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 ) = ln ( x 2 + 4 ) . The function is defined at all points. So, the domain of the function is ( − ∞ , ∞ ) . The function never becomes zero. So there is no x -intercept for the function. Substitute x = 0 for the function to obtain y -intercepts. f ( 0 ) = ln ( 0 + 4 ) = ln ( 4 ) Thus the y -intercept is ( 0 , ln 4 ) . Since, lim x → ∞ ( ln ( x 2 + 4 ) ) = ∞ , it has no horizontal asymptote Since the function has no undefined at the points, it has no vertical asymptote. The derivative of the function f ( x ) is, f ′ ( x ) = d d x ( ln ( x 2 + 4 ) ) = 2 x x 2 + 4 Equate f ′ ( x ) to 0 to obtain the critical number. 2 x x 2 + 4 = 0 2 x = 0 x = 0 The critical point of f ( x ) is x = 0 . The sign chart of f ′ ( x ) is as shown below in Table 1. Interval Test numbers f ′ ( x ) − ∞ < x < 0 −1 − 2 5 decreasing 0 < x < ∞ 1 2 5 increasing Table 1 From Table 1, it is observed that f ( x ) is decreasing on ( − ∞ , 0 ) and increasing on ( 0 , ∞ ) . Thus, the graph of the function has a Local minimum at x = 0 . Substitute x = 0 in the function to obtain its minimum value, f ( 0 ) = ln ( 0 2 + 4 ) = ln ( 4 ) Local minimum of f ( x ) is ( 0 , ln ( 4 ) ) . The second derivative of the function f ( x ) is, f ′ ′ ( x ) = d d x ( 2 x x 2 + 4 ) = 2 ( − x 2 + 4 ) ( x 2 + 4 ) 2 Equate f ′ ′ ( x ) to 0 to obtain the critical number

Pearson eText for College Mathematics for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

14th Edition

ISBN: 9780137553341

Author: Raymond Barnett, Michael Ziegler

Publisher: PEARSON+

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Differential Equations

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Chapter 11.4, Problem 34E

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To sketch: The graph of the function f(x)=ln(x2+4) by using graphing strategy.

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Chapter 11.4, Problem 33E

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Chapter 11 Solutions

Pearson eText for College Mathematics for Business, Economics, Life Sciences, and Social Sciences -- Instant Access (Pearson+)

Ch. 11.1 - Matched Problem 1 Repeat Example 1 for f(x) = x2 –...Ch. 11.1 - Matches Problem 2 Find the critical numbers of f,...Ch. 11.1 - Matched Problem 3 Find the critical numbers of f,...Ch. 11.1 - Prob. 4MP Ch. 11.1 - Prob. 5MP Ch. 11.1 - Prob. 6MP Ch. 11.1 - Prob. 7MP Ch. 11.1 - Prob. 8MP Ch. 11.1 - Prob. 9MP Ch. 11.1 - Skills Warm-up Exercises In Problems 1-8, inspect...

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

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Chapter 11 Solutions