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Math Help - Calculus Antiderivative Question

  1. #1
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    Calculus Antiderivative Question

    Here is the question:

    Find a function f which satisfies both of the following properties:
    f ' (x) = x^3
    The line x + y = 0 is tangent to the graph of f.

    I figured out that f(x) is 1/4x^4 + C. Now I don't know what to do. I know I need to figure out C but i'm stuck. I isolated x+y=0 for y to get y= -x, and the derivative of that is -1, so the slope of the tangent line is -1. I don't know if im supposed to do this or how to relate this to C. Please help!
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  2. #2
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    Good job ndiggity, you're almost there, and everything you have done so far is on the right track.

    All you need to do now is find the value of x for which f'(x) = -1. This must be the point at which x+y=0 is a tangent, as the derivative is the same there. You can then get the y value for that point from x+y = 0 then solve your other equation for C.
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  3. #3
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    Ok, so the x value would be -1 then, and therefore the y value is 1. So the point is (-1,1). So do I plug in -1 in the anti-derivative I came up with to find C?
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  4. #4
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    Exactly right

    Ok, so the x value would be -1 then, and therefore the y value is 1. So the point is (-1,1). So do I plug in -1 in the anti-derivative I came up with to find C?
    Exactly right
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