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61. Find the points on the graph of y2 = x³ – 3x +1 (Figure 10) where the tangent line is horizontal. (a) First show that 2yy' = 3x² – 3, where y' = dy/dx. (b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values of x where the slope may be zero. (c) Show that the positive value of x does not correspond to a point on the graph. (d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates. FIGURE 10 Graph of y² = x – 3x +1.

61. Find the points on the graph of y2 = x³ – 3x +1 (Figure 10) where the tangent line is horizontal. (a) First show that 2yy' = 3x² – 3, where y' = dy/dx. (b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values of x where the slope may be zero. (c) Show that the positive value of x does not correspond to a point on the graph. (d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates. FIGURE 10 Graph of y² = x – 3x +1.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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61. Find the points on the graph of y2 = x³ – 3x +1 (Figure 10) where the tangent line is horizontal. (a) First show that 2yy' = 3x² – 3, where y' = dy/dx. (b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values of x where the slope may be zero. (c) Show that the positive value of x does not correspond to a point on the graph. (d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates. FIGURE 10 Graph of y² = x – 3x +1.
Transcribed Image Text:61. Find the points on the graph of y2 = x³ – 3x +1 (Figure 10) where the tangent line is horizontal. (a) First show that 2yy' = 3x² – 3, where y' = dy/dx. (b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values of x where the slope may be zero. (c) Show that the positive value of x does not correspond to a point on the graph. (d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates. FIGURE 10 Graph of y² = x – 3x +1.
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