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Thread: tangent line/derivative

  1. #1
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    tangent line/derivative

    Find the values of $\displaystyle x$ for all points on the graph of $\displaystyle f(x)= x^3 + 2x^2 + 5x -16$ at which the slope of the tangent line is 5.

    my work:

    $\displaystyle f(x)= x^3 + 2x^2 + 5x -16$
    $\displaystyle f`(x) = 3x^2 + 4x + 5$
    $\displaystyle 5 = 3x^2 + 4x + 5$
    $\displaystyle 0 = 3x^2 + 4x$

    stuck after this...
    Last edited by algebra2; Oct 23rd 2008 at 05:14 PM.
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  2. #2
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    Quote Originally Posted by algebra2 View Post
    Find the values of $\displaystyle x$ for all points on the graph of $\displaystyle f(x)= x^3 + 2x^2 + 5x -16$ at which the slope of the tangent line is 4.

    my work:

    $\displaystyle f(x)= x^3 + 2x^2 + 5x -16$
    $\displaystyle f`(x) = 3x^2 + 4x + 5$
    $\displaystyle 5 = 3x^2 + 4x + 5$
    $\displaystyle 0 = 3x^2 + 4x$

    stuck after this...
    $\displaystyle 5 = 3x^2 + 4x + 5$ ??? look again at your problem statement.

    should be ...

    $\displaystyle 4 = 3x^2 + 4x + 5$

    $\displaystyle 0 = 3x^2 + 4x + 1$

    $\displaystyle 0 = (3x + 1)(x + 1)$

    finish.



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  3. #3
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    oh sorry i typed it wrong, the slope should be 5.

    ...and how do you factor polynomials with a number in front like $\displaystyle 3x^2 + 4x$, there's a three in front of the x^2. is there a technique to this?
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  4. #4
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    first rule of factoring ... pull out whatever factors each term has in common.

    $\displaystyle 3x^2 + 4x = 0$

    $\displaystyle x(3x + 4) = 0$

    $\displaystyle x = 0$

    $\displaystyle x = -\frac{4}{3}$
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  5. #5
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    i see, thank you.
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