Calculus

Find a function f such that f '(x) = 3x3 and the line 3x + y = 0 is tangent to the graph of f.
f(x) =

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  1. dy/dx = 3 x^3
    then
    y = (3/4) x^4 + c

    slope of line y = -3x + 0 = -3
    so where does our slope = -3?
    3 x^3 = -3
    x = 1
    so make y the same for the function and the line at x = 1
    line at x = 1 is y = -3
    so
    (3/4)x^4 + c = -3 at x = 1
    3/4 + c = -3 = -12/4
    c = -15/4
    so
    f(x) = (3/4) x^4 -15/4
    or
    4 y = 3 x^4 - 15

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