Explanation Given: The given curve is as below. f ( x ) = y = e x x 2 (1) Calculation: (a) Calculate the domain. Equate the denominator of equation (1) to zero. x 2 = 0 x = 0 Hence, the domain is all real numbers except zero. Therefore, the domain of the function is ( − ∞ , 0 ) ∪ ( 0 , ∞ ) . (b) Calculate the intercepts. Calculate the y –intercept. Substitute 0 for the value of x in equation (1). f ( x ) = e x x 2 f ( 0 ) = e 0 0 2 = ∞ The y –intercept does not exist. Calculate the value of x -intercept. Substitute 0 for y in the equation (1). 0 = e x x 2 x = ∞ The x intercept does not exist. (c) Calculate the symmetry. Substitute − x for x in the equation (1). f ( − x ) = e − x ( − x ) 2 = e − x x 2 ≠ f ( x ) ≠ − f ( x ) The function is neither even nor odd. Therefore, the curve has is not symmetric about y axis and origin. (d) Calculate asymptotes. Apply limit of x tends to ∞ ( x → ∞ ) in the equation (1). lim x → ∞ f ( x ) = e x x 2 = e ∞ ∞ 2 = ∞ Apply limit of x tends to 0 ( x → 0 ) in the equation (1). lim x → 0 f ( x ) = e x x 2 = e 0 0 2 = ∞ From either sides of zero, infinity occurs due to positive function. It has the vertical asymptote x = 0 . Apply limit of x tends to − ∞ ( x → − ∞ ) in the equation (1). lim x → − ∞ f ( x ) = e − ∞ − ∞ 2 = 0 It has horizontal asymptote y = 0 going to the left. (e) Calculate the intervals. f ( x ) = e x x 2 Differentiate the equation (1) with respect to x . f ' ( x ) = e x ( x 2 ) − 2 x ( e x ) x 4 = x e x − 2 e x x 3 f ' ( x ) = e x ( x − 2 ) x 3 (2) Equate equation (2) to zero. e x ( x − 2 ) x 3 = 0 e x ( x − 2 ) = 0 e x = 0 or ( x − 2 ) = 0 x = 2 Consider the limit ( − ∞ , 0 ) . Substitute − 1 for x in equation (2). f ' ( 1 ) = e − 1 ( − 1 − 2 ) − 1 3 = 1.103 > 0 The function f shows an increasing behavior for ( − ∞ , 0 ) . Consider the limit ( 0 , 2 ) . Substitute 1 for x in equation (2). f ' ( 1 ) = e 1 ( 1 − 2 ) 1 3 = − 2.71 < 0 The function f shows a decreasing behavior for ( 0 , 2 ) . Consider the limit ( 2 , ∞ ) . Substitute 3 for x in equation (2). f ' ( 3 ) = e 3 ( 3 − 2 ) 3 3 = 0

Single Variable Calculus: Early Transcendentals, Volume I

8th Edition

ISBN: 9781305270343

Author: James Stewart

Publisher: Cengage Learning

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Chapter 4.5, Problem 48E

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Chapter 4.5, Problem 47E

Chapter 4.5, Problem 49E

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Chapter 4 Solutions

Single Variable Calculus: Early Transcendentals, Volume I

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - Prob. 9E Ch. 4.1 - Prob. 10E

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