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Thread: Tangents

  1. #1
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    Tangents

    Let $\displaystyle f(x) = 12 - x^2$ for $\displaystyle x > 0$ and $\displaystyle f(x)$ is greater than or equal to 0. The line tangent to the graph of $\displaystyle f$ at the point $\displaystyle (k,f(k))$ intercepts the x-axis at $\displaystyle x = 4$. What is the value of k?

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    My attempt:

    $\displaystyle f'(x) = -2x$

    $\displaystyle y = -2(4) + b = 0, b= 8$

    $\displaystyle y = -2x +8$ --> tangent line


    I know it's wrong because the tangent line is $\displaystyle y = -4x + 16$ and is tangent at $\displaystyle x = 2$.

    How would one go about doing it correctly?
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  2. #2
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    Sorry for the bump, but I really need help.
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  3. #3
    Jen
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    Consider what you know about the derivative.

    You know that you are looking for a tangent line at the point $\displaystyle (k, f(k))$. You also know that the derivative of your function is linear so it will have a constant slope. Consider what is happening between the points $\displaystyle (k,f(k))$ and $\displaystyle (4,0)$ where it intersects the x-axis.



    Tangents-capture.jpg

    Use what you know about the slope between $\displaystyle (k,f(k))$ and $\displaystyle (4,0)$ and the fact that the derivative represents a slope to help you solve for k.

    Hope these hints help.
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