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Math Help - Tangent Line

  1. #1
    Super Member Aryth's Avatar
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    Tangent Line

    Find the equation of the line tangent to the graph of f(x) = -4x^3 + 3\sqrt[5]{x} +4 at the point (-1,f(-1)). Leave your answer in the form y = mx + b.

    I just want to verify if what I'm doing is correct... It has been awhile since I have done one of these.

    I assumed I should use the equation y = f(a) + f'(a)(x-a)

    Which gave me:

    f(-1) = 8 + 3\sqrt[5]{-1}
    f'(-1) = 12 + \frac{3}{5}\sqrt[5]{(-1)^4} = \frac{63}{5}

    y = 8 + 3\sqrt[5]{-1} + \frac{63}{5}(x+1)
    y = \frac{63}{5}x + (8 + \frac{63}{5} + 3\sqrt[5]{-1})
    y = \frac{63}{5}x + (\frac{103}{5} + 3\sqrt[5]{-1})

    Here is where I have trouble... I am unsure of the fifth root of -1. I have seen that it could be complex, or it could be 1... And I just can't verify it myself. I'm sure it is easier than I am making it out to be, but any help would be appreciated.
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  2. #2
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    Re: Tangent Line

    In the real numbers \sqrt[5]{-1}=-1.
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