Answer
$(-1,\sqrt{3})$ 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 $(-1,-\sqrt{3})$.
Work Step by Step
(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 $(-1,-\sqrt{3})$.
