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2x2 Use the guidelines to sketch the curve y = - 1 A. The domain is {x | x? - 1 + 0} = {x | x * ±1} (using interval notation). B. The x- and y-intercepts are both o. c. Since f(-x) = f(x), the function f is ---Select--- O . The curve is symmetric about the y-axis. 2 2x2 lim D. X+ t00 x - 1 lim X+ t0 1 - 1 Therefore, the line y = is a horizontal asymptote.

2x2 Use the guidelines to sketch the curve y = - 1 A. The domain is {x | x? - 1 + 0} = {x | x * ±1} (using interval notation). B. The x- and y-intercepts are both o. c. Since f(-x) = f(x), the function f is ---Select--- O . The curve is symmetric about the y-axis. 2 2x2 lim D. X+ t00 x - 1 lim X+ t0 1 - 1 Therefore, the line y = is a horizontal asymptote.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 15T
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2x2 Use the guidelines to sketch the curve y = - 1 A. The domain is {x | x2 – 1 + 0} = {x | x + ±1} %3D (using interval notation). В. The x- and y-intercepts are both 0. C. Since f(-x) = f(x), the function f is ---Select--- The curve is symmetric about the y-axis. 2 2x2 D. lim lim 1 X → t00 1 - X → ±0 x2 1 Therefore, the line y = is a horizontal asymptote. Since the denominator is 0 when x = ±1, we compute the following limits. 2x2 lim x - 1+ x2 - 1 2x2 lim X → 1- x² 1 2x2 lim X → -1+ x? 1 2x2 lim X → -1- x 1 The re, the lines x = 1 and x = are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the owing figure. y= 2 | x=-1 I = x||
Transcribed Image Text:2x2 Use the guidelines to sketch the curve y = - 1 A. The domain is {x | x2 – 1 + 0} = {x | x + ±1} %3D (using interval notation). В. The x- and y-intercepts are both 0. C. Since f(-x) = f(x), the function f is ---Select--- The curve is symmetric about the y-axis. 2 2x2 D. lim lim 1 X → t00 1 - X → ±0 x2 1 Therefore, the line y = is a horizontal asymptote. Since the denominator is 0 when x = ±1, we compute the following limits. 2x2 lim x - 1+ x2 - 1 2x2 lim X → 1- x² 1 2x2 lim X → -1+ x? 1 2x2 lim X → -1- x 1 The re, the lines x = 1 and x = are vertical asymptotes. This information about limits and asymptotes enables us to draw the asymptotes in the owing figure. y= 2 | x=-1 I = x||
4x(x2 - 1) – 2x2 . 2x (x² – 1)2 Е. f'(x) = Since f'(x) > 0 when x < 0 (x + -1) and f'(x) < 0 when x > 0 (x # 1), f is increasing on the intervals (-0, -1) and and decreasing on the intervals (0, 1) and The only critical number is x = 0. Since f' changes from positive to negative at ) = F. is a local maximum by the First Derivative Test. -4(x2 – 1)2 + 4x· 2(x² – 1)2x (x² – 1)4 G. f"(x) = Since the simplified numerator, > 0 for all x, we have the following. f"(x) > 0 x2 - 1 > 0 |x| > and f"(x) < 0 |x| < Thus. the curve is concave upward on the intervals (-∞, –1) and and concave downward on It has no point of inflection since 1 and are not in the domain of f. Н. Using the information in parts (E)-(G), we finish the sketch in the following figure. y=2 x=-1 :=1
Transcribed Image Text:4x(x2 - 1) – 2x2 . 2x (x² – 1)2 Е. f'(x) = Since f'(x) > 0 when x < 0 (x + -1) and f'(x) < 0 when x > 0 (x # 1), f is increasing on the intervals (-0, -1) and and decreasing on the intervals (0, 1) and The only critical number is x = 0. Since f' changes from positive to negative at ) = F. is a local maximum by the First Derivative Test. -4(x2 – 1)2 + 4x· 2(x² – 1)2x (x² – 1)4 G. f"(x) = Since the simplified numerator, > 0 for all x, we have the following. f"(x) > 0 x2 - 1 > 0 |x| > and f"(x) < 0 |x| < Thus. the curve is concave upward on the intervals (-∞, –1) and and concave downward on It has no point of inflection since 1 and are not in the domain of f. Н. Using the information in parts (E)-(G), we finish the sketch in the following figure. y=2 x=-1 :=1
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