Explanation Graph: The given function is, g ( x ) = x x 2 − 4 (1) The domain of the function is the set of all real numbers except the number that is undefined on the function. As ± 2 are undefined on the given function. Thus, the domain of the given function is ( − ∞ , − 2 ) ∪ ( − 2 , 2 ) ∪ ( 2 , ∞ ) . It is known that y = g ( x ) . Substitute 0 for x in the given function to calculate the y -intercept. y = 0 ( 0 ) 2 − 4 y = 0 − 4 y = 0 Substitute 0 for y in the function to calculate the x -intercept. 0 = x x 2 − 4 0 = x x = 0 This shows that x -intercept is 0 and y -intercept is 0 . To determine the behavior of the function for large absolute values of x , take the limit of x to ∞ and − ∞ . Take the limit x approaches to ∞ . lim x → ∞ = x x 2 − 4 = ∞ Take the limit x approaches to − ∞ . lim x → − ∞ = x x 2 − 4 = ∞ Thus, the graph of g approaches to ∞ as x becomes arbitrarily large. Since, the function tends to ∞ for x = ± 2 . Thus the vertical asymptote is x = 2 and x = − 2 . Since, the function has no line for any value of x . Thus, there is no horizontal asymptote. Differentiate equation (1) with respect to x . g ′ ( x ) = ( x 2 − 4 ) d d t ( x ) − ( x ) d d t ( x 2 − 4 ) ( x 2 − 4 ) 2 g ′ ( x ) = ( x 2 − 4 ) ( 1 ) − ( x ) ( 2 x ) ( x 2 − 4 ) 2 g ′ ( x ) = x 2 − 4 − 2 x 2 ( x 2 − 4 ) 2 g ′ ( x ) = − x 2 − 4 ( x 2 − 4 ) 2 (2) This shows that for g ′ ( x ) = 0 , the critical point comes out to be ± 2 . Thus, the intervals are ( − ∞ , − 2 ) , ( − 2 , 2 ) and ( 2 , ∞ ) . Take the test points for different intervals in the table below to determine the signs of g ′ ( x ) . Intervals Test point c g ′ ( c ) Sign of g ′ ( x ) ( − ∞ , − 2 ) − 3 − 13 25 − ( − 2 , 2 ) 1 − 5 9 − ( 2 , ∞ ) 3 13 25 − Thus, the function decreases on the interval ( − ∞ , − 2 ) , ( − 2 , 2 ) and ( 2 , ∞ ) . The sign diagram of g ′ ( x ) is shown below in Figure 1. Figure 1 shows the sign diagram of g ′ ( x ) . Since the sign of g ′ ( x ) do not change from left to right in the whole interval. Thus, there is no critical number of g . Hence, there is no relative extrema. Differentiate equation (2) with respect to x . g ″ ( x ) = ( x 2 − 4 ) 2 d d t ( − x 2 − 4 ) − ( − x 2 − 4 ) d d t ( x 2 − 4 ) 2 ( x 2 − 4 ) 4 = ( x 2 − 4 ) 2 ( − 2 x ) − ( − x 2 − 4 ) ( 2 ) ( x 2 − 4 ) 2 − 1 d d t ( x 2 − 4 ) ( x 2 − 4 ) 4 = − 2 x ( x 2 − 4 ) 2 − 2 ( − x 2 − 4 ) ( x 2 − 4 ) 1 ( 2 x ) ( x 2 − 4 ) 4 = − 2 x ( x 2 − 4 ) 2 − 4 x ( − x 2 − 4 ) ( x 2 − 4 ) ( x 2 − 4 ) 4 Take ( x 2 − 4 ) common from numerator and cancel the common factors

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 4.3, Problem 54E

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Chapter 4.3, Problem 53E

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Chapter 4 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

Ch. 4.1 - Explain each of the following: a. f is increasing...Ch. 4.1 - Prob. 2CQ Ch. 4.1 - Prob. 3CQ Ch. 4.1 - Prob. 4CQ Ch. 4.1 - Prob. 5CQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E

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The graph of the function g ( x ) = x x 2 − 4 by curve-sketching guide.

Solution Summary: The author illustrates the graph of the function g(x)=x 2-4 by curve-sketching guide.

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