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 ??x + 2

Solution Step 1 3 In this problem we need to make a complete graph of f(x) = x x2in ts domain or in the given interval. Since the interval is not mentioned, we take the domain. The domain of any polynomial is [0,). 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 3 Given f(x) = x +2 There is no symmetry because there are points x in the domain for which -x is not in the domain. Now let us find the critical points. 3 Consider f(x) = x x2 Now by the definition of critical points given in step 1, we equate f’(x) to 0. Putting f (x) = 0we get, Thus the critical point is 0.0878. Step 3 Now let us find the inflection points. By definition the necessary condition is f (x) = 0 1 2 1 1 We have f (x) = x3 x 2 2 Now f (x) = 0, we get Thus the inflection point is 0.4933