91Ó°ÊÓ

First, we need to find the first derivative of the given equation. We will differentiate each term of the equation with respect to \(x\) using the Implicit Differentiation method: \(\frac{d}{dx}(x) + \frac{d}{dx}(y^3 - y) = \frac{d}{dx}(1)\). By applying the Chain Rule for the term containing "y": \(1 + [3y^2\frac{dy}{dx} - \frac{dy}{dx}] = 0\) Now, solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx}(3y^2 - 1) = -1\) \(\frac{dy}{dx} = \frac{-1}{3y^2 - 1}\)

Vertical tangent lines occur where the derivative is undefined. In this case, the derivative will be undefined when the denominator is zero: \(3y^2 - 1 = 0\) \(3y^2 = 1\) \(y^2 = \frac{1}{3}\) \(y = \pm\frac{1}{\sqrt{3}}\) Now we have two possible values for \(y\). We must plug them back into the original equation to find the corresponding \(x\)-values: For \(y = \frac{1}{\sqrt{3}}\): \(x + \left(\frac{1}{\sqrt{3}}\right)^3 - \frac{1}{\sqrt{3}} = 1\) \(x = -\frac{2}{3\sqrt{3}}\) Thus, one of the points with a vertical tangent line is \(\left(-\frac{2}{3\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\). For \(y = -\frac{1}{\sqrt{3}}\): \(x + \left(-\frac{1}{\sqrt{3}}\right)^3 + \frac{1}{\sqrt{3}} = 1\) \(x = \frac{2}{3\sqrt{3}}\) So the other point with a vertical tangent line is \(\left(\frac{2}{3\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)\).

Horizontal tangent lines happen when the first derivative equals zero: \(\frac{-1}{3y^2 - 1} = 0\) However, this equation has no solution since the fraction can never equal zero, as the numerator is always non-zero and the denominator never has a zero value. Therefore, there are no horizontal tangent lines on the curve. In conclusion, the curve \(x+y^3-y=1\) has vertical tangent lines at the points \(\left(-\frac{2}{3\sqrt{3}}, \frac{1}{\sqrt{3}}\right)\) and \(\left(\frac{2}{3\sqrt{3}}, -\frac{1}{\sqrt{3}}\right)\). It does not have any horizontal tangent lines.