Chapter 2: Q. 84 (page 211)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given Information

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

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

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

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

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

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

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

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

$Consider the equation : y^{3} - 3 y - x = 1 Substitute y = 0 in above equation {(0)}^{3} - 3 (0) - x = 1 x = - 1 Thus the points are : (- 1, 0)$

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Short Answer

Step by step solution

Step 1. Given Information

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

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

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

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

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