problem 3.\n(a) find the points on the graph of $y^{2}=x^{3}-3x + 1$ where the tangent line is horizontal.

Explanation:

Step1: Differentiate implicitly

Differentiate $y^{2}=x^{3}-3x + 1$ with respect to $x$. Using the chain - rule on the left - hand side and power rule on the right - hand side. $2y\frac{dy}{dx}=3x^{2}-3$. Then solve for $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{3x^{2}-3}{2y}$.

Step2: Set the derivative equal to 0

A horizontal tangent line has a slope of 0. So set $\frac{dy}{dx}=0$. $\frac{3x^{2}-3}{2y}=0$. This implies $3x^{2}-3 = 0$ (since the denominator $2y\neq0$ for non - degenerate cases; if $y = 0$, the fraction is undefined). Solve $3x^{2}-3 = 0$ for $x$: $3x^{2}=3$, then $x^{2}=1$, so $x=\pm1$.

Step3: Find the corresponding y - values

When $x = 1$, substitute into the original equation $y^{2}=x^{3}-3x + 1$. $y^{2}=1^{3}-3\times1 + 1=1 - 3+1=-1$. Since $y^{2}=-1$ has no real solutions, we discard $x = 1$. When $x=-1$, substitute into the original equation $y^{2}=x^{3}-3x + 1$. $y^{2}=(-1)^{3}-3\times(-1)+1=-1 + 3+1=3$. So $y=\pm\sqrt{3}$.

Answer:

The points are $(-1,\sqrt{3})$ and $(-1,-\sqrt{3})$.