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Thread: Rational average rate of change problem

  1. #1
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    Rational average rate of change problem

    Given the function$\displaystyle f(x) = \frac{\ 2x}{x-4}$ , determine the coordinates of a point on f(x) where the slope of the tangent line equals the slope of the secant line that passes through A(5,10) and B(8,4)
    Last edited by mr fantastic; Apr 10th 2011 at 02:18 PM. Reason: Replaced 'instantaneous' with 'average' in title.
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  2. #2
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    First find the slope of the secant line that passes through A(5,10) and B(8,4)

    Hint: $\displaystyle \displaystyle m = \frac{y_2-y_1}{x_2-x_1}$

    What do you get?
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  3. #3
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    m= (4 - 10)/(8-5)
    m= -2
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  4. #4
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    So now solve $\displaystyle \displaystyle -2 = \left(\frac{2x}{x-4}\right)'$
    Last edited by pickslides; Apr 10th 2011 at 02:40 PM.
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  5. #5
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    $\displaystyle -2x + 8 = 2x$

    $\displaystyle 4x=8$

    $\displaystyle x=2$
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  6. #6
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    Quote Originally Posted by Devi09 View Post
    $\displaystyle -2x + 8 = 2x$

    $\displaystyle 4x=8$

    $\displaystyle x=2$
    Nope, $\displaystyle \displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$
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  7. #7
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    Quote Originally Posted by pickslides View Post
    Nope, $\displaystyle \displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$
    How did you get that?

    when you cross multiply $\displaystyle -2= (2x)/(x-4)$

    you get x = 2
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  8. #8
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    You need to find $\displaystyle \displaystyle \left(\frac{2x}{x-4}\right)'$ which is the gradient function of $\displaystyle \displaystyle \frac{2x}{x-4}$

    $\displaystyle \displaystyle \left(\frac{2x}{x-4}\right)'= \left(\frac{2(x-4)-2x\times 1}{(x-4)^2}\right)$ via the quotient rule for differentiation.
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