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. Result used: The function f ( x ) is a rational function for which the degree of numerator is 1 more than the degree of denominator. Then, using long division method f ( x ) can be written as f ( x ) = m x + b + r ( x ) d ( x ) , where degree of r ( x ) is less than degree of d ( x ) . The line y = m x + b is an oblique asymptote for the graph of f ( x ) . Calculation: Domain: The given function is f ( x ) = x − 4 x 2 . The function can be rewritten as f ( x ) = x 3 − 4 x 2 The denominator becomes zero at x = 0 . So the function is not defined at 0. Thus, the domain of the function is all real numbers except 0. y- intercepts: The function f ( x ) = x 3 − 4 x 2 is not defined at the point x = 0 . So, there is no y -intercept for the given function. x- intercepts: Equate the given function f ( x ) to 0 and obtain its x -intercepts. x 3 − 4 x 2 = 0 x 3 − 4 = 0 x = 1.59 Thus, the x -intercept is ( 1.59 , 0 ) . Asymptotes: To prove that y = x is an oblique asymptote. Apply long division method to solve f ( x ) = x 3 − 4 x 2 . That is, f ( x ) = x − 4 x 2 As x tends to infinity, the function becomes y = f ( x ) = x . Thus, y = x is an oblique asymptote of the function f ( x ) = x − 4 x 2 . That is, the oblique asymptote is y = x The function is not defined at 0. So, as x tends to 0, the value of the function will tends to infinity. Thus, the vertical asymptote is x = 0 . First derivative test: The derivative of the function f ( x ) is, f ′ ( x ) = d d x ( x − 4 x 2 ) = 1 − 4 ( − 2 x 3 ) = 1 + 8 x 3 Equate f ′ ( x ) to 0 and obtain the critical numbers. x 3 + 8 = 0 x 3 = − 8 x = − 2 The critical number of f ′ ( x ) is x = − 2 . Since the function is undefined at x = 0 , its partition numbers are x = − 2 and x = 0 . The sign chart of f ′ ( x ) is as shown below in Table 1. Interval Test numbers f ′ ( x ) − ∞ < x < − 2 −3 f ′ ( x ) = 1 + 8 ( − 3 ) 3 = 0

Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

14th Edition

ISBN: 9780134668574

Author: Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen, Christopher J. Stocker

Publisher: PEARSON

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Chapter 4.4, Problem 65E

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To show: The line y=x is an oblique asymptote for the function f(x)=x−4x2 and sketch the graph.

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Chapter 4.4, Problem 64E

Chapter 4.4, Problem 66E

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Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)

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The line y = x is an oblique asymptote for the function f ( x ) = x − 4 x 2 and sketch the graph.

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To show: The line y=x is an oblique asymptote for the function f(x)=x−4x2 and sketch the graph.

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