Explanation Definition used: If f ′ ( x ) > 0 on an interval ( a , b ) , then the graph of f ( x ) increases on that interval and if f ′ ( x ) < 0 , then f ( x ) decreases on that interval. Concavity: For the interval ( a , b ) , if f ″ ( x ) > 0 , then the function f is concave upward and if f ″ ( x ) < 0 , then f is concave downward. Calculation: The given function is f ( x ) = 2 x 2 + 5 4 − x 2 . The domain of the function is obtained as follows, Since, the given f ( x ) is a rational function, its domain is the set of all real numbers other than the roots of its denominator. Equate the denominator to 0 in order to obtain its undefined points, 4 − x 2 = 0 x 2 = 4 x = ± 2 The function is not defined at x = ± 2 . Thus, the domain of the function is all real numbers except −2 and 2. Substitute x = 0 for the function to obtain y -intercepts. f ( 0 ) = 2.0 + 5 4 − 0 = 5 4 Thus, the y -intercept is ( 0 , 5 4 ) . From the given function, it is noted that the numerator and the denominator have same degree. Thus, the horizontal asymptote can be obtained using a m x m b n x n . That is, 2 − 1 = − 2 . Thus, the horizontal asymptote is the line y = − 2 . Obtain the vertical asymptote of the function f ( x ) = 2 x 2 + 5 4 − x 2 by equating the denominator to zero. 4 − x 2 = 0 x = ± 2 Thus, the vertical asymptotes are x = − 2 and x = 2 . The derivative of the function f ( x ) is, f ′ ( x ) = d d x ( 2 x 2 + 5 4 − x 2 ) = ( 4 − x 2 ) ( 4 x ) − ( 2 x 2 + 5 ) ( − 2 x ) ( 4 − x 2 ) 2 = 16 x − 4 x 3 + 4 x 3 + 10 x ( 4 − x 2 ) 2 = 26 x ( 4 − x 2 ) 2 To obtain the critical points, find the points where the first derivative become zero or undefined. Equate f ′ ( x ) to 0 to obtain the critical number. 26 x ( 4 − x 2 ) 2 = 0 26 x = 0 x = 0 The derivative is zero when x = 0 and is not defined when x = ± 2 . Thus, the partition numbers are x = − 2 , x = 0 and x = 2 . The sign chart of f ′ ( x ) is as shown below in Table 1. Interval Test numbers f ′ ( x ) − ∞ < x < − 2 −4 f ′ ( − 4 ) = 26 ( − 4 ) ( 4 − ( − 4 ) 2 ) 2 = − 0.722 f ′ ( x ) < 0 decreasing − 2 < x < 0 −1 f ′ ( − 1 ) = 26 ( − 1 ) ( 4 − ( − 1 ) 2 ) 2 = − 2.888 f ′ ( x ) < 0 decreasing 0 < x < 2 1 f ′ ( 1 ) = 26 ( 1 ) ( 4 − ( 1 ) 2 ) 2 = 2.888 f ′ ( x ) > 0 increasing 2 < x < ∞ 4 f ′ ( 4 ) = 26 ( 4 ) ( 4 − ( 4 ) 2 ) 2 = 0

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

15th Edition

ISBN: 9781323047620

Author: Barnett, Ziegler, Byleen

Publisher: Pearson Education

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Chapter 4.4, Problem 48E

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Chapter 4.4, Problem 47E

Chapter 4.4, Problem 49E

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Chapter 4 Solutions

Calculus for Business, Economics, Life Sciences, and Social Sciences - Boston U.

Ch. 4.1 - Matched Problem 1 Repeat Example 1 for f(x) = x2 ...Ch. 4.1 - Matches Problem 2 Find the critical numbers of f,...Ch. 4.1 - Matched Problem 3 Find the critical numbers of f,...Ch. 4.1 - Matched Problem 4 Find the critical numbers of f,...Ch. 4.1 - Matched Problem 5 Find the critical numbers of f,...Ch. 4.1 - Prob. 6MP Ch. 4.1 - Matched Problem 7 Given f(x) = x3 9x2 + 24x 10,...Ch. 4.1 - Matched Problem 8 The graph in Figure 13...Ch. 4.1 - Matched Problem 9 The graph of the total revenue...Ch. 4.1 - Skills Warm-up Exercises In Problems 1-8, inspect...

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The graph of the function f ( x ) = 2 x 2 + 5 4 − x 2 by using graphing strategy.

Solution Summary: The author illustrates the graph of the function f(x) by using graphing strategy. Since it is a rational function, its domain is the set of all real numbers other than the roots of its de

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