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Solution: Curve sketching Use the guidelines of this chapter

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 13RE Chapter 4

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 13RE

Curve sketching? Use the guidelines of this chapter to make a complete graph of the following functions on their domains or on the given interval. Use a graphing utility to check your work. f(x) = x4? 3x + 4x + 1 2

Step-by-Step Solution:

Solution Step 1 x 2 In this problem we need to make a complete graph of f(x) = 2 3x + 4x + 1in its domain or in the given interval. Since the interval is not mentioned, we take the domain. Here given function is a polynomial, therefore the domain is ( ,). In order to sketch the complete graph, we need to find the critical points, inflection points, local maximum and local minimum if possible. First let us see the definitions: Critical point: An interior point cof the domain of a function f at which f (c) = 0or f c)fails to exist is called a critical point of f Inflection Point: An inflection point is a point on a curve at which the sign of the curvature (i.e., the concavity) changes. Inflection points may be stationary points, but are not local maxima or local minima. A necessary condition for x to be an inflection point is f (x) = Local maximum: Let f be function defined on an interval [a,b]and let pbe a point in the open interval (a,b). Then the function f has local maximum at pif f(p) f(x)for all xin the neighborhood of the point p. Local minimum: Let f be function defined on an interval [a,b]and let pbe a point in the open interval (a,b). Then the function f has local minimum at pif f(p) f(x)for all xin the neighborhood of the point p. Step 2 Given f(x) = x4 3x + 4x + 1 2 Since this function consists of odd and even terms, the graph is symmetric about origin. Now let us find the critical points. Consider f(x) = x4 3x + 4x + 1 2 3 f (x) = 4x 6x + 4 = 2x 6x + 4 2 Now by the definition of critical points given in step 1, we equate f’(x) to 0. f x) = 0 2x 6x + 4 = 0 Now we have to make the above equation to a quadratic equation. By substituting 1 for x, we get 2(1) 6(1) + 4 = 2 6 + 4 = 0 Therefore (x 1) is a factor. Hence factorize by inspection we get, 2x 6x + 4 = (x 1)(x 1)(2x + 4) That is when f (x = 0we get x = 1,1, 2 Thus the critical points are x = 2,1 Step 3 Now let us find the inflection points. By definition the necessary condition is f (x) 0 3 We have f (x)= 2x 6x + 4 2 f ( = 6x 6 f ) = 0 6x 6 = 0 6(x 1) = 0 x 1= 0 x = 1 x = ± 1 Thus the inflection points are x = 1,1

Step 4 of 6

Chapter 4, Problem 13RE is Solved
Step 5 of 6

Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Solution: Curve sketching Use the guidelines of this chapter

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