Explanation Given Information: The provided function is, f ( x ) = x 3 + 6 x 2 + 9 x + 3 Formula used: A vertical asymptote of a function f ( x ) is a line x = a such that f ( a ) = ∞ . A vertical asymptote of a rational function h ( x ) = f ( x ) g ( x ) is x = a where g ( a ) = 0 and f ( a ) ≠ 0 . A horizontal asymptote of a function f ( x ) is a line y = b such that lim x → ∞ f ( x ) = b or lim x → − ∞ f ( x ) = b . A horizontal asymptote of a rational function h ( x ) = f ( x ) g ( x ) is 1. A line y = 0 if the degree of the numerator is less than the degree of the denominator. 2. The line y is ratio of the leading coefficients if the degree of the numerator is equal to the degree of the denominator. 3. Does not exist if the degree of the numerator is greater than the degree of the denominator. To find relative maxima and minima of a function, 1. Set the first derivative of the function to zero, f ' ( x ) = 0 , to find the critical values of the function. 2. Substitute the critical values into f ( x ) and calculate the critical points. 3. Calculate the sign of the function to the left and right of the critical values. (a) If f ′ ( x ) < 0 and f ′ ( x ) > 0 on the left and right side respectively of a critical value, then the point is relative minimum. (b) If f ′ ( x ) > 0 and f ′ ( x ) < 0 on the left and right side respectively of a critical value, then the point is relative maximum. Calculation: Consider the provided function, f ( x ) = x 3 + 6 x 2 + 9 x + 3 Now, calculate the first derivative of above with respect to x and solve: f ′ ( x ) = 3 x 2 + 12 x + 9 = 3 ( x 2 + 4 x + 3 ) = 3 ( x 2 + 3 x + x + 3 ) = 3 ( x + 3 ) ( x + 1 ) Now, to obtain the critical values, set f ′ ( x ) = 0 as, 3 ( x + 3 ) ( x + 1 ) = 0 Thus, either ( x + 1 ) = 0 or ( x + 3 ) = 0 . First consider ( x + 1 ) = 0 . Subtract 1 from both sides. x + 1 − 1 = 0 − 1 x = − 1 And, Subtract 3 from both sides. x + 3 − 3 = 0 − 3 x = − 3 Thus, the critical values of the function are at x = − 1 and x = − 3 . Now, substitute − 1 for x in the equation f ( x ) = x 3 + 6 x 2 + 9 x + 3 , f ( x ) = ( − 1 ) 3 + 6 ( − 1 ) 2 + 9 ( − 1 ) + 3 = − 1 + 6 − 9 + 3 = − 1 Substitute − 3 for x in the equation f ( x ) = x 3 + 6 x 2 + 9 x + 3 , f ( x ) = ( − 3 ) 3 + 6 ( − 3 ) 2 + 9 ( − 3 ) + 3 = − 27 + 54 − 27 + 3 = 3 Thus, the critical points are ( − 1 , − 1 ) and ( − 3 , 3 ) . Now, calculate the sign of the y ′ on either side of the critical point. First consider x = − 1 . Calculate f ′ ( x ) at x = − 2 . f ′ ( x ) = 3 ( x + 3 ) ( x + 1 ) = 3 ( − 2 + 3 ) ( − 2 + 1 ) = 3 ( 1 ) ( − 1 ) = − 3 Thus, f ′ ( x ) < 0 on left of x = − 1

Mathematical Applications for the Management, Life, and Social Sciences

11th Edition

ISBN: 9781305108042

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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To calculate: The relative minimum, relative maximum, points of inflection and asymptotes for the function f(x)=x3+6x2+9x+3 and sketch its graph.

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Chapter 10 Solutions

Mathematical Applications for the Management, Life, and Social Sciences

Ch. 10.1 - 1. The x- values of critical points are found...Ch. 10.1 - Prob. 2CP Ch. 10.1 - Prob. 3CP Ch. 10.1 - Prob. 4CP Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - 3. Use the graph of in Problem 1 to identify at...Ch. 10.1 - 4. Use the graph of in Problem 2 to identify at...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...

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The relative minimum, relative maximum , points of inflection and asymptotes for the function f ( x ) = x 3 + 6 x 2 + 9 x + 3 and sketch its graph.

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To calculate: The relative minimum, relative maximum, points of inflection and asymptotes for the function f(x)=x3+6x2+9x+3 and sketch its graph.

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