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# Curve sketching Use the guidelines of this | Ch 4 - 14RE ISBN: 9780321570567 2

## Solution for problem 14RE Chapter 4

Calculus: Early Transcendentals | 1st Edition

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Problem 14RE

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) = x +3

Step-by-Step Solution:

Solution Step 1 In this problem we need to make a complete graph of f(x) = 2x in its domain or in the x +3 given interval. Since the interval is not mentioned, we take the domain. Here in the given function equate denominator to 0, we get x =± 3i, therefore the domain is {x; x = / ± 3i} . 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) = 0 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 x in the neighborhood of the point p. Step 2 Given f(x) = x +3 Since this function consists of odd function divided by even function, and we know that the product of even and odd function is odd. Therefore the graph is symmetric about origin. Now let us find the critical points. Consider f(x) = x +3 (x +3)3(3x)(2x) f (x) = 2 2 (x +3) = 3x +96x2= 3x +9 (x+3)2 (x+3)2 3x +9 Thus f (x) = (x +3)2 Now by the definition of critical points given in step 1, we equate f’(x) to 0. f(x) = 0 3x +9 (x +3) = 0 3x +9 = 0 3x = 9 x = 3 x = ± 3 Thus the critical points are x = 3, 3 Step 3 Now let us find the inflection points. By definition the necessary condition is f (x) = 0 3x +9 We have f (x) = (x +3)2 2 (x+3) (6x) (3x +9)(x+3) f (x) = (x +3) Now f (x) = 0, we get 2 2 2 2 (x +3) (6x) (3x +9)2(x +3)= 0 (x +3) 3 6x2543 = 0 (x +3) 3 6x 54x = 0 6x(x 9) = 0 2 x(x 9) = 0 x = 0and x = ±3 Thus the inflection points are x = 3,0,3

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##### ISBN: 9780321570567

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