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anonymous

  • one year ago

I'll post what I have already solved, but I'm not sure if there's more to the question.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    f(0) = (0)^3 = 0 f(1) = (1)^3 = 1 \[\frac{ 1-0 }{ 1-0 }=\frac{1}{1}=1\] The derivative of the function is 3x^2 3x^2 = 1 x^2= 1/3 x = sqrt(1/3) or -sqrt(1/3) Since only sqrt(1/3) is found in [0,1], then it is the only answer.

  3. anonymous
    • one year ago
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    Is there anything else this question is asking me to do? (It's just worded confusingly to me)

  4. anonymous
    • one year ago
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    @amistre64

  5. amistre64
    • one year ago
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    do you have to define how the mean value thrm applies to it? i think it has to do with being continuous along the interval

  6. amistre64
    • one year ago
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    other than that, you did well at finding x=c that has a slope equal to the slope between x=a and x=b

  7. anonymous
    • one year ago
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    What do mean by define how it applies?

  8. amistre64
    • one year ago
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    what are the required conditions for the MVT to be applicable?

  9. anonymous
    • one year ago
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    The function must be continuous and differentiable

  10. amistre64
    • one year ago
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    ..." if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b)" ... http://www.sosmath.com/calculus/diff/der11/der11.html correct, and since this function is defined, continuous, and diffy-able then yada yada

  11. amistre64
    • one year ago
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    one of the common examples of a fail for this is f(x) = 1/x, on the interval (-1,1)

  12. amistre64
    • one year ago
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    a slope can be defined across the interval, but the function has no value in the interval that matches that slope ... it is not continuous or diffy-able at x=0 within the interval

  13. anonymous
    • one year ago
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    Ah okay. Could you interpret this by chance? ... "Find any tangent lines to the graph of f that are parallel to the secant line."

  14. anonymous
    • one year ago
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    I could post the problem if it would help

  15. amistre64
    • one year ago
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    more info usually helps, but the idea is to find the slope ofthe secant line in order to evaluate the derivative tha tmatches it

  16. anonymous
    • one year ago
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    Is there a formula to finding the slope of the secant line?

  17. anonymous
    • one year ago
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    Oh nevermind! It's 2/3. I've already solved for it.

  18. amistre64
    • one year ago
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    :) f(b)-f(a) ------- b-a

  19. anonymous
    • one year ago
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    So how would I find the parallel lines?

  20. amistre64
    • one year ago
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    parallel lines have the same slope ....

  21. anonymous
    • one year ago
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    So it would be any function with the slope 2/3?

  22. amistre64
    • one year ago
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    the derivative of the function (if one is given) define the slope at any single point

  23. amistre64
    • one year ago
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    any value x=c of which f'(c) = 2/3

  24. amistre64
    • one year ago
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    this is just another wording of the MVT

  25. anonymous
    • one year ago
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    Okay thank you!

  26. amistre64
    • one year ago
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    youre welcome was stuck in OS limbo for a bit :/

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