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 for-mulas into their calculators. It is still important tounderstand what derivatives tell us about the qual-itative and geometrical behavior of functions andtheir graphs.)1Here are the things to look for when analy z-ing a function (or sketching its graph).Domain; any points of discontinuityInterceptsSymmetryAsymptotesIntervals of increase and d ecreaseLocal extrema (maxima and minima)Inflection points and intervals of concavity2I will approach this subject by d oi ng exam-ples, each of which illustrates t he fu l l set of con-cepts and theorems involved, rather than givinga th eoretical d i scussion of each concept in turnas the book needs to do.Example 1Discuss f(x) = −x3+ 3x (with respect to what-ever c on cepts apply from th e list on the p rev i ou sslide).3y = f(x) = −x3+ 3x = −x(x2− 3).It’s a polynomial, so it’s defi ned and continuouseverywhere, with no asymptotes. All the pow-ers are odd, so the functi on is odd [f(−x) =−f(x)], which implies that the graph is symmet-ric through the origin. The horizontal interceptsoccur when 0 = y = x(3 − x2), or x = 0 andx = ±√3 ≈ ±1.7. The vertical intercept is j ustf(0) = 0.4So far I haven’t mentioned derivatives. Let’scalculate th em 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 hap-pen in this example). Here the critical numbersare ±1. The only zero of the second derivative isx = 0 ; this is a potential inflec ti on point.5It is usefu l to make a table listing the signsof f, f′, and f′′on each of the intervals intowhich the real line is divided by the horizontalintercepts, critical numbers, potential inflectionpoints, and vertical asymptotes. To see the signseasily, it’s good to write th e fun ctions in factoredform:f(x) = −x(x −√3)(x +√3),f′(x) = −3(x − 1)(x + 1), f′′(x) = −6x.A func ti on ch an ges sign where a factor doe s:6xff′f′′−√3−10 1√3+ + ++ ++ + + − − −− − − −− − −7We see that f is decreasing as x varies from−∞ to −1, increasing from −1 to +1, then de-creasing again. Calculate f(−1) = −2, f(1) = 2,to see where to plot the minimum and maximumpoints.We also see that f′′ch an ges from positi ve tonegative at x = 0, so that is indeed an i nflectionpoint. The graph is concave u p on the left sideand down on the right.8With this information you can easily s ketchthe graph freehand ( much more easily than I cantype it).Example 2Discuss f(x) =2x2x2− x − 2.9f(x) =2x2(x + 1)(x − 2).The most obv i ou s f eature is the vertical asy mp-totes at x = −1 an d x = 2. (Those are theonly two points where f is undefined or discon-tinu ous.) The re is also a horizontal asymptote ,y = 2. There are no obvious symmetries.10Use the quotient rule and simpl i fy to findf′(x) =−2x(x + 4)(x + 1)2(x − 2)2.So the critical nu mbers (not counting the asymp-totes) are x = 0 and x = −4.f′′is a mess in th i s p roblem, so le t’ s see howmuch we can do without it. Again make a tab l e:11xff′−4 −10 2+ + ++ +− − −− −Minimum: f(−4) = 16/9 ≈ 1.8.Maximum: f(0) = 0.12The function is increasing on the interval[−4, −1] an d on the interval [−1, 0]. At x = −1(a vertical asymptote) it jumps from +∞ to −∞.(Therefore, i t is not correct to say that it is in-creasing f rom x = −4 to x = 0, even though ourtable h as only p l us signs for f′in that range.)Similarly, there is a ju mp from −∞ to +∞ atthe other asymptote.13Clearly, the graph must be mostly concavedown between the asymptotes and mostly con-cave up outside them. However, because theminimum value of 1.8 is below the horizontalasymptote at 2, the concavity mu st be downwardas x → −∞. Therefore, there must be an in-flection point somewhere to the lef t of x = −4.There may be othe r infl ection points (an evennumber in each of the three intervals d elimitedby the vertical asymptotes), but probabl y not.14xy•••........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................15Example 3Discuss f(x) = x2/3(x − y)2+ 2.16f(x) = x2/3(x − 7)2+ 2.This function is positive everywhere. It is con-tinu ous, but the derivative will be discontinuousat 0.f′(x) =23x−1/3(x − 7)2+ 2x2/3(x − 7).Remember th at our strategy i s to write f′as aproduct (factor it) so its zeros will be obvious.17Use x2/3= x−1/3x+1:f′(x) = 2(x − 7)x − 7 + 3x3x1/3=23x−1/3(4x − 7)(x − 7).So the critical nu mbers are x = 7 and x =74(the zeros of f′) an d x = 0 (a cusp: f′is notcont i nuous there but f is).To get the sec ond derivative it is useful to18multipl y f′out again:f′(x) =23x−1/3(4x2− 35x + 49).f′′(x)= −29x−4/3(4x2− 35x + 49) +23x−1/3(8x − 35)=29x−4/3(20x2− 70x − 49).The roots of f′′arex =70 ±√702+ 80 × 4940=70 ± 42√540.19How did I do that arithmetic? Break thingsinto prime factors:702+ 80 × 49 = 722252+ 72245 = 72225(5 + 4)= 7222325 = (42)25.Conclusion: f′′(x) = 0 at two pl aces, x ≈−0.6 an d x ≈ 4.1. Furthermore, f′′is undefinedat x = 0, so that also is a potential inflectionpoint.20We ne ed to find the sign of f′in the i nter-vals between critic al numbers. Note that at eachsuch number (0,74, 7), on e of the thre e factors(x−1/3, 4x − 7, x − 7) changes from ne gativeto positive. Thus f i s increasing on (0,74) and(7, ∞) and d ecreasing on (−∞, 0) and (74, 7). Ittherefore has minima at 0 and 7 and a maximumat74.Now note that f′′has the f orm x−4/3(ax2+bx + c) with a > 0, so f′′is negative precise l y21in the i nterval between its two roots. (Since4 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
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