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 ) = ( x − 1 ) 3 ( x + 3 ) . The given function is a polynomial and thus defined on all real values of x . So the domain of the function f ( x ) = ( x − 1 ) 3 ( x + 3 ) is all real numbers. Equate f ( x ) to 0 in order to obtain its x -intercept. ( x − 1 ) 3 ( x + 3 ) = 0 x − 1 = 0 or x + 3 = 0 x = 1 or x = − 3 Thus, the x -intercept are ( 1 , 0 ) and ( − 3 , 0 ) . Substitute x = 0 in the given function in the given function to obtain its y -intercept. f ( 0 ) = ( 0 − 1 ) 3 ( 0 + 3 ) = − 1 ( 3 ) = − 3 Thus, the y - intercept is ( 0 , − 3 ) . As x tends to infinity, the value of the function also tends to infinity and for every finite value the function also attains finite value. So, the given function f ( x ) = ( x − 1 ) 3 ( x + 3 ) has no asymptotes. Obtain the first derivative of the function f ( x ) = ( x − 1 ) 3 ( x + 3 ) . f ' ( x ) = d d x ( ( x − 1 ) 3 ( x + 3 ) ) = ( x − 1 ) 3 + 3 ( x + 3 ) ( x − 1 ) 2 = 3 x 3 + 3 x 2 − 15 x + 9 + x 3 − 3 x 2 + 3 x − 1 = 4 x 3 − 12 x + 8 Equate the first derivative to zero to obtain the partition points. 4 x 3 − 12 x + 8 = 0 4 ( x − 1 ) 2 ( x + 2 ) = 0 x = 1 or x = − 2 Partition number of f ′ ( x ) is x = 1 and x = − 2 . Thus, the critical number is x = 1 and x = − 2 . Sign chart of f ′ ( x ) is as shown below in Table 1. Interval Test numbers f ' ( x ) − ∞ < x < − 2 -3 − 64 decreasing − 2 < x < 1 0 8 increasing 1 < x < ∞ 2 16 increasing Table 1 From Table 1, it is observed that f ( x ) is decreasing on ( − ∞ , − 2 ) and increasing on ( − 2 , 1 ) and ( 1 , ∞ )

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, Problem 24RE

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To sketch: The graph of the function f(x)=(x−1)3(x+3) using graphing strategy.

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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 ) = ( x − 1 ) 3 ( x + 3 ) using graphing strategy.

Solution Summary: The author illustrates the graphing strategy of the function f(x): it is a polynomial and thus defined on all real values of x.

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To sketch: The graph of the function f(x)=(x−1)3(x+3) using graphing strategy.

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