 = 9-144 = -135

  15. myininaya
    • 5 years ago
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    yay thats what i got except differently

  16. myininaya
    • 5 years ago
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    -12<=[f(2)-f(14)]/[2-14]

  17. anonymous
    • 5 years ago
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    Does that make sense Chap?

  18. anonymous
    • 5 years ago
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    How do you figure what f(14) is?

  19. myininaya
    • 5 years ago
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    chap you can also solve the above for f(14) resulting in f(14)>=(-135)

  20. myininaya
    • 5 years ago
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    I used the mean value thm just like you said

  21. anonymous
    • 5 years ago
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    Right. But I have -12<= 9 - f(14)

  22. anonymous
    • 5 years ago
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    Oops wait

  23. anonymous
    • 5 years ago
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    I have -12 <= 9 - f(14) ----------- -12

  24. myininaya
    • 5 years ago
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    multipliy -12 on both sides don't forget to flip the inequality when you multipliy or divide by a negative

  25. anonymous
    • 5 years ago
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    Sorry )= I don't understand how to do that.

  26. anonymous
    • 5 years ago
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    \[-12 \le \frac{9- f(14)}{-12} \implies 144 \ge 9-f(14)\]

  27. myininaya
    • 5 years ago
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    how do you solve x/5=4

  28. anonymous
    • 5 years ago
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    \[-4a < b \implies a \gt \frac{b}{-4}\]

  29. anonymous
    • 5 years ago
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    Okay. But how does 135 >= - f(14) solve for f(14)?

  30. anonymous
    • 5 years ago
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    Multiply by -1

  31. myininaya
    • 5 years ago
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    now multipliy both sides by negative 1 don't forget to flip

  32. anonymous
    • 5 years ago
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    135 <= f(14)

  33. myininaya
    • 5 years ago
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    right!

  34. myininaya
    • 5 years ago
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    wait where's the negative?

  35. anonymous
    • 5 years ago
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    That's the answer?

  36. anonymous
    • 5 years ago
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    -135 <= f(14)

  37. myininaya
    • 5 years ago
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    ok so we have f(14)>=-135 this means the smallest that f(14) can be is -135

  38. anonymous
    • 5 years ago
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    OH! I thought f(14) would be positive. Thank you so much!

  39. anonymous
    • 5 years ago
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    It might be positive!

  40. anonymous
    • 5 years ago
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    We only know that the smallest it can be is -135. It could be very very large.

  41. myininaya
    • 5 years ago
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    it could be 1111111111111111111111

  42. anonymous
    • 5 years ago
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    over 9000!

  43. myininaya
    • 5 years ago
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    5 trillion billion if that means anything

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