Lecture for Week 8 (Secs. 4.1–4.2)Exponentials and Inverse Functions1Let’s get right to the point : The main rea-son for s t udying exponential functions is to solveproblems like those in Sec. 4.5, where the growthof some “s t uff” is proportional to the amount ofstuff already present:dP (t)dt= kP (t). (∗)(Compound interest, popu lation growth , radioac-tive decay all fall into this category.)2Suppose we knew a function whose deriva-tive is itself:ddxexp(x) = exp(x).Then P (t) = C exp(kt) would s ol ve (∗)[P′= kP ], for any constant C.Normalization: Suppose also thatexp(0) = 1. And suppose that we know P (0).Then P (t) = P (0) exp(kt) is the correct solutionof (∗) for our probl em.3With this, we have s ol ved our first nontrivialdifferential equation. The t rou ble with this ap-proach to ex p(x) is that it may not be obviousthat such a fu nction exis t s , or that there are notmore than one of them satisfying exp(0) = 1.Here is a di ffe re nt app roach (summarizedfrom Stewart):1. From elementary algebra we unde rstandwhat axmeans, for any a > 1 and any ra-tional nu mber x.42. We can d efine axfor irrational x by continu-ity ( “filling in the holes in the graph”).3. Define e as the number such thatlimh→0eh− 1h= 1.4. Prove thatddxex= ex.5. Obs erve that a0= 1 for all a, hence e0= 1.5So we can defi ne exp(x) as ex. You canthink of exeither as a number, e, raised to apower, or as a special function , exp(x), analo-gous to sin(x) and the other trig functions.The numerical value of e is a f amous “tran-scend ental ” number li ke π:e = 2.7182818 ···.6Algebraic properties of exponentials(“the laws of exponents”)ex+y= exey.(ex)y= exy.e−x=1ex.e0= 1.e1= e.7These laws also hold with a in place of eeverywhere. Also,(ab)x= axbx.And we also have ex> 0 andlimx→+∞ex= +∞,limx→−∞ex= 0.These als o hold for axif a > 1 ; if 0 < a < 1, thelimits reve rse, si nce ax= (1/a)−x.8Exercise 4.1.29Differentiate y = xe2x.Exercise 4.1.51If f(x) = e−2x, fi nd f(8)(x).9y = xe2x.Use t he product rule, the chain rule, and the ba-sic exponential derivative formuladdueu= eu:y′= e2x+ xddxe2x= e2x+ 2xe2x.10y = e−2x.What is it s 8th derivative? Every time I differen-tiate, I j ust get a factor −2. S of(8)(x) = (−2)8e−2x= 256e−2x.Exponential and logarithmic functions are in-verses of each other, so at this point we digress todiscuss inverse functions in general.11Remember the formu la for the volume of asphere:V =43πr3≡ f(r).Solve it t o get a formul a for the radius as a func-tion of the volume:r =3r3V4π≡ g(v) ≡ f−1(V ).The functions f and g are inverse to e ach other.Note that f−1does not mean1fin this con-text.12In a p hysical app lication like that, the vari-ables have natural names (r and V ), becausethey re present physical quantities. Bu t in genericmath, we u s ually write y = f(x). Should we thenwrite x = f−1(y) = g(y), or y = f−1(x) = g(x) ?Both Stewart and Maple insis t on t he latter,so th at x is always the independent variableand y the dependent one. To avoid c onfusion,I try to us e neut ral letters, say u = f(w) andw = f−1(u).13In obtainin g an inverse for a given fun ction,two complications may arise, the domain prob-lem an d th e branch problem. You know thatthey both occur for t he square root — the in-ve rse of a very simple f unction, u = w2≡ f(w).1. f−1(u) may not be de fined for some valuesof u. Example: If u < 0, th en u is not equalto w2for any real w. So the domain of t hesquare root function contains only nonnega-tive numbers.142. To make f−1single-valued ( as required bythe definition of a function ), we may needto exclu de some values of w that satisfyf(w) = u. We must choose just one w foreach u. This is call ed “ch oosing a branch ofthe inverse function”. Exampl e: We define√u t o be the n onnegative sq uare root. Thegraph of√u is th e right-hand hal f of thegraph of w2, fl ipped over so that th e u andw positive axes are interchanged.15A condition that assures that the branchproblem does not arise is the horizont al lin e test— which say s that t he graph of the inverse wil lpass the verti cal line test without our having tothrow part of the graph away. The original fu nc-tion is then cal led one -to-one. (We don’t havetwo points w mappi ng into the same u.)The domain problem does not arise if thefunct i on i s onto R — that is, eve ry u ∈ R ap-pears as f(w) for s ome w (whi ch will be f−1(u)).16Exercise 4.2.13Show that f(x) = 4x + 7 is one-to-one and findits inverse function.17We need to s ol ve y = 4x+7. That is elemen -tary:x =14(y − 7) =y4−74.(This makes se nse for all y, so th e function fis onto. And the solution for x is un i que, so fis one-to-one. Altern at i vely, you could sketchthe graph of f and see that every horizontal linecrosses it exactly once.)Therefore, we cou ld write f−1(y) =y4−74;18but to match the textbook ’ s notation we mustswitch the variable s :f−1(x) =x4−74.Finally we ge t to the main point: W hat isthe derivative of f−1(x) ? (We’re assuming weknow th e derivative of f.)19This quest ion could have bee n an s weredback in t he sec tion on implicit diffe re ntiation.To say that w = f−1(u) ≡ g(u) is to say thatu = f(w) (and that a branch has been chosen, ifneces s ary). So1 =dudu= f′(w)dwdu= f′(w)g′(u).Sog′(u) = [f′(w)]−1=1f′f−1(u).20Back in our original example , we might wantto write t his re lation asdrdV=dVdr−1.(Here the exponent −1 does mean to take th ereciprocal (“one over” the number).) Like thechain rule,dVdr=dVdCdCdr,21the theorem looks trivial in the Leibniz notation.(But be carefu l where the functions are evalu-ated. We need f′(f−1(u)), not f′(u).)Exercise 4.2.31Suppose g = f−1and f(4) = 5, f′(4) =23. Findg′(5).22We want g′(5). S ince g is t he inverse of f,g′is t he re ciprocal of f′. So we look for f′(5) inthe given inf ormation and don’t find it. (Evenworse, it might be given but be irrelevant!)What is wrong? You have to remember to evalu-ate f′at th e number f−1(5), wh ich is 4.g′(5) =1f′(g(5))=1f′(4)=32.23Exercise 4.2.25Find g′(1) if g = f−1and f(x) = x3+ x + 1.24f′(x) = 3x2+ 1.We need to e valu ate at a number x where f(x) =1. That requires0 = f(x) − 1 = x3+ x = x(x2+ 1).Since the q uadratic factor has no real roots, onlyx = 0 qualifies. ( If we did have more than oneroot, f−1would not exist!)g′(1) =1f′(0)=11=
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