![]() Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License. Use the information below to generate a citation. Then you must include on every digital page view the following attribution: If you are redistributing all or part of this book in a digital format, Then you must include on every physical page the following attribution: Show that the difference between the maximum and mean shear stress in the web of an I-section beam is Q h 2 24 I where Q is the shear force on the cross-section, h is the depth of the web and I is the second moment of area of the cross-section about the neutral axis of bending. If you are redistributing all or part of this book in a print format, ![]() Want to cite, share, or modify this book? This book uses the Inertia for an Area or Second Moment of Area for typical cross section profiles. Likewise, because the function will have a vertical asymptote where each factor of the denominator is equal to zero, we can form a denominator that will produce the vertical asymptotes by introducing a corresponding set of factors. (An exception occurs in the case of a removable discontinuity.) As a result, we can form a numerator of a function whose graph will pass through a set of x-intercepts by introducing a corresponding set of factors. A rational function written in factored form will have an x-intercept where each factor of the numerator is equal to zero. Now that we have analyzed the equations for rational functions and how they relate to a graph of the function, we can use information given by a graph to write the function. Given the function f ( x ) = ( x + 2 ) 2 ( x − 2 ) 2 ( x − 1 ) 2 ( x − 3 ), f ( x ) = ( x + 2 ) 2 ( x − 2 ) 2 ( x − 1 ) 2 ( x − 3 ), use the characteristics of polynomials and rational functions to describe its behavior and sketch the function. Since the graph has no x-intercepts between the vertical asymptotes, and the y-intercept is positive, we know the function must remain positive between the asymptotes, letting us fill in the middle portion of the graph as shown in Figure 20. To sketch the graph, we might start by plotting the three intercepts. Calculus 7.7 Approximate Integration Asher Roberts 3.68K subscribers Subscribe 1.4K views 2 years ago Calculus My notes are available at (so you can write along with. This means there are no removable discontinuities.įinally, the degree of denominator is larger than the degree of the numerator, telling us this graph has a horizontal asymptote at y = 0. There are no common factors in the numerator and denominator. This occurs when x + 1 = 0 x + 1 = 0 and when x – 2 = 0, x – 2 = 0, giving us vertical asymptotes at x = –1 x = –1 and x = 2. To find the vertical asymptotes, we determine when the denominator is equal to zero. Scribd is the worlds largest social reading and publishing site. At each, the behavior will be linear (multiplicity 1), with the graph passing through the intercept. Section-7 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. ![]() Setting each factor equal to zero, we find x-intercepts at x = –2 x = –2 and x = 3. To find the x-intercepts, we determine when the numerator of the function is zero. ![]()
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