problem 3. (a) find the points on the graph of y² = x³ - 3x + 1 where the tangent line is horizontal…

problem 3. (a) find the points on the graph of y² = x³ - 3x + 1 where the tangent line is horizontal. solution: differentiating the equation of the graph implicitly gives 2yy = 3x² - 3. setting y = 0 gives 3x² - 3 = 0 so that x² = 1, and either x = -1 or x = 1. when x = 1, x³ - 3x + 1 = 1 - 3 + 1 = -1 < 0 and so there is no corresponding y - value. when x = -1, x³ - 3x + 1 = -1 + 3 + 1 = 3 and we get either y = √3 or y = -√3. there are two points: (-1,√3) and (-1,-√3).

problem 3. (a) find the points on the graph of y² = x³ - 3x + 1 where the tangent line is horizontal. solution: differentiating the equation of the graph implicitly gives 2yy = 3x² - 3. setting y = 0 gives 3x² - 3 = 0 so that x² = 1, and either x = -1 or x = 1. when x = 1, x³ - 3x + 1 = 1 - 3 + 1 = -1 < 0 and so there is no corresponding y - value. when x = -1, x³ - 3x + 1 = -1 + 3 + 1 = 3 and we get either y = √3 or y = -√3. there are two points: (-1,√3) and (-1,-√3).

Answer

Explanation:

Step1: Implicit differentiation

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, we get $2yy'=3x^{2}-3$.

Step2: Find x - values for horizontal tangent

Set $y' = 0$. Then $3x^{2}-3 = 0$. Solving for $x$: [ \begin{align*} 3x^{2}-3&=0\ x^{2}&=1\ x&=\pm1 \end{align*} ]

Step3: Find corresponding y - values

When $x = 1$, $y^{2}=x^{3}-3x + 1=1 - 3+1=-1<0$, so no real $y$ - values. When $x=-1$, $y^{2}=x^{3}-3x + 1=-1 + 3+1 = 3$, then $y=\pm\sqrt{3}$.

Answer:

$(-1,\sqrt{3})$ and $(-1,-\sqrt{3})$