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Thread: antiderivatives

  1. November 6th 2012, 12:53 PM #1
    pnfuller
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    antiderivatives

    find a function f such that f'(x)=x3 and the line x+y=0 is tangent to the graph of f
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  2. November 6th 2012, 02:10 PM #2
    skeeter
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    Re: antiderivatives

    f'(x) = x^3 \implies f(x) = \frac{x^4}{4} + C

    x+y = 0 \implies y = -x ... slope of the tangent line is m = -1

    at the point of tangency (x_0,y_0) ...

    f'(x_0) = x_0^3 = -1 \implies x_0 = -1

    finally ...

    y_0 = -x_0 = -(-1) = \frac{(-1)^4}{4} + C

    ... solve for C to finish it.
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