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Thread: To find steeper tangents....please help!

  1. #1
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    To find steeper tangents....please help!

    Find all values x such that the tangent line to the graph of f(x)=4x^2 + 11x + 2 is steeper than the tangent line to g(x)=x^3

    I think i get stuck while solving quadratic inequalities...

    nybody know how to do this??? please help
    Thanks!
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  2. #2
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    $\displaystyle 8x + 11 > 3x^2$

    $\displaystyle 0 > 3x^2 - 8x - 11$

    $\displaystyle 0 > (3x - 11)(x + 1)$

    two critical values ... $\displaystyle x = \frac{11}{3}$ and $\displaystyle x = -1$

    the quadratic is negative between the two roots ...

    $\displaystyle -1 < x < \frac{11}{3}$
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