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Sketch the graph of f(x)= A. The domain is R. B. C. D. The x- and y-intercepts are both Since f(-x) = -f(x), fis ---Select--- and its graph is symmetric about the origin. f(x) = Since x² + 7 is never 0, there is no vertical asymptote. Since f(x) → ∞o as x → ∞o and f(x)→-coas x→∞o, there is no horizontal asymptote. But long division gives us the following. f(x) - = - X = = x² + 7 x³ +7 So the line y = = X- x² + 7 1 + x² + 7 as x ±0o is a slant asymptote.

Sketch the graph of f(x)= A. The domain is R. B. C. D. The x- and y-intercepts are both Since f(-x) = -f(x), fis ---Select--- and its graph is symmetric about the origin. f(x) = Since x² + 7 is never 0, there is no vertical asymptote. Since f(x) → ∞o as x → ∞o and f(x)→-coas x→∞o, there is no horizontal asymptote. But long division gives us the following. f(x) - = - X = = x² + 7 x³ +7 So the line y = = X- x² + 7 1 + x² + 7 as x ±0o is a slant asymptote.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 64E
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f(x) = x^3/x^2+7

 
 
Sketch the graph of f(x)= A. The domain is R. B. C. D. E. F. G. The x- and y-intercepts are both Since f(-x) = -f(x), fis--Select--- and its graph is symmetric about the origin. f(x) = Since x² + 7 is never 0, there is no vertical asymptote. Since f(x) → ∞o as x → ∞o and f(x) →-coas x→-co, there is no horizontal asymptote. But long division gives us the following. f(x) - - X = = = So the line y = f'(x) = x² + 7 x³ +7 f"(x) = = X- x² + 7 1 + x² + 7 3x²(x² + 7) x³. 2x (x² + 7)² x²(x² +21) (x² + 7)² Since f'(x) > 0 for all x (except 0), f is increasing on (-00, 00). Although f'(0) = as x ±0o is a slant asymptote. = ,f'(x) does not change sign at 0, so there is no local maximum or minimum. (4x³ + 42x)(x² + 7)² − (x4 + 21x²) · 2(x² + 7)2x (+²+7)4 14x(21-x²) (+²+7)³
Transcribed Image Text:Sketch the graph of f(x)= A. The domain is R. B. C. D. E. F. G. The x- and y-intercepts are both Since f(-x) = -f(x), fis--Select--- and its graph is symmetric about the origin. f(x) = Since x² + 7 is never 0, there is no vertical asymptote. Since f(x) → ∞o as x → ∞o and f(x) →-coas x→-co, there is no horizontal asymptote. But long division gives us the following. f(x) - - X = = = So the line y = f'(x) = x² + 7 x³ +7 f"(x) = = X- x² + 7 1 + x² + 7 3x²(x² + 7) x³. 2x (x² + 7)² x²(x² +21) (x² + 7)² Since f'(x) > 0 for all x (except 0), f is increasing on (-00, 00). Although f'(0) = as x ±0o is a slant asymptote. = ,f'(x) does not change sign at 0, so there is no local maximum or minimum. (4x³ + 42x)(x² + 7)² − (x4 + 21x²) · 2(x² + 7)2x (+²+7)4 14x(21-x²) (+²+7)³
H. Since f"(x) = 0 when x = 0 or x = ± √/21, we set up the following chart. Interval x < -√21 -√21 < x < 0 0 < x < √21 √21 < x and (x, y) = 7x 21 − x²(x² + 7)³ - - The points of inflection are (x, y): ----- + + + (larger x-value). f"(x) + + The graph of f is sketched in the following figure. 5 * -5 5 -5 X CU on (-∞, -√21) @ CD on (-√21, 0) (smaller x-value), (0, 0), CU on (0, √21) CD on (√21, ∞)
Transcribed Image Text:H. Since f"(x) = 0 when x = 0 or x = ± √/21, we set up the following chart. Interval x < -√21 -√21 < x < 0 0 < x < √21 √21 < x and (x, y) = 7x 21 − x²(x² + 7)³ - - The points of inflection are (x, y): ----- + + + (larger x-value). f"(x) + + The graph of f is sketched in the following figure. 5 * -5 5 -5 X CU on (-∞, -√21) @ CD on (-√21, 0) (smaller x-value), (0, 0), CU on (0, √21) CD on (√21, ∞)
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