y+1 = -1/f'(1) (x-1)
3(x-y-1)^2 (1-y') = 1
at (1,-1)
-3(1-y') = 1
y' = 4/3
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)
y =
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