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anonymous

  • one year ago

Graph the function f(x)=x^3−2x and it's secant line through the points (-2,-4) and (2,4). Use the graph to estimate the x-coordinate of the points where the tangent line is parallel to the secant line. Find the exact value of the numbers c that satisfy the conclusion of the mean value theorem for the interval [-2,2].

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  1. anonymous
    • one year ago
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    I need help finding 'c'

  2. anonymous
    • one year ago
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    I know that the Mean Value Theorem says... \[\frac{ f(b)-f(a) }{ b-a }=f'(c)\]

  3. anonymous
    • one year ago
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    just apply that...

  4. anonymous
    • one year ago
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    But that gives me f'(c)

  5. anonymous
    • one year ago
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    lemme check

  6. anonymous
    • one year ago
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    here a=-2 b=2

  7. anonymous
    • one year ago
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    Right, plugging all that in do you get f'(c) = 2 ?

  8. anonymous
    • one year ago
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    i get 8

  9. anonymous
    • one year ago
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    How?

  10. anonymous
    • one year ago
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    f(b)=2^3-2(2)..rite?

  11. anonymous
    • one year ago
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    f(a)=(-2)^3-2(-2)

  12. anonymous
    • one year ago
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    Doesn't it already give us the coordinates for the points? (2,4) and (-2,-4)

  13. anonymous
    • one year ago
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    Is it wrong to say \[\frac{ 4+4 }{ 2+2}=\frac{ 8 }{ 4 } = 2\] ? Or am I totally wrong right now? Lol

  14. anonymous
    • one year ago
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    what did u do just now?

  15. myininaya
    • one year ago
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    \[(x^3-2x)'|_{x=c}=\frac{f(2)-f(-2)}{2-(-2)} \\ (x^3-2x)'|_{x=c}=\frac{4-(-4)}{2-(-2)} \\ (x^3-2x)'|_{x=c}=\frac{4+4}{2+2}\] that is right for the right hand side

  16. anonymous
    • one year ago
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    Well I plugged in f(a) = -4, f(b) = 4, and a=-2, b=2 in the mean value theorem formula

  17. anonymous
    • one year ago
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    what happens to the left hand side?

  18. anonymous
    • one year ago
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    do i just derive it?

  19. myininaya
    • one year ago
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    you still have to differentiate the left hand side and plug in c afterwards as the equation above suggests

  20. anonymous
    • one year ago
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    3x^2 - 2 right?

  21. myininaya
    • one year ago
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    yep and plug in c (though that part doesn't matter to much; it is just what they call it in the formula above) yep now solve 3x^2-2=2

  22. myininaya
    • one year ago
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    or you named it c 3c^2-2=2

  23. anonymous
    • one year ago
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    Ohh so then its +/- 2/sqrt3 right?

  24. myininaya
    • one year ago
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    are both of those in the given interval?

  25. anonymous
    • one year ago
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    hmmm... yes?

  26. myininaya
    • one year ago
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    ok then cool stuff

  27. anonymous
    • one year ago
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    Thats it? :O

  28. myininaya
    • one year ago
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    that is it

  29. myininaya
    • one year ago
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    though there are two problems

  30. myininaya
    • one year ago
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    did you do the first problem?

  31. anonymous
    • one year ago
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    Yay!!! Thats correct! :D Thank you so much!

  32. myininaya
    • one year ago
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    we did the second problem we found the exact solutions

  33. anonymous
    • one year ago
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    Yeah Im always better in graphing haha

  34. myininaya
    • one year ago
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    alright

  35. anonymous
    • one year ago
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    Thanks again for the help! :)

  36. myininaya
    • one year ago
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    np

  37. anonymous
    • one year ago
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    tracy?got the answer for the second part?

  38. anonymous
    • one year ago
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    Yeah! I got it, its + or - 2/sqrt3

  39. anonymous
    • one year ago
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    Thank you too @chethus!

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