Explanation Given: The function is f ( x ) = − 8 x 2 − 6 x − 7 . Definition used: (1) If the function gets larger and larger in magnitude without bound as x approaches the number k, then the line x = k is the vertical asymptote. (2) A function f ( x ) said to be symmetric about the origin if f ( − x ) = − f ( x ) . Calculation: The given function is f ( x ) = − 8 x 2 − 6 x − 7 . Domain: Consider the given function f ( x ) = − 8 x 2 − 6 x − 7 . Simplify the function as follows, f ( x ) = − 8 x 2 − 6 x − 7 = − 8 ( x + 1 ) ( x − 7 ) The function f ( x ) = − 8 x 2 − 6 x − 7 is defined at all the real number except the points x = − 1 and x = 7 . Thus, the domain of the function f ( x ) = − 8 x 2 − 6 x − 7 is ( − ∞ , − 1 ) ∪ ( − 1 , 7 ) ∪ ( 7 , ∞ ) . Symmetry: The given function is f ( x ) = − 8 x 2 − 6 x − 7 . Substitute x = − x in f ( x ) , f ( − x ) = − 8 ( − x ) 2 − 6 × ( − x ) − 7 = − 8 x 2 + 6 x − 7 Thus, f ( − x ) ≠ − f ( x ) . Therefore, from the definition stated above, the function does not have symmetry about the origin. Intercept: The function f ( x ) intercepts y -axis at x = 0 . Substitute the value x = 0 in f ( x ) , f ( 0 ) = − 8 ( 0 ) 2 − 6 × ( 0 ) − 7 = − 8 − 7 = 8 7 That is the function f ( x ) intercepts y -axis at the point ( 0 , 8 7 ) . The value of the function f ( x ) = − 8 x 2 − 6 x − 7 is never zero. Thus, the function f ( x ) does not intercept x -axis at any point. Asymptote: The function f ( x ) = − 8 x 2 − 6 x − 7 is not defined at the points x = − 1 and x = 7 . Thus, the function f ( x ) = − 8 x 2 − 6 x − 7 has vertical asymptotes at x = − 1 and x = 7 . The value of the function f ( x ) = − 8 x 2 − 6 x − 7 is never zero. Thus, the function f ( x ) has horizontal asymptotes at y = 0 . Critical point: The given function is f ( x ) = − 8 x 2 − 6 x − 7 . Differentiate f ( x ) with respect to x , f ′ ( x ) = 0 − ( − 8 ) ( 2 x − 6 ) ( x 2 − 6 x − 7 ) 2 = 16 ( x − 3 ) [ ( x + 1 ) ( x − 7 ) ] 2 For critical points, f ′ ( x ) = 0 . 16 ( x − 3 ) [ ( x + 1 ) ( x − 7 ) ] 2 = 0 16 ( x − 3 ) = 0 x = 3 Thus, the critical point for the function f ( x ) = − 8 x 2 − 6 x − 7 is x = 3 . Substitute x = 3 in the function f ( x ) = − 8 x 2 − 6 x − 7 to obtain the critical point as ( 3 , 1 2 ) . Relative Maxima and Relative Minima : The critical point for the function f ( x ) = − 8 x 2 − 6 x − 7 is at x = 3 . Substitute x = − 2 in f ′ ( x ) = 16 ( x − 3 ) [ ( x + 1 ) ( x − 7 ) ] 2 , f ′ ( − 2 ) = 16 ( − 2 − 3 ) [ ( − 2 + 1 ) ( − 2 − 7 ) ] 2 = 16 ( − 5 ) [ ( − 1 ) ( − 9 ) ] 2 = − 80 81 < 0 Thus, the function f ( x ) is decreasing on the interval ( − ∞ , − 1 ) . Substitute x = 0 in f ′ ( x ) = 16 ( x − 3 ) [ ( x + 1 ) ( x − 7 ) ] 2 , f ′ ( 0 ) = 16 ( 0 − 3 ) [ ( 0 + 1 ) ( 0 − 7 ) ] 2 = 16 ( − 3 ) [ ( 1 ) ( − 7 ) ] 2 = − 48 49 < 0 Thus, the function f ( x ) is decreasing on the interval ( − 1 , 3 ) . Substitute x = 6 in f ′ ( x ) = 16 ( x − 3 ) [ ( x + 1 ) ( x − 7 ) ] 2 , f ′ ( 6 ) = 16 ( 6 − 3 ) [ ( 6 + 1 ) ( 6 − 7 ) ] 2 = 16 ( 3 ) [ ( 7 ) ( − 1 ) ] 2 = 48 49 > 0 Thus, the function f ( x ) is increasing on the interval ( 3 , 7 )

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 13.4, Problem 16E

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Chapter 13 Solutions

Finite Mathematics and Calculus with Applications (10th Edition)

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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