Explanation The given function is f ( x ) = 2 x 3 − 9 x 2 + 12 x − 1 . The above function is polynomial function. Therefore, the domain of the function is ( − ∞ , ∞ ) . Substitute 0 for x in the given function to evaluate the y -intercept. f ( 0 ) = 0 − 0 + 0 − 1 = − 1 Since, the function is cubic polynomial, x intercept is not used. Since the function is polynomial, there are no vertical or horizontal asymptotes. Differentiate the function f and equate to 0, to evaluate the critical point. d d x ( 2 x 3 − 9 x 2 + 12 x − 1 ) = 0 6 x 2 − 18 x + 12 = 0 x 2 − 3 x + 2 = 0 ( x − 1 ) ( x − 2 ) = 0 The critical points are x = 1 and x = 2 , so the intervals are ( − ∞ , 1 ) , ( 1 , 2 ) and ( 2 , ∞ ) . To check the sign of f ′ , check the test point of the function f ′ . The table below shows the test value. Intervals Test point c f ′ ( c ) Sign of f ′ ( − ∞ , 1 ) 0 12 + ( 2 , ∞ ) 3 12 + ( 1 , 2 ) 1.5 − 1.5 − Sign diagram of f ′ is shown below in Figure 1. So, the sign diagram of f ′ shows that f is decreasing on interval ( 1 , 2 ) and increasing on interval ( − ∞ , 1 ) ∪ ( 2 , ∞ ) . Figure 1 shows the function changes its sign from negative to positive at x = 2 . So, the relative minima occurs at x = 2 and function changes its sign form positive to negative at x = 1 . So, the relative maxima occurs at x = 1 . Substitute 2 for x in f ( x ) = 2 x 3 − 9 x 2 + 12 x − 1 , to evaluate the value of relative minima. f ( 2 ) = 2 ⋅ ( 2 ) 3 − 9 ⋅ ( 2 ) 2 + 12 ( 2 ) − 1 = 16 − 36 + 24 − 1 = 3 So, the relative minima of f is f ( 2 ) = 3 . Substitute 1 for x in f ( x ) = 2 x 3 − 9 x 2 + 12 x − 1 , to evaluate the value of relative maxima. f ( 1 ) = 2 ⋅ ( 1 ) 3 − 9 ⋅ ( 1 ) 2 + 12 ( 1 ) − 1 = 2 − 9 + 12 − 1 = 4 So, the relative maxima of f is f ( 1 ) = 4

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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Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter | (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) ☐ 1. For all n > 1, seriesΣ In(n) In(n) converges. 2, 1, arctan(n) the series arctan(n) n³ ☐ 4. For all n > 1, 123 converges. 1 n ln(n) series In(n) diverges. 2n . and the seriesΣconverges, so by the Comparison Test, 2, 3, and the series converges, so by the Comparison Test, the series-3 1 converges. ☐ 6. For all n > 2, In(n) >, and the series Σ converges, so by the Comparison Test, the seriesΣ In(n) converges.

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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 f ( x ) = 2 x 3 − 9 x 2 + 12 x − 1 .

Solution Summary: The author illustrates the graph of the function f(x). Since it is a cubic polynomial function, x intercept is not used.

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To sketch: The graph of the function f(x)=2x3−9x2+12x−1.

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