Q. 84 - Calculus

Question

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{graph} \mathrm{of} \mathrm{the} \mathrm{solutions} \mathrm{of} \mathrm{the} \mathrm{equation} {\mathrm{y}}^3 - 3\mathrm{y} - \mathrm{x} = 1. \\ \left(\mathrm{a}\right) \mathrm{Find} \mathrm{all} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph} \mathrm{with} \mathrm{an} \mathrm{x}-\mathrm{coordinate} \mathrm{of} \mathrm{x} = -1, \\ \mathrm{and} \mathrm{then} \mathrm{find} \mathrm{the} \mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{at} \mathrm{each} \mathrm{of} \mathrm{these} \mathrm{points}. \\ \left(\mathrm{b}\right) \mathrm{Find} \mathrm{all} \mathrm{points} \mathrm{on} \mathrm{the} \mathrm{graph} \mathrm{with} \mathrm{a} \mathrm{y}-\mathrm{coordinate} \mathrm{of} \mathrm{y} = 2, \\ \mathrm{and} \mathrm{then} \mathrm{find} \mathrm{the} \mathrm{slope} \mathrm{of} \mathrm{the} \mathrm{tangent} \mathrm{line} \mathrm{at} \mathrm{each} \mathrm{of} \mathrm{these} \mathrm{points}. \\ \left(\mathrm{c}\right) \mathrm{Find} \mathrm{all} \mathrm{points} \mathrm{where} \mathrm{the} \mathrm{graph} \mathrm{has} \mathrm{a} \mathrm{horizontal} \mathrm{tangent} \mathrm{line}. \\ \left(\mathrm{d}\right) \mathrm{Find} \mathrm{all} \mathrm{points} \mathrm{where} \mathrm{the} \mathrm{graph} \mathrm{has} \mathrm{a} \mathrm{vertical} \mathrm{tangent} \mathrm{line}. \end{gathered}$$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(\mathrm{a}\right). \mathrm{The} \mathrm{points} \mathrm{are} : (-1,0),(-1,\sqrt{3}),(-1,-\sqrt{3}). \\ \left(\mathrm{b}\right). \mathrm{The} \mathrm{point} \mathrm{is} (1,2). \\ \left(\mathrm{c}\right). \mathrm{The} \mathrm{points} \mathrm{are} : (0,1),(0,2),(0,-2). \\ \left(\mathrm{d}\right). \mathrm{The} \mathrm{point} \mathrm{is} : (-1,0). \end{gathered}$$

Step 1. Given Information

$\mathrm{Equation} : {\mathrm{y}}^{3 }-3\mathrm{y}-\mathrm{x}=1$

Step 2. Solution (a) : The   points   on   the   graph   with   x   coordinate   of   x   =   - 1 .

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : {\mathrm{y}}^3-3\mathrm{y}-\mathrm{x}=1 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} : \\ 3{\mathrm{y}}^2\frac{d\mathrm{y}}{d\mathrm{x}}-3\frac{d\mathrm{y}}{d\mathrm{x}}-1 = 0 \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{1}{3{\mathrm{y}}^2-3}. \\ \mathrm{Substitute} \mathrm{x} =-1 \mathrm{in} \mathrm{above} \mathrm{equation} \\ {\mathrm{y}}^3-3\mathrm{y}-(-1) = 1 \\ \mathrm{y}=0,\pm \sqrt{3} \\ \mathrm{Thus} \mathrm{the} \mathrm{points} \mathrm{are} : (-1,0),(-1,\sqrt{3}),(-1,-\sqrt{3}) \\ \mathrm{The} \mathrm{slope} \mathrm{is}, \\ {\overline{)\frac{d\mathrm{y}}{d\mathrm{x}}}}_{\mathrm{x}=-1,\mathrm{y}=0} = \frac{1}{3{\left(0\right)}^2-3} =-\frac{1}{3} \\ \mathrm{The} \mathrm{slope} \mathrm{is}, \\ {\overline{)\frac{d\mathrm{y}}{d\mathrm{x}}}}_{\mathrm{x}=-1,\mathrm{y}=\sqrt{3}}=\frac{1}{3{\left(\sqrt{3}\right)}^2-3} = \frac{1}{6} \\ \mathrm{The} \mathrm{slope} \mathrm{is}, \\ {\overline{)\frac{d\mathrm{y}}{d\mathrm{x}}}}_{\mathrm{x}=-1,\mathrm{y}=-\sqrt{3}}=\frac{1}{3{(-\sqrt{3})}^2-3} = \frac{1}{6} \end{gathered}$$

Step 3. Solution (b) : The   points   on   the   graph   with   y - coordinate   of   y   = 2 .

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : {\mathrm{y}}^3-3\mathrm{y}-\mathrm{x}=1 \\ \mathrm{Differentiate} \mathrm{above} \mathrm{equation} : \\ 3{\mathrm{y}}^2\frac{d\mathrm{y}}{d\mathrm{x}}-3\frac{d\mathrm{y}}{d\mathrm{x}}-1 = 0 \\ \frac{d\mathrm{y}}{d\mathrm{x}} = \frac{1}{3{\mathrm{y}}^2-3}. \\ \mathrm{Substitute} \mathrm{y} =2 \mathrm{in} \mathrm{above} \mathrm{equation} \\ {\left(2\right)}^3-3\left(2\right)-\mathrm{x}=1 \\ \mathrm{x}=1 \\ \mathrm{Thus} \mathrm{the} \mathrm{point} \mathrm{is} (1,2) \\ \mathrm{Thus} \mathrm{slope} \mathrm{is}, \\ {\overline{)\frac{d\mathrm{y}}{d\mathrm{x}}}}_{\mathrm{x}=1,\mathrm{y}=2}=\frac{1}{3{\left(2\right)}^2-3} = \frac{1}{9} \end{gathered}$$

Step 4. Solution (c) : The   points   where   the   tanget   line   is   horizontal .

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : {\mathrm{y}}^3-3\mathrm{y}-\mathrm{x}=1 \\ \mathrm{Substitute} \mathrm{x} =0 \mathrm{in} \mathrm{above} \mathrm{equation} \\ {\mathrm{y}}^3-3\mathrm{y}-\left(0\right) = 1 \\ \mathrm{y}=1,\pm 2 \\ \mathrm{Thus} \mathrm{the} \mathrm{points} \mathrm{are} : (0,1),(0,2),(0,-2). \end{gathered}$$

Step 5. Solution (d) : The   points   where   tangent   line   is   vertical

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{equation} : {\mathrm{y}}^3-3\mathrm{y}-\mathrm{x}=1 \\ \mathrm{Substitute} \mathrm{y}=0 \mathrm{in} \mathrm{above} \mathrm{equation} \\ {\left(0\right)}^3-3\left(0\right)-\mathrm{x}=1 \\ \mathrm{x}=-1 \\ \mathrm{Thus} \mathrm{the} \mathrm{points} \mathrm{are} : (-1,0) \end{gathered}$$