Explanation Given: The given equation is as below. y = x 3 + 1 3 (1) Calculation: a) Calculate the domain as follows. From the equation (1), the function is defined for all real values of x (there is no restriction on the value of x ). Hence, the domain of f ( x ) is all real values of x . For the polynomial the domain is ( − ∞ , ∞ ) . b) Calculate the intercepts as below. Calculate the value of x -intercept. Substitute 0 for y in the equation (1). x 3 + 1 3 = 0 x 3 + 1 = 0 x 3 = − 1 x = − 1 Hence, x -intercept is ( − 1 , 0 ) . Calculate the y -intercept. Substitute 0 for x in the equation (1). f ( 0 ) = 0 3 + 1 3 = 1 Therefore, the y -intercept is ( 0 , 1 ) . c) Calculate the symmetry as follows. Apply negative and positive values for x in the equation (1). Substitute − 1 for x in the equation (1). f ( − 1 ) = ( − 1 ) 3 + 1 3 = 0 Substitute + 1 for x in the equation (1). f ( − 1 ) = ( 1 ) 3 + 1 3 = 2 3 = 1.26 Here f ( − x ) ≠ f ( x ) , hence it has no symmetry. d) Calculate asymptotes as follows. Apply limit x → ∞ in the equation (1). lim x → ∞ y = ∞ Apply limit x → − ∞ in the equation (1). lim x → − ∞ y = − ∞ Here, the value of limit gets infinity; this implies there is no horizontal asymptote or vertical asymptote. e) Calculate the intervals as below. Differentiate the equation (1) with respect to x Apply the chain rule. d f ( u ) d x = d f d u d u d x Substitute u 3 for f and x 3 + 1 for u in the above equation. d f ( u ) d x = d d u ( u 3 ) d d x ( x 3 + 1 ) = d d u ( u 1 3 ) d d x ( x 3 + 1 ) = 1 3 u ( 1 3 − 1 ) ( 3 x 2 ) = 1 3 u − 2 3 ( 3 x 2 ) = x 2 u 2 3 Substitute x 3 + 1 for u in the above equation. f ' ( x ) = x 2 ( x 3 + 1 ) 2 3 (2) Substitute 0 for f ' ( x ) in the equation (2). x 2 ( x 3 + 1 ) 2 3 = 0 x 2 = 0 x = 0 Take the interval as ( − ∞ , ∞ ) . Here, f ' ( x ) > 0 for all values of x except for x = − 1 and x = 0 where the function f is not differentiable. Hence the function f ( x ) is increasing in the interval ( − ∞ , ∞ ) except for x = − 1 and x = 0 where the function f is not differentiable. f) The function f ( x ) is increasing on the entire domain and there are no local maxima or minima. g) Calculate the concavity as follows. Differentiate the equation (2) with respect to x using the quotient rule. ( u v ) ' = u ' v − v ' u v 2 Substitute 2 x for u and 3 ( x 2 − 1 ) 2 3 for v in the above formula. ( x 2 ( x 3 + 1 ) 2 3 ) ' = ( ( 2 x ) ( x 3 + 1 ) 2 3 − ( 2 x 2 x 3 + 1 3 ) ( x 2 ) ( ( x 3 + 1 ) 2 3 ) 2 ) = ( ( 2 x ) ( x 3 + 1 ) 2 3 − ( 2 x 4 x 3 + 1 3 ) ( x 3 + 1 ) 4 3 ) = ( ( 2 x ) ( x 3 + 1 ) 2 3 ( x 3 + 1 ) 1 3 − 2 x 4 ( x 3 + 1 ) 1 3 ( x 3 + 1 ) 4 3 ) f ' ' ( x ) = ( ( 2 x ) ( x 3 + 1 ) − 2 x 4 ( x 3 + 1 ) 5 3 ) = ( 2 x 4 + 2 x − 2 x 4 ( x 3 + 1 ) 5 3 ) = 2 x ( x 3 + 1 ) 5 3 f ' ' ( x ) = 2 x ( x 3 + 1 ) 5 3 (3) Here f ' ' ( x ) > 0 , when x < − 1 or x > 0 and f ' ' ( x ) < 0 when − 1 < x < 0

Single Variable Calculus: Early Transcendentals, Volume I

8th Edition

ISBN: 9781305270343

Author: James Stewart

Publisher: Cengage Learning

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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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