Explanation Given: The given curve is as below. f ( x ) = y = ln sin x (1) Calculation: (a) Calculate the domain. The function f ( x ) is undefined for sin x =0 . Therefore, the domain is x ∈ ( 2 n π , π + 2 n π ) . (b) Calculate the intercepts. Calculate the y intercept. Substitute 0 for the value of x in equation (1). f ( x ) = ln sin x f ( 0 ) = ln sin 0 = ln 0 = ∞ The y intercept does not exist. Calculate the value of x intercept. Substitute 0 for y in the equation (1). ln sin x = 0 Here the term ln ( 1 ) = 0 . So sin x = 1 Therefore, x intercept occurs when the condition x = π 2 + 2 n π is true. (c) Calculate the symmetry. Substitute − x for x in the equation (1). f ( − x ) = ln sin ( − x ) ≠ f ( x ) ≠ − f ( x ) The function is neither even nor odd. Therefore, the curve is not symmetric about y axis and origin. (d) Calculate asymptotes. Apply limit of x tends to n π ( x → n π ) in the equation (1). lim x → n π f ( x ) = ln sin ( n π ) = ln sin ( n π ) = − ∞ It has the vertical asymptotes x = n π . Horizontal asymptotes do not exist. (e) Calculate the intervals. f ( x ) = ln sin x Differentiate the equation (1) with respect to x . f ' ( x ) = d ln ( sin x ) d sin x × d sin x d x = d ( sin x ) sin x f ' ( x ) = cot x (2) So equate equation (2) to zero. cot x = 0 x = cot − 1 ( 0 ) = π 2 + n π Consider the limit ( 0 , π )

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 49E

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Chapter 4.5, Problem 48E

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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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The curve by plotting x and y coordinates using guideline.

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