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Math Help - Derivative & Average Rate of Change

  1. #1
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    Derivative & Average Rate of Change

    Let f(1)= 1/t for t>0. For what value of t is f '(t) equal to the average rate of change on f on [a,b]

    A) -sqrt(ab) B) sqrt(ab) C) -1/(sqrt[ab]) D) 1/(sqrt[ab])
    E) sqrt [(1/2)(1/b-1/a)]

    Please Help
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  2. #2
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    Quote Originally Posted by rawkstar View Post
    Let f(1)= 1/t for t>0. For what value of t is f '(t) equal to the average rate of change on f on [a,b]

    A) -sqrt(ab) B) sqrt(ab) C) -1/(sqrt[ab]) D) 1/(sqrt[ab])
    E) sqrt [(1/2)(1/b-1/a)]

    Please Help
    I'm sure you meant f(t) = \frac{1}{t}


    f'(t) = -\frac{1}{t^2}

    f'(t) = \frac{f(b) - f(a)}{b-a}

    -\frac{1}{t^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b - a}

    -\frac{1}{t^2} = \frac{\frac{a-b}{ab}}{b - a}

    finish from here?
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  3. #3
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    I don't know how you got from
    to
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  4. #4
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    Skeeter was merely writing the numerator as a single fraction,
    to simplify further in your remaining steps, rawkstar.

    (1/b)-(1/a)=(a/a)(1/b)-(b/b)(1/a)=a/(ab)-b/(ab)=(a-b)/(ab)
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  5. #5
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    Quote Originally Posted by rawkstar View Post
    I don't know how you got from
    to
    \frac{1}{b} - \frac{1}{a}

    common denominator is ab ...

    \frac{1}{b} \cdot \frac{a}{a} - \frac{1}{a} \cdot \frac{b}{b}

    \frac{a}{ab} - \frac{b}{ab}

    put them together ...

    \frac{a-b}{ab}
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  6. #6
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    ok i think I got it, anyways the answer i got was B) sqrt(ab)
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  7. #7
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    As\ t>0,\ 'answer\ A)'\ is\ invalid
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