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Calculus 2nd Post please help

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Find an equation of the tangent line to the graph of the function f defined by the following equation at the indicated point.
(x - y - 1)3 = x; (1, -1)

  • Calculus 2nd Post please help - ,

    I will assume your equation is
    (x - y - 1)^3 = x

    then 3(x-y-1)^2(1 - dy/dx) = 1
    at the given point...
    3(1+1-1)^2(1-dy/dx) = 1
    3 - 3dy/dx = 1
    dy/dx = 2/3

    so now you have the slope m=2/3 and the point (1,-1)

    Use the method you learned in grade 9 to find the equation of that line.

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