Lecture for Week 11 Secs 5 1 3 Analysis of Functions We used to call this topic curve sketching before students could sketch curves by typing formulas into their calculators It is still important to understand what derivatives tell us about the qualitative and geometrical behavior of functions and their graphs 1 Here are the things to look for when analyzing a function or sketching its graph Domain any points of discontinuity Intercepts Symmetry Asymptotes Intervals of increase and decrease Local extrema maxima and minima Inflection points and intervals of concavity 2 I will approach this subject by doing examples each of which illustrates the full set of concepts and theorems involved rather than giving a theoretical discussion of each concept in turn as the book needs to do Example 1 Discuss f x x3 3x with respect to whatever concepts apply from the list on the previous slide 3 y f x x3 3x x x2 3 It s a polynomial so it s defined and continuous everywhere with no asymptotes All the powers are odd so the function is odd f x f x which implies that the graph is symmetric through the origin The horizontal intercepts occur when 0 y x 3 x2 or x 0 and x 3 1 7 The vertical intercept is just f 0 0 4 So far I haven t mentioned derivatives Let s calculate them now f x 3x2 3 3 x2 1 3 x 1 x 1 f x 6x The critical numbers are the places where f equals 0 or doesn t exist but that can t happen in this example Here the critical numbers are 1 The only zero of the second derivative is x 0 this is a potential inflection point 5 It is useful to make a table listing the signs of f f and f on each of the intervals into which the real line is divided by the horizontal intercepts critical numbers potential inflection points and vertical asymptotes To see the signs easily it s good to write the functions in factored form f x x x 3 x 3 f x 3 x 1 x 1 f x 6x A function changes sign where a factor does 6 x 3 0 1 1 3 f f f 7 We see that f is decreasing as x varies from to 1 increasing from 1 to 1 then decreasing again Calculate f 1 2 f 1 2 to see where to plot the minimum and maximum points We also see that f changes from positive to negative at x 0 so that is indeed an inflection point The graph is concave up on the left side and down on the right 8 With this information you can easily sketch the graph freehand much more easily than I can type it Example 2 2x2 Discuss f x 2 x x 2 9 2x2 f x x 1 x 2 The most obvious feature is the vertical asymptotes at x 1 and x 2 Those are the only two points where f is undefined or discontinuous There is also a horizontal asymptote y 2 There are no obvious symmetries 10 Use the quotient rule and simplify to find 2x x 4 f x x 1 2 x 2 2 So the critical numbers not counting the asymptotes are x 0 and x 4 f is a mess in this problem so let s see how much we can do without it Again make a table 11 4 x 0 1 2 f f Minimum f 4 16 9 1 8 Maximum f 0 0 12 The function is increasing on the interval 4 1 and on the interval 1 0 At x 1 a vertical asymptote it jumps from to Therefore it is not correct to say that it is increasing from x 4 to x 0 even though our table has only plus signs for f in that range Similarly there is a jump from to at the other asymptote 13 Clearly the graph must be mostly concave down between the asymptotes and mostly concave up outside them However because the minimum value of 1 8 is below the horizontal asymptote at 2 the concavity must be downward as x Therefore there must be an inflection point somewhere to the left of x 4 There may be other inflection points an even number in each of the three intervals delimited by the vertical asymptotes but probably not 14 y 15 x Example 3 Discuss f x x2 3 x y 2 2 16 f x x2 3 x 7 2 2 This function is positive everywhere It is continuous but the derivative will be discontinuous at 0 f x 23 x 1 3 x 7 2 2x2 3 x 7 Remember that our strategy is to write f as a product factor it so its zeros will be obvious 17 Use x2 3 x 1 3 x 1 x 7 3x 3x1 3 23 x 1 3 4x 7 x 7 f x 2 x 7 So the critical numbers are x 7 and x 47 the zeros of f and x 0 a cusp f is not continuous there but f is To get the second derivative it is useful to 18 multiply f out again f x 23 x 1 3 4x2 35x 49 f x 92 x 4 3 4x2 35x 49 32 x 1 3 8x 35 92 x 4 3 20x2 70x 49 The roots of f are 2 70 42 5 70 70 80 49 x 40 40 19 How did I do that arithmetic Break things into prime factors 2 2 2 2 2 4 2 2 70 80 49 7 2 5 7 2 5 7 2 5 5 4 2 2 2 2 7 2 3 5 42 5 Conclusion f x 0 at two places x 0 6 and x 4 1 Furthermore f is undefined at x 0 so that also is a potential inflection point 20 We need to find the sign of f in the intervals between critical numbers Note that at each such number 0 47 7 one of the three factors x 1 3 4x 7 x 7 changes from negative to positive Thus f is increasing on 0 47 and 7 and decreasing on 0 and 47 7 It therefore has minima at 0 and 7 and a maximum at 47 Now note that f has the form x 4 3 ax2 bx c with a 0 so f is negative precisely 21 in the interval between its two roots Since 4 is even x 4 3 0 So f is concave up on 0 6 and 4 1 and concave down on 0 6 0 and 0 4 1 Put the slope and concavity information together At x 0 the graph has a sharp point cusp The graph reverses itself there from decreasing to increasing but on both sides the concavity is down not up as it would be at a typical smooth minimum 22 Let s plot the points corresponding to the critical numbers and the possible inflections …
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