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Find an equation of the tangent line to (x+y)^3 -5x+y=1 at the point where the curve intersects the line x+y=1 How am I supposed to determine what the point is? Im guessing system of equations but i can only think of one equation..
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The system would be \[\begin{cases}(x+y)^3-5x+y=1\\ x+y=1\end{cases}\] Solve by substitution: \(x+y=1~~\iff~~x=1-y\) \[(1-y+y)^3-5(1-y)+y=1\\ 1-5(1-y)+y=1\\ y=5-5y\\ 6y=5\\ y=\frac{5}{6}~~\Rightarrow~~x=\frac{1}{6}\] The point of intersection is thus \(\left(\dfrac{1}{6},\dfrac{5}{6}\right)\).
So from here, you would find the derivative of the given curve at this point, then find the equation of the tangent line.
Oh thank you very much! I feel stupid I totally overlooked that. Thankss!
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