## Calculus (3rd Edition)

$(-1,\sqrt{3})$ and $(-1,-\sqrt{3})$ are the points where tangent line is horizontal. (a) By chain rule, $2yy'=3x^2-3$ (b) $x=\pm1$ (c) Since $y^2$ can not be negative. The positive value of $x$ does not correspond to a point on the graph. (d) The coordinates of the two points are $(-1,\sqrt{3})$ and $(-1,-\sqrt{3})$.
(a) Differentiate $y^2 = x^3 − 3x + 1$ with respect to $x$. We get, $\dfrac{d}{dx}(y^2)=3x^2-3$ Now apply the chain rule. We get, $2y\dfrac{dy}{dx}=3x^2-3$ or $2yy'=3x^2-3$ (b) Substitute $y'=0$ in $2yy'=3x^2-3$. We get, $3x^2-3=0$ $\implies x^2-1=0$ $\implies x^2=1$ $\implies x=\pm1$ (c) Substitute $x=1$ in $y^2 = x^3 − 3x + 1$. We get, $y^2 = 1^3 − 3(1) + 1=-1$ Since $y^2$ can not be negative. The positive value of $x$ does not correspond to a point on the graph. (d) Now substitute $x=-1$ in $y^2 = x^3 − 3x + 1$. We get, $y^2 = (-1)^3 − 3(-1) + 1=3$ Thus, $y=\pm\sqrt{3}$ So, the coordinates of the two points are $(-1,\sqrt{3})$ and $(-1,-\sqrt{3})$.