Explanation It is given that, the function is y = x 3 − 6 x 2 − 15 x + 4 . Domain: It is known that, domain is the set of values of x for which f ( x ) is defined. It is observed that, the function has no undefined points nor domain constraints. Therefore, the domain of the function is ( − ∞ , ∞ ) . Intercepts: Set y = 0 and solve for x to find the x -intercepts as, x 3 − 6 x 2 − 15 x + 4 = 0 x = 0.2438 , x = − 2.0905 , x = 7.8467 Thus, the x -intercept is ( 0.2438 , 0 ) , ( − 2.0905 , 0 ) and ( 7.8467 , 0 ) The y -intercept is f ( 0 ) that is where the curve intersects the y -axis. Set x = 0 and solve for y to find the y -intercepts as, y = ( 0 ) 3 − 6 ( 0 ) 2 − 15 ( 0 ) + 4 = 4 Thus, the y -intercept is ( 0 , 4 ) . Symmetry: It is known that, the function is an even function if f ( − x ) = f ( x ) and the curve is symmetric about the y-axis and the function is an odd function if f ( − x ) = − f ( x ) and the curve is symmetric about the origin. Let the given function be represented as, y = f ( x ) that is y = x 3 − 6 x 2 − 15 x + 4 Replace x by – x in y = x 3 − 6 x 2 − 15 x + 4 as, y = ( − x ) 3 − 6 ( − x ) 2 − 15 ( − x ) + 4 = − x 3 − 6 x 2 + 15 x + 4 It is observed that, f ( − x ) ≠ − f ( x ) or f ( x ) . Therefore, the curve has no symmetry. Asymptotes: It is known that, the line y = L is an horizontal asymptote of the curve if lim x → ∞ f ( x ) = L and the line x = a is a vertical asymptote if one of the following is true, lim x → a + f ( x ) = ± ∞ and lim x → a − f ( x ) = ± ∞ . Note that, polynomial function of degree 1 or higher can’t have asymptotes. Thus, there are no vertical or horizontal asymptotes. Intervals of increase or decrease: Compute the first derivative f ′ ( x ) as, f ′ ( x ) = d d x ( x 3 − 6 x 2 − 15 x + 4 ) = 3 x 2 − 12 x − 15 Equate the first derivative f ′ ( x ) to 0 as, 3 x 2 − 12 x − 15 = 0 x = − 1 , x = 5 The points where f ′ ( x ) is 0 or undefined that is x = − 1 , x = 5 . Thus, the domain is split into intervals ( − ∞ , − 1 ) , ( − 1 , 5 ) , ( 5 , ∞ ) . Check the sign of f ′ ( x ) in the above intervals as shown in the below table. Intervals f ′ ( x ) = 3 x 2 − 12 x − 15 Increasing or decreasing ( − ∞ , − 1 ) f ′ ( − 2 ) = 3 ( − 2 ) 2 − 12 ( − 2 ) − 15 > 0 Increasing ( − 1 , 5 ) f ′ ( − 0 ) = 3 ( 0 ) 2 − 12 ( 0 ) − 15 < 0 Decreasing ( 5 , ∞ ) f ′ ( 6 ) = 3 ( 6 ) 2 − 12 ( 6 ) − 15 > 0 Increasing Table 1 Thus, the function is increasing on the interval ( − ∞ , − 1 ) , ( 5 , ∞ ) and the function is decreasing on the interval ( − 1 , 5 )

Essential Calculus: Early Transcendentals

2nd Edition

ISBN: 9781133112280

Author: James Stewart

Publisher: Cengage Learning

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Essential Calculus: Early Transcendentals

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Suppose f is a continuous function defined on a...Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...Ch. 4.1 - 710 Sketch the graph of a function f that is...

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The graph of the function y = x 3 − 6 x 2 − 15 x + 4 .

Solution Summary: The author describes the graph of the function y = x3-6 X2-15x+4. The function has no undefined points or domain constraints.

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