C h a p t e r 1. Functions, Graphs, and Lines. 1.1 Functions

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1 C h p t e r Functions, Grphs, nd Lines Trying to do clculus without using functions would e one of the most pointless things you could do. If clculus hd n ingredients list, functions would e first on it, nd y some mrgin too. So, the first two chpters of this ook re designed to jog your memory out the min fetures of functions. This chpter contins review of the following topics: functions: their domin, codomin, nd rnge, nd the verticl line test; inverse functions nd the horizontl line test; composition of functions; odd nd even functions; grphs of liner functions nd polynomils in generl, s well s rief survey of grphs of rtionl functions, eponentils, nd logrithms; nd how to del with solute vlues. Trigonometric functions, or trig functions for short, re delt with in the net chpter. So, let s kick off with review of wht function ctully is.. Functions A function is rule for trnsforming n oject into nother oject. The oject you strt with is clled the input, nd comes from some set clled the domin. Wht you get ck is clled the output; it comes from some set clled the codomin. Here re some emples of functions: Suppose you write f() =. You hve just defined function f which trnsforms ny numer into its squre. Since you didn t sy wht the domin or codomin re, it s ssumed tht they re oth, the set of ll rel numers. So you cn squre ny rel numer, nd get rel numer ck. For emple, f trnsforms into ; it trnsforms / into /; nd it trnsforms into. This lst one isn t much of chnge t ll, ut tht s no prolem: the trnsformed oject doesn t hve to e different from the originl one. When you write f() =, wht you relly men

2 Functions, Grphs, nd Lines is tht f trnsforms into. By the wy, f is the trnsformtion rule, while f() is the result of pplying the trnsformtion rule to the vrile. So it s techniclly not correct to sy f() is function ; it should e f is function. Now, let with domin consisting only of numers greter thn or equl to. (Such numers re clled nonnegtive.) This seems like the sme function s f, ut it s not: the domins re different. For emple, f( /) = /, ut g( /) isn t defined. The function g just chokes on nything not in the domin, refusing even to touch it. Since g nd f hve the sme rule, ut the domin of g is smller thn the domin of f, we sy tht g is formed y restricting the domin of f. Still letting f() =, wht do you mke of f(horse)? Oviously this is undefined, since you cn t squre horse. On the other hnd, let s set h() = numer of legs hs, where the domin of h is the set of ll nimls. So h(horse) =, while h(nt) = 6 nd h(slmon) =. The codomin could e the set of ll nonnegtive integers, since nimls don t hve negtive or frctionl numers of legs. By the wy, wht is h()? This isn t defined, of course, since isn t in the domin. How mny legs does hve, fter ll? The question doesn t relly mke sense. You might lso think tht h(chir) =, since most chirs hve four legs, ut tht doesn t work either, since chir isn t n niml, nd so chir is not in the domin of h. Tht is, h(chir) is undefined. Suppose you hve dog clled Junkster. Unfortuntely, poor Junkster hs indigestion. He ets something, then chews on it for while nd tries to digest it, fils, nd hurls. Junkster hs trnsformed the food into... something else ltogether. We could let j() = color of rf when Junkster ets, where the domin of j is the set of foods tht Junkster will et. The codomin is the set of ll colors. For this to work, we hve to e confident tht whenever Junkster ets tco, his rf is lwys the sme color (sy, red). If it s sometimes red nd sometimes green, tht s no good: function must ssign unique output for ech vlid input. Now we hve to look t the concept of the rnge of function. The rnge is the set of ll outputs tht could possily occur. You cn think of the function working on trnsforming everything in the domin, one oject t time; the collection of trnsformed ojects is the rnge. You might get duplictes, ut tht s OK. So why isn t the rnge the sme thing s the codomin? Well, the rnge is ctully suset of the codomin. The codomin is set of possile outputs, while the rnge is the set of ctul outputs. Here re the rnges of the functions we looked t ove:

3 Section..: Intervl nottion If f() = with domin nd codomin, the rnge is the set of nonnegtive numers. After ll, when you squre numer, the result cnnot e negtive. How do you know the rnge is ll the nonnegtive numers? Well, if you squre every numer, you definitely cover ll nonnegtive numers. For emple, you get y squring (or ). If, where the domin of g is only the nonnegtive numers ut the codomin is still ll of, the rnge will gin e the set of nonnegtive numers. When you squre every nonnegtive numer, you still cover ll the nonnegtive numers. If h() is the numer of legs the niml hs, then the rnge is ll the possile numers of legs tht ny niml cn hve. I cn think of nimls tht hve,,, 6, nd 8 legs, s well s some creepy-crwlies with more legs. If you include individul nimls which hve lost one or more legs, you cn lso include,, 5, nd 7 in the mi, s well s other possiilities. In ny cse, the rnge of this function isn t so cler-cut; you proly hve to e iologist to know the rel nswer. Finlly, if j() is the color of Junkster s rf when he ets, then the rnge consists of ll possile rf-colors. I dred to think wht these re, ut proly right lue isn t mong them... Intervl nottion In the rest of this ook, our functions will lwys hve codomin, nd the domin will lwys e s much of s possile (unless stted otherwise). So we ll often e deling with susets of the rel line, especilly connected intervls such s { : < 5}. It s it of pin to write out the full set nottion like this, ut it sure ets hving to sy ll the numers etween nd 5, including ut not 5. We cn do even etter using intervl nottion. We ll write [, ] to men the set of ll numers etween nd, including nd themselves. So [, ] mens the set of ll such tht. For emple, [, 5] is the set of ll rel numers etween nd 5, including nd 5. (It s not just the set consisting of,,, nd 5: don t forget tht there re lods of frctions nd irrtionl numers etween nd 5, such s 5/, 7, nd π.) An intervl such s [, ] is clled closed. If you don t wnt the endpoints, chnge the squre rckets to prentheses. In prticulr, (, ) is the set of ll numers etween nd, not including or. So if is in the intervl (, ), we know tht < <. The set (, 5) includes ll rel numers etween nd 5, ut not or 5. An intervl of the form (, ) is clled open. You cn mi nd mtch: [, ) consists of ll numers etween nd, including ut not. And (, ] includes ut not. These intervls re closed t one end nd open t the other. Sometimes such intervls re clled hlf-open. An emple is the set { : < 5} from ove, which cn lso e written s [, 5). There s lso the useful nottion (, ) for ll the numers greter thn not including ; [, ) is the sme thing ut with included. There re three other possiilities which involve ; ll in ll, the sitution looks like this:

4 PSfrg replcements Functions, Grphs, nd Lines Sfrg replcements (, ) [, ] (, ] [, ) (, ) [, ) (, ) (, ] : < < } : } : < } replcements : < } { : (, } ) { : > [, } ] { : (, } ] { : [, < } ) (, ) [, ) (, ) (, ] : < < } : } : < } : < } { : } { : > } { : } { : < } (, ) { : < < } [, ] { : } (, ] { : < } [, ) { : < } (, ) { : > } [, ) (, ) { : } { : < } (, ] { : }.. Finding the domin Sometimes the definition of function will include the domin. (This ws the cse, for emple, with our function g from Section. ove.) Most of the time, however, the domin is not provided. The sic convention is tht the domin consists of s much of the set of rel numers s possile. For emple, if k() =, the domin cn t e ll of, since you cn t tke the squre root of negtive numer. The domin must e [, ), which is just the set of ll numers greter thn or equl to. OK, so squre roots of negtive numers re d. Wht else cn cuse screw-up? Here s list of the three most common possiilities:. The denomintor of frction cn t e zero.. You cn t tke the squre root (or fourth root, sith root, nd so on) of negtive numer.. You cn t tke the logrithm of negtive numer or of. (ememer logs? If not, see Chpter 9!) You might recll tht tn(9 ) is lso prolem, ut this is relly specil cse of the first item ove. You see, sin(9 ) tn(9 ) = =, cos(9 ) so the reson tn(9 ) is undefined is relly tht hidden denomintor is zero. Here s nother emple: if we try to define log ( + 8) 6 f() =, ( )( + 9) then wht is the domin of f? Well, for f() to mke sense, here s wht needs to hppen: We need to tke the squre root of (6 ), so this quntity hd etter e nonnegtive. Tht is, 6. This cn e rewritten s.

5 Section..: Finding the rnge using the grph 5 We lso need to tke the logrithm of ( + 8), so this quntity needs to e positive. (Notice the difference etween logs nd squre roots: you cn tke the squre root of, ut you cn t tke the log of.) Anywy, we need + 8 >, so > 8. So fr, we know tht 8 <, so the domin is t most ( 8, ]. The denomintor cn t e ; this mens tht () = nd (+9) =. In other words, = nd = 9. This lst one isn t prolem, since we lredy know tht lies in ( 8, ], so cn t possily e 9. We do hve to eclude, though. So we hve found tht the domin is the set ( 8, ] ecept for the numer. This set could e written s ( 8, ]\{}. Here the ckslsh mens not including... Finding the rnge using the grph Let s define new function F y specifying tht its domin is [, ] nd tht F () = on this domin. (ememer, the codomin of ny function we look t will lwys e the set of ll rel numers.) Is F the sme function s f, where f() = for ll rel numers? The nswer is no, since the two functions hve different domins (even though they hve the sme rule). As in the cse of the function g from Section. ove, the function F is formed y restricting the domin of f. Now, wht is the rnge of F? Well, wht hppens PSfrg if you replcements squre every numer etween nd inclusive? You should e le to work (, this ) out directly, ut this is good opportunity to see how to use grph to find [, ] the rnge of function. The ide is to sketch the grph of the function, (, ] then imgine two rows of lights shining from the fr left nd fr right of the [, grph ) horizontlly towrd the y-is. The curve will cst two shdows, one (, on ) the left side nd one on the right side of the y-is. The rnge is the [, union ) of oth shdows: tht is, if ny point on the y-is lies in either the left-hnd (, ) or the right-hnd shdow, it is in the rnge of the function. Let s see (, how ] this works with our function F : { : < < } { : } { : < } { : < } { : } { : > } { : } { : < } shdow

6 6 Functions, Grphs, nd Lines replcements (, ) [, ] (, ] [, ) (, ) [, ) (, ) (, ] : < < } : } : < } : < } { : } { : > } { : } { : < } shdow The left-hnd shdow covers ll the points on the y-is etween nd inclusive, which is [, ]; on the other hnd, the right-hnd shdow covers the points etween nd inclusive, which is [, ]. The right-hnd shdow doesn t contriute nything etr: the totl coverge is still [, ]. This is the rnge of F... The verticl line test In the lst section, we used the grph of function to find its rnge. The grph of function is very importnt: it relly shows you wht the function looks like. We ll e looking t techniques for sketching grphs in Chpter, ut for now I d like to remind you out the verticl line test. You cn drw ny figure you like on coordinte plne, ut the result my not e the grph of function. So wht s specil out the grph of function? Wht is the grph of function f, nywy? Well, it s the collection of ll points with coordintes (, f()), PSfrg where replcements is in the domin of f. Here s nother wy of looking t this: strt with some numer (, ). If is in the domin, you plot the point (, f()), which of course [, ] is t height of f() units ove the point on the -is. If isn t in the (, domin, ] you don t plot nything. Now repet for every rel numer to uild [, ) up the grph. Here s the key ide: you cn t hve two points with (, ) the sme -coordinte. In other words, no two points on the grph cn lie [, on ) the sme verticl line. Otherwise, how would you know which of the two (, or more ) heights ove the point on the -is corresponds to the vlue of (, f()? ] So, this leds us to the verticl line test: if you hve some grph nd (, you ) wnt to know whether it s the grph of function, see whether ny { : verticl < < line } intersects the grph more thn once. If so, it s not the grph { of : function; } ut if no verticl line intersects the grph more thn once, you { re : indeed < deling } with the grph of function. For emple, the circle of rdius { : units < } centered t the origin hs grph like this: { : } { : > } { : } { : < } shdow Such commonplce oject should e function, right? No, check the verticl lines tht re shown in the digrm. Sure, to the left of or to the right of, there s no prolem the verticl lines don t even hit the grph, which is fine. Even t or, the verticl lines only intersect the curve in one point ech, which is lso fine. The prolem is when is in the intervl (, ). For

7 PSfrg replcements Section.: Inverse Functions 7 (, ) [, ] ny of these vlues of, the verticl line through (, (, ) ] intersects the circle twice, which screws up the circle s potentil function-sttus. [, ) You just don t know whether f() is the top point or the ottom (, point. ) The est wy to slvge the sitution is to chop [, ) the circle in hlf horizontlly nd choose only the top or the ottom (, hlf. ) The eqution for the whole circle is + y = 9, wheres the eqution (, for ] the top semicircle is y = 9. The ottom semicircle hs eqution (, y = ) 9. These lst two re functions, oth with domin [, { ]. : If < you < } felt like chopping in different wy, you wouldn t ctully hve { to tke : semicircles you } could chop nd chnge etween the upper nd lower { semicircles, : < s } long s you don t violte the verticl line test. For emple, here s { : the grph < } of function which lso hs domin [, ]: { : } { : > } { : } { : < } shdow The verticl line test checks out, so this is indeed the grph of function.. Inverse Functions Let s sy you hve function f. You present it with n input ; provided tht is in the domin of f, you get ck n output, which we cll f(). Now we try to do things ll ckwrd nd sk this question: if you pick numer y, wht input cn you give to f in order to get ck y s your output? Here s how to stte the prolem in mth-spek: given numer y, wht in the domin of f stisfies f() = y? The first thing to notice is tht y hs to e in the rnge of f. Otherwise, y definition there re no vlues of such tht f() = y. There would e nothing in the domin tht f would trnsform into y, since the rnge is ll the possile outputs. On the other hnd, if y is in the rnge, there might e mny vlues tht work. For emple, if f() = (with domin ), nd we sk wht vlue of trnsforms into 6, there re oviously two vlues of : 8 nd 8. On the other hnd, if g() =, nd we sk the sme question, there s only one vlue of, which is. The sme would e true for ny numer we give to g to trnsform, ecuse ny numer hs only one (rel) cue root. So, here s the sitution: we re given function f, nd we pick y in the rnge of f. Idelly, there will e ectly one vlue of which stisfies f() = y. If this is true for every vlue of y in the rnge, then we cn define new

8 8 Functions, Grphs, nd Lines function which reverses the trnsformtion. Strting with the output y, the new function finds the one nd only input which leds to the output. The new function is clled the inverse function of f, nd is written s f. Here s summry of the sitution in mthemticl lnguge:. Strt with function f such tht for ny y in the rnge of f, there is ectly one numer such tht f() = y. Tht is, different inputs give different outputs. Now we will define the inverse function f.. The domin of f is the sme s the rnge of f.. The rnge of f is the sme s the domin of f.. The vlue of f (y) is the numer such tht f() = y. So, if f() = y, then f (y) =. The trnsformtion f cts like n undo utton for f: if you strt with nd trnsform it into y using the function f, then you cn undo the effect of the trnsformtion y using the inverse function f on y to get ck. This rises some questions: how do you see if there s only one vlue of tht stisfies the eqution f() = y? If so, how do you find the inverse, nd wht does its grph look like? If not, how do you slvge the sitution? We ll nswer these questions in the net three sections... The horizontl line test For the first question how to see tht there s only one vlue of tht works for ny y in the rnge perhps the est wy is to look PSfrg t the replcements grph of your function. We wnt to pick y in the rnge of f nd hopefully only hve (, one ) vlue [, ] of such tht f() = y. Wht this mens is tht the horizontl line through (, ] the point (, y) should intersect the grph ectly once, t some point [, ) (, y). Tht is the one we wnt. If the horizontl line intersects the (, curve ) more thn once, there would e multiple potentil inverses, which is d. [, ) In tht cse, the only wy to get n inverse function is to restrict the domin; (, ) we ll (, ] come ck to this very shortly. Wht if the horizontl line doesn t intersect the curve t ll? Then y isn t in the rnge fter ll, which { is : OK. < < } So, we hve just descried the horizontl line test: if every { : horizontl } line intersects the grph of function t most once, the function { : hs < n } inverse. If even one horizontl line intersects the grph more thn once, { : there < } isn t n { : } inverse function. For emple, look t the grphs of f() = nd : { : > } { : } { : < } f() = shdow

9 Section..: Finding the inverse 9 PSfrg replcements No horizontl line hits y = f() more thn once, so f (, hs ) n inverse. On the other hnd, some of the horizontl lines hit the curve [, y ] = g() twice, so g hs no inverse. Here s the prolem: if you wnt to solve (, ] y = for, where y is positive, then there re two solutions, = y nd [, ) = y. You don t know which one to tke! (, ) [, ) (, ).. Finding the inverse (, ] Now let s move on to the second of our questions: (, how do ) you find the inverse of function f? Well, you write down y = { f() : < nd < try } to solve for. In { our emple of f() =, we hve y =, so : = } y. This mens tht f (y) = y. If the vrile y here offends { you, : < y ll } mens switch it to : you cn write f () = if you prefer. { : Of course, < } solving for is not lwys esy nd in fct is often impossile. On { the : other } hnd, if you know wht the grph of your function looks like, the { grph : > of } the inverse function is esy to find. The ide is to drw the line y = { on : the } grph, then pretend tht this line is two-sided mirror. The inverse { function : < } is the reflection of the originl function in this mirror. When f() =, here s wht f looks like: f() = shdow mirror (y = ) f () = f() = The originl function f is reflected in the mirror y = to get the inverse function. Note tht the domin nd rnge of oth f nd f re the whole rel line... estricting the domin Finlly, we ll ddress our third question: if the horizontl line test fils nd there s no inverse, wht cn e done? Our prolem is tht there re multiple vlues of tht give the sme y. The only wy to get round the prolem is to throw wy ll ut one of these vlues of. Tht is, we hve to decide which one of our vlues of we wnt to keep, nd throw the rest wy. As we sw in Section. ove, this is clled restricting the domin of our function. Effectively, we ghost out prt of the curve so tht wht s left no longer fils the horizontl line test. For emple, if, we cn ghost out the left hlf of the grph like this:

10 Functions, Grphs, nd Lines shdow f() = f() = mirror (y = ) f () = [, ) (, ) [, ) (, ) (, ] { : < < } { : } g() { = : < } { : < } { : } { : > } { : } { : < } The new (unghosted) curve hs the reduced domin [, shdow ) nd stisfies the horizontl line test, so there is n inverse function. More precisely, the function h, which hs domin [, ) nd is defined y h() = on this domin, hs n inverse. Let s ply the reflection PSfrg gme replcements to see wht it looks like: (, ) [, ] y = h() mirror g() (y = ) f() = y = h g() = () f() = (, ] [, ) (, ) [, ) (, ) (, ] { : < < } { : } { : < } { : < } f () = To find the eqution of the inverse, we { hve : to } solve for in the eqution y =. Clerly the solution is = y { or : = > } y, ut which one do we need? We know tht the rnge of the inverse { : function } is the sme s the domin of the originl function, which we { hve : < restricted } to e [, ). So we need nonnegtive numer s our nswer, nd tht hs to e = y. Tht is, h (y) = y. Of course, we could hve ghosted out the right hlf of the originl grph to restrict the domin to (, ]. In tht cse, we d get function j which hs domin (, ] nd gin shdow stisfies j() =, ut only on this domin. This function lso hs n inverse, ut the inverse is now the negtive squre root: j (y) = y. By the wy, if you tke the originl function g given y with domin, which fils the horizontl line test, nd try to reflect it in the mirror y =, you get the following picture: f() = f() = mirror (y = ) f () = y = h() y = h ()

11 { } { : < } Section..: Inverses of inverse functions shdow Notice tht the grph fils the verticl line test, so it s not the grph of Sfrg replcements function. This illustrtes the connection etween the verticl nd horizontl (, ) line tests when horizontl lines re reflected in the mirror y =, they ecome [, ] verticl lines. (, ] [, ).. Inverses of inverse functions (, ) f() [, = ) One more thing out inverse functions: if f hs n inverse, it s true tht g() (, = ) f (f()) = for ll in the domin of f, nd lso tht f(f (y)) = y for f() (, = ] ll y in the rnge of f. (ememer, the rnge of f is the sme s the domin mirror (, (y = ) ) of f, so you cn indeed tke f (y) for y in the rnge of f without cusing f : () < = < } ny screwups.) : y = h() } For emple, if f() =, then f hs n inverse given y f () =, : y = < h () } nd so f (f()) = = for ny. ememer, the inverse function is : < } like n undo utton. We use s n input to f, nd then give the output to { : } f ; this undoes the trnsformtion nd gives us ck, the originl numer. { : > } Similrly, f(f (y)) = ( y) = y. So f is the inverse function of f, nd { : } f is the inverse function of f. In other words, the inverse of the inverse is { : < } the originl function. Now, you hve to e creful in the cse where you restrict the domin. Let ; we ve seen tht you need to restrict the domin to get n inverse. Let s sy we restrict the domin to [, ) nd crelessly continue to refer to shdow the function s g insted of h, s in the previous section. We would then sy tht g () =. If you clculte g(g ()), you find tht this is ( ), which equls, provided tht. (Otherwise you cn t tke the squre root in the first plce.) On the other hnd, if you work out g (g()), you get, which is not lwys the sme thing s. For emple, if =, then = nd so = =. So it s not true in generl tht g (g()) =. The prolem is tht isn t in the restricted-domin version of g. Techniclly, you cn t f() = even compute g(), since is no longer in the domin of g. We relly should e working with h, not g, so tht we rememer to e more creful. f() = Nevertheless, in prctice, mthemticins will often restrict the domin without chnging letters! So it will e useful to summrize the sitution s follows: mirror (y = ) f () = y = h() If the domin of function f cn e restricted so tht f hs n inverse y = h () f, then f(f (y)) = y for ll y in the rnge of f; ut f (f()) my not equl ; in fct, f (f()) = only when is in the restricted domin. We ll e revisiting these importnt points in the contet of inverse trig functions in Section..6 of Chpter.. Composition of Functions Let s sy we hve function g given y. You cn replce y nything you like, s long s it mkes sense. For emple, you cn write

12 (, ] [, ) (, ) replcements [, ) Functions, Grphs, nd Lines (, (, ) (, [, ] g(y) = y, or g( + 5) = ( + 5). This lst emple shows tht you need to (, (, ) ] e very creful with prentheses. It would e wrong to write g(+5) = +5, : < [, < } ) since this is just + 5, which is not the sme thing s ( + 5). If in dout, : (, ) } use prentheses. Tht is, if you need to write out f(something), replce every : < [, ) } instnce of y (something), mking sure to include the prentheses. Just : (, < } ) out the only time you don t need to use prentheses is when the function is { :(, } ] n eponentil function for emple, if h() =, then you cn just write {(, : > ) } h( + 6) = +6. You don t need prentheses since you re lredy writing : { < : < } the + 6 s superscript. : { : < } Now consider the function f defined y f() = cos( ). If I give you : < } numer, how do you compute f()? Well, first you squre it, then you tke : < } the cosine of the result. Since we cn decompose the ction of f() into these { : } two seprte ctions which re performed one fter the other, we might s { : shdow > } well descrie those ctions s functions themselves. So, let nd { : } h() = cos(). To simulte wht f does when you use s n input, you { : < } could first give to g to squre it, nd then insted of tking the result ck you could sk g to give its result to h insted. Then h spits out numer, which is the finl nswer. The nswer will, of course, e the cosine of wht cme out of g, which ws the squre of the originl. This ehvior ectly shdow mimics f, so we cn write f() = h(g()). Another wy of epressing this is g() = to write f = h g; here the circle mens composed with. Tht is, f is h f() = composed with g, or in other words, f is the composition of h nd g. Wht s g() = tricky is tht you write h efore g (reding from left to right s usul!) ut f() = you pply g first. I gree tht it s confusing, ut wht cn I sy you just mirror (y = ) hve to del with it. f () It s useful to prctice composing two or more functions together. For g() y = = h() emple, if g() =, h() = 5, nd j() =, wht is formul for f() y = h = () the function f = g h j? Well, just replce one thing t time, strting with j, then h, then g. So: f() = mirror (y = ) f () = y = h() y = h () f() = g(h(j())) = g(h( )) = g(5( ) ) = 5( ). You should lso prctice reversing the process. For emple, suppose you strt off with f() = tn(5 log ( + )). How would you decompose f into simpler functions? Zoom in to where you see the quntity. The first thing you do is dd, so let g() = +. Then you hve to tke the se logrithm of the resulting quntity, so set h() = log (). Net, multiply y 5, so set j() = 5. Then tke the tngent, so put k() = tn(). Finlly, tke reciprocls, so let m() = /. With ll these definitions, you should check tht f() = m(k(j(h(g())))). Using the composition nottion, you cn write f = m k j h g.

13 Sfrg replcements (, ) [, ] (, ] replcements [, ) (, (, ) ) [, [, ) ] (, (, ) ] (, [, ) ] (, (, ) : < [, < ) } : (, } ) : < (, } ] : (, < ) } : { < : < } } : { : > } } : { < : } : { : < } { : } { : > } { : } { : < } shdow shdow g() = f() = g() = f() g() = mirror (y ) f f() = g() y = = h() y f() = h = () mirror (y = ) f () = y = h() y = h () Section.: Composition of Functions This isn t the only wy to rek down f. For emple, we could hve comined h nd j into nother function n, where n() = 5 log (). Then you should check tht n = j h, nd f = m k n g. Perhps the originl decomposition (involving j nd h) is etter ecuse it reks down f into more elementry steps, ut the second one (involving n) isn t wrong. After ll, n() = 5 log () is still pretty simple function of. Bewre: composition of functions isn t the sme thing s multiplying them together. For emple, if f() = sin(), then f is not the composition of two functions. To clculte f() forpsfrg ny given replcements, you ctully hve to find oth nd sin() (it doesn t mtter which one you (, ) find first, unlike with composition) nd then multiply these two things together. [, ] If nd h() = sin(), then we d write f() = g()h(), or f (, = ] gh. Compre this to the composition of the two functions, j = g h, which [, is ) given y (, ) j() = g(h()) = g(sin()) = (sin()) [, ) or simply j() = sin (). The function j is completely (, ) different function from the product sin(). It s lso different from (, the] function k = h g, which is lso composition of g nd h ut in the (, other ) order: { : < < } k() = h(g()) = h({ ): = sin( ). } { : < } This is yet nother completely different function. { : The < } morl of the story is tht products nd compositions re not the sme { : thing, } nd furthermore, the order of the functions mtters when you compose { : them, > } ut not when you multiply them together. { : } One simple ut importnt emple of composition { : < } of functions occurs when you compose some function f with g() =, where is some constnt numer. You end up with new function h given y h() = f( ). A useful point to note is tht the grph of y = h() is the sme s the grph of y = f(), ecept tht it s shifted over units toshdow the right. If is negtive, then the shift is to the left. (The wy to think of this, for emple, is tht shift of units to the right is the sme s shift of units to the left.) So, how would you sketch the grph of y = ( )? This is the sme s y =, ut with replced y. So the grph of y = needs to e shifted to the right y unit, nd looks like this: f() = g() y = ( ) f() = mirror (y = ) f () = y = h() y = h ()

14 [, ) (, ) [, ) (, ) Functions, Grphs, nd Lines (, ] : < < } : } : < } : < } { : } { : > } { : } { : < } shdow f() = f() = mirror (y = ) f () = y = h() y = h () y = ( ) Similrly, the grph of y = ( + ) is the grph of y = shifted to the left y units, since you cn interpret ( + ) s ( ()).. Odd nd Even Functions Some functions hve some symmetry properties tht mke them esier to del with. Consider the function f given y f() =. Pick ny positive numer you like (I ll choose ) nd hit it with f (I get 9). Now tke the negtive of tht numer, in my cse, nd hit tht with f (I get 9 gin). You should get the sme nswer oth times, s I did, regrdless of which numer you chose. You cn epress this phenomenon y writing f( ) = f() for ll. Tht is, if you give to f s n input, you get ck the sme nswer s if you used the input insted. Notice tht g() = nd h() = 6 lso hve this property in fct, j() = n, where n is ny even numer (n could in fct e negtive), hs the sme property. Inspired y this, we sy tht function f is even if f( ) = f() for ll in the domin of f. It s not good enough for this eqution to e true for some vlues of ; it hs to e true for ll in the domin of f. Now, let s sy we ply the sme gme with f() =. Tke your fvorite positive numer (I ll stick with ) nd hit tht with f (I get 7). Now try gin with the negtive of your numer, in my cse; I get 7, nd you should lso get the negtive of wht you got efore. You cn epress this mthemticlly s f( ) = f(). Once gin, the sme property holds for j() = n when n is ny odd numer (nd once gin, n could e negtive). So, we sy tht function f is odd if f( ) = f() for ll in the domin of f. In generl, function might e odd, it might e even, or it might e neither odd nor even. Don t forget this lst point! Most functions re neither odd nor even. On the other hnd, there s only one function tht s oth odd nd even, which is the rther oring function given y f() = for ll (we ll cll this the zero function ). Why is this the only odd nd even function? Let s convince ourselves. If the function f is even, then f( ) = f() for ll. But if it s lso odd, then f( ) = f() for ll. Tke the first of these equtions nd sutrct the second from it. You should get = f(), which mens tht f() =. This is true for ll, so the function f must just e the zero function. One other nice oservtion is tht if function f is odd, nd the numer is in its domin, then f() =. Why is it so? Becuse f( ) = f() is true for ll in the domin of f, so let s try it for =. You get f( ) = f(). But is the sme thing s, so we hve f() = f(). This simplifies to f() =, or f() = s climed. Anywy, strting with function f, how cn you tell if it is odd, even, or neither? And so wht if it is odd or even nywy? Let s look t this second question efore coming ck to the first one. One nice thing out knowing tht function is odd or even is tht it s esier to grph the function. In fct, if you cn grph the right-hnd hlf of the function, the left-hnd hlf is piece of cke! Let s sy tht f is n even function. Then since f() = f( ), the grph of y = f() is t the sme height ove the -coordintes nd. This is true for ll, so the sitution looks something like this:

15 replcements (, ) [, ] (, ] [, ) (, ) [, ) (, ) (, ] : < < } : } : < } : < } { : } { : > } { : } { : < } shdow f() = f() = mirror (y = ) f () = y = h() y = h () y = ( ) Sme height Sme length, opposite signs (, ) [, ) (, f() = Section.: Odd nd Even ) (, Functions ] 5 g() = (, ) f() { : = < < } Sme height mirror {(y: = ) f () = } { : < } y { = : h() < } y = h {() : } y = ( { ) : > } { : } { : < } shdow We cn conclude tht the grph of n even function hs mirror symmetry out the y-is. So, if you grph the right hlf of function which you know is even, you cn get the left hlf y reflecting the right hlf out the y-is. Check the grph of y = to mke sure tht it hs this mirror symmetry. On the other hnd, let s sy tht f is n odd function. g() = Since we hve f( ) = f(), the grph of y = f() is t the sme f() height = ove the -coordinte s it is elow the -coordinte. (Of g() course, = if f() is negtive, then you hve to switch the words ove nd f() elow. ) = In ny cse, the picture looks like this: mirror (y = ) f () = y = h() Sme length, opposite signs y = h swq () y = ( ) Sme height The symmetry is now point symmetry out the origin. Tht is, the grph of n odd function hs 8 point symmetry out the origin. This mens tht if you only hve the right hlf of function which you know is odd, you cn get the left hlf s follows. Pretend tht the curve is sitting on top of the pper, so you cn pick it up if you like ut you cn t chnge its shpe. Insted of picking it up, put pin through the curve t the origin (rememer, odd functions must pss through the origin if they re defined t ) nd then spin the whole curve round hlf revolution. This is wht the left-hnd hlf of the grph looks like. (This doesn t work so well if the curve isn t continuous, tht is, if the curve isn t ll in one piece!) Check to see tht the ove grph nd lso the grph of y = hve this symmetry. Now, suppose f is defined y the eqution f() = log 5 ( 6 6 +). How do you tell if f is odd, even, or neither? The technique is to clculte f( ) y replcing every instnce of with ( ), mking sure not to forget the prentheses round, nd then simplifying the result. If you end up with the originl epression f(), then f is even; if you end up with the negtive of the originl epression f( ), then f is odd; if you end up with mess tht isn t either f() or f(), then f is neither (or you didn t simplify enough!).

16 f () [, = ) y (, = h() ) y = (, h () ] y = (, ( ) ) 6 Functions, Grphs, nd Lines : < < } : } In the emple ove, you d write Sme : < height } : < } f( ) = log 5 (( ) 6 6( ) + ) = log 5 ( ), { : } PSfrg Sme replcements { : length, > } which is ctully equl to the originl f(). So the function f is even. How opposite { : signs } (, ) out { : < } [, ] g() = + nd h() = ? (, ] [, ) Well, for g, we hve (, ) shdow [, ) g( ) = ( ) + ( ) ( ) = (, ) (, ] Now you hve to oserve tht you cn tke the minus sign out front nd write { : < < } { : } g( ) = + { : < + 5, } { : < f() = } { : } which, you notice, is equl to g(). Tht is, prt from the minus sign, we { g() : = > } get the originl function ck. So, g is n odd function. How out h? We f() : = } hve mirror {(y : = < ) h( ) = ( ) + ( ) ( ) = + 5. f () = } + 5 y = h() Once gin, we tke out the minus sign to get y = h () y = ( ) shdow h( ) = Sme height Hmm, this doesn t pper to e the negtive of the originl function, ecuse of the + term in the numertor. It s not the originl function either, so the function h is neither odd nor even. Sme length, Let s look t one more emple. Suppose you wnt to prove tht the opposite g() signs = product of two odd functions is lwys n even function. How would you go f() = out doing this? Well, it helps to hve nmes for things, so let s sy we hve two odd functions f nd g. We need to look t the product of these functions, f() = so let s cll the product h. Tht is, we define h() = f()g(). So, our tsk is mirror (y = ) to show tht h is even. We ll do this y showing tht h( ) = h(), s usul. f () = It will e helpful to note tht f( ) = f() nd g( ) = g(), since f y = h() nd g re odd. Let s strt with h( ). Since h is the product of f nd g, we y = h () hve h( ) = f( )g( ). Now we use the oddness of f nd g to epress y = ( ) this lst term s ( f()) ( g()). The minus signs now come out front nd cncel out, so this is the sme thing s f()g() which of course equls h(). We could (nd should) epress ll this tet mthemticlly like this: Sme height h( ) = f( )g( ) = ( f()) ( g()) = f()g() = h(). Sme length, opposite signs Anywy, since h( ) = h(), the function h is even. Now you should try to prove tht the product of two even functions is lwys even, nd lso tht the product of n odd nd n even function must e odd. Go on, do it now!

17 { } shdow Section.5: Grphs of Liner Functions 7 Functions of the form f() = m + re clled liner. There s good reson for this: the grphs of these functions re g() lines. = (As fr s we re concerned, the word line lwys mens stright line. ) f() The = slope of the line is given y m. Imgine for moment tht you re in g() the = pge, climing the line s if it were mountin. You strt t the left f() side = of PSfrg replcements the pge nd hed to the right, like this: mirror (y = ) f () = (, ) y = h() [, ] y = h () (, ] y = ( ) [, ) (, ) [, ) Sme height (, ) (, ] (, ) { Sme : length, < < } opposite { : signs } { : < } If the slope m is positive, s it is in the ove { : picture, < } then you re heding uphill. The igger m is, the steeper the clim. { : On} the other hnd, if the slope is negtive, then you re heding downhill. { : > } The more negtive the slope, the steeper the downhill grde. If the slope { : is zero, } then the line is flt, { : < } or horizontl you re going neither uphill nor downhill, just trudging long flt line. To sketch the grph of liner function, you only need to identify two points on the grph. This is ecuse there s only shdow one line tht goes through two different points. You just put your ruler on the points nd drw the line. One point is esy to find, nmely, the y-intercept. Set = in the eqution y = m +, nd you see tht y = m + =. Tht is, the y-intercept is equl to, so the line goes through (, ). To find nother point, you could find the -intercept y setting y = nd finding wht is. This works pretty well ecept in two cses. The first cse is when =, in which cse we re g() = just deling with y = m. This goes through the origin, so the -intercept f() = nd the y-intercept re oth zero. To get nother point, you ll just hve to g() = sustitute in = nd see tht y = m. So, the line f() = y = m goes through the origin nd (, m). For emple, the line mirror y = (y = ) goes through the origin nd lso through (, ), so it looks like this: f () = y = h() y = h () y y = = ( ) Sme height.5 Grphs of Liner Functions Sme length, opposite signs

18 { : } [, ) { : > } (, ) Sme { : height } [, ) { : < } (, ) 8 Functions, Grphs, nd Lines f() = (, ] Sme length, opposite signs The other d cse is when m =. But then f() we just = hve y =, which is : < < } y = horizontl line through (, ). mirror (y = ) : shdow } f For more interesting emple, consider () y = =. The y-intercept is : < }, nd the slope is. To sketch the line, find y = the h() -intercept y setting : < } y =. We get =, which simplifies to y = h =. () So, the line looks like { : } this: y = ( ) { : > } { : } { : < } y Sme = height f() = Sme length, f() = shdow opposite signs mirror (y = ) y = f () = y = h() y = h () y = ( ) g() = Sme height f() = Now, let s suppose you know tht you hve line in the plne, ut you don t know its eqution. If you know it goes through certin point, nd you know f() = wht its slope is, then you cn find the eqution of the line. You relly, relly, Sme length, mirror (y = ) relly need to know how to do this, since it comes up lot. This formul, opposite signs f () = clled the point-slope form of liner function, is wht you need to know: y = h() y = h () If line goes through (, y ) nd hs slope m, y then its eqution is y y = m( ). y = = ( ) For emple, wht is the eqution of the line through (, 5) which hs slope Sme height? It is y 5 = ( ()), which you cn epnd nd simplify down to y =. Sometimes you don t know the slope of the line, ut you do know two points tht it goes through. How do you find the eqution? The technique is to find the slope, then use the previous ide with one of the points (your choice) to find the eqution. First, you need to know this: Sme length, opposite signs y = y = If line goes through (, y ) nd (, y ), its slope is equl to y y. So, wht is the eqution of the line through (, ) nd (, 6)? Let s find the slope first: slope = 6 ( ) = =. 5 We now know tht the line goes through (, ) nd hs slope, so its eqution is y = ( ( )), or fter simplifying, y =. Alterntively, we could hve used the other point (, 6) with slope to see tht the eqution of the line is y ( 6) = ( ), which simplifies to y =. Thnkfully this is the sme eqution s efore it doesn t mtter which point you pick, s long s you hve used oth points to find the slope.

19 .6 Common Functions nd Grphs PSfrg replcements (, ) Section.6: Common Functions nd Grphs [, 9 ] (, ] [, ) (, ) [, ) Here re the most importnt functions you should know out. (, ) (, ]. Polynomils: these re functions uilt out of nonnegtive integer { : < powers < } of. You strt with the uilding locks,,,, nd so on, { nd : you re } llowed to multiply these sic functions y numers nd dd { finite : < numer } of them together. For emple, the polynomil f() = 5 { + : is formed < } y tking 5 times the uilding lock, nd times the uilding{ lock : }, nd times the uilding lock, nd dding them together. You { : might > } lso wnt to include the intermedite uilding locks nd, ut { since : they } { : < } don t pper, you need to tke times of ech. The mount tht you multiply the uilding lock n y is clled the coefficient of n. For emple, in the polynomil f ove, the coefficient of is 5, the coefficient of is, the coefficients of nd re oth, nd the coefficient of is. (Whyshdow llow nd, y the wy? They seem different from the other locks, ut they re not relly: = nd =.) The highest numer n such tht n hs nonzero coefficient is clled the degree of the polynomil. For emple, the degree of the ove polynomil f is, since no power of greter thn is present. The mthemticl wy to write generl polynomil of degree n is p() = n n + n n , f() = where n is the coefficient of n, n is the coefficient of n, nd f() so = on down to, which is the coefficient of. mirror (y = ) Since the functions n re the uilding locks of ll polynomils, f () = you should know wht their grphs look like. The even powers mostly looky similr = h() to ech other, nd the sme cn e sid for the odd powers. Here sy wht = h () the grphs look like, from up to 7 y = ( ) : y = Sme height y = y = y = Sme length, opposite signs y = y = y = y = 5 y = 6 y = 7

20 (, ) [, ) (, ) (, ] : < < } : } : < } : < } { : } { : > } { : } { : < } shdow f() = f() = mirror (y = ) f () = y = h() y = h () y = ( ) Sme height Sme length, opposite signs y = y = f() = Sketching the grphs of more generl polynomils is more difficult. Even f() finding the -intercepts is often impossile unless the polynomil ismirror very simple. (y = ) = There is one spect of the grph tht is firly strightforwrd, which f () is = wht hppens t the fr left nd right sides of the grph. This is determined y = h() y the so-clled leding coefficient, which is the coefficient of the highest-degree y = h () term. This is siclly the numer n defined ove. For emple, y = ( in our ) polynomil f() = 5 + from ove, the leding coefficient is 5. In fct, it only mtters whether the leding coefficient is positive or negtive. It Sme height lso mtters whether the degree of the polynomil is odd or even; so there re four possiilities for wht the edges of the grph cn look like: (, ) [, ] (, ] [, ) Functions, Grphs, nd Lines Sme length, opposite signs y = y = n even, n > n odd, n > n even, n < n odd, n < The wiggles in the center of these digrms ren t relevnt they depend on the other terms of the polynomil. The digrm is just supposed to show wht the grphs look like ner the left nd right edges. In this sense, the grph of our polynomil f() = 5 + looks like the leftmost picture ove, since n = is even nd n = 5 is positive. Let s spend little time on degree polynomils, which re clled qudrtics. Insted of writing p() = + +, it s esier to write the coefficients s,, nd c, so we hve p() = + + c. Qudrtics hve two, one, or zero (rel) roots, depending on the sign of the discriminnt. The discriminnt, which is often written s, is given y = c. There re three possiilities. If >, then there re two roots; if =, there is one root, which is clled doule root; nd if <, then there re no roots. In the first two cses, the roots re given y ± c. Notice tht the epression in the squre root is just the discriminnt. An importnt technique for deling with qudrtics is completing the squre. Here s how it works. We ll use the emple of the qudrtic +. The first step is to tke out the leding coefficient s fctor. So our qudrtic ecomes ( + 5). This reduces the sitution to deling with monic qudrtic, which is qudrtic with leding coefficient equl to. So, let s worry out + 5. The min technique now is to tke the coefficient of, which in our emple is, divide it y to get, nd squre it. We get We wish tht the constnt term were 6 insted of 5, so let s do some

21 [, ] (, ] [, ) (, ) Section.6: Common Functions nd Grphs [, ) (, ) mentl gymnstics: (, ] + 5 = { : < < } { : } { : < } Why on erth would we wnt to dd nd sutrct 9 6? Becuse { the : first three < } terms comine to form ( ). So, we hve ( + 5 = + 9 ) ( 6 6 = ) { : } { : 6. < } Now we just hve to work out the lst little it, which is just rithmetic: = 7 6. Putting it ll together, nd restoring the fctor of, we hve shdow + = ( ) ( ( + 5 = ) ) tionl functions: these re functions of the form p() q(), { : } { : > } ( = ) f() = It turns out tht this is much nicer form to del with in numerg() of situtions. Mke sure you know how to complete the squre, since we ll f() e using = = this technique lot in Chpters 8 nd 9. mirror (y = ) f () = y = h() y = h () y = ( ) Sme height where p nd q re polynomils. tionl functions will pop up in mny different contets, nd the grphs cn look very different depending on the polynomils p nd q. The simplest emples of rtionl functions Sme relength, polynomils themselves, which rise when q() is the constnt polynomil opposite. signs The net simplest emples re the functions / n, where n is positive integer. Let s look t some of the grphs of these functions: y = y = y = y = y = y = The odd powers look similr to ech other, nd the even powers look similr to ech other. It s worth knowing wht these grphs look like.

22 g() (, = ] f() (, = ) { mirror : < (y = < ) } { Functions, Grphs, nd Lines f: () = } { : < y = h() } { : y = h < () }. Eponentils nd logrithms: you need y { = ( : to know ) } wht grphs of eponentils look like. For emple, here is y = { : : > } { : } Sme { : height < } y = Sme length, opposite shdow signs y = y = f() y = f() = The grph of y = for ny other semirror > looks (y = ) similr to this. Things to notice re tht the domin is the wholef rel () line, = the y-intercept is, the rnge is (, ), nd there is horizontl symptote y = h() on the left t y =. In prticulr, the curve y = does not, I y repet, = h () not touch the -is, no mtter wht it looks like on your grphingy = clcultor! ( ) (We ll e looking t symptotes gin in Chpter.) The of y = is just the reflection of y = in the y-is: Sme height Sme length, y = opposite signs y = y = y = y = How out when the se is less thn? For emple, consider the grph of y = ( ). Notice tht ( ) = / =, so the ove grph of y = is lso the grph of y = ( ), since nd ( ) re equl for ny. The sme sort of thing hppens for y = for ny < <, not just =. Now, notice tht the grph of y = stisfies the horizontl line test, so there is n inverse function. This is in fct the se logrithm, which is written y = log (). Using the line y = s mirror, the grph of y = log () looks like this:

23 } { : } { : > } { : } { : < } shdow f() = f() = mirror (y = ) f () = y = h() y = h () y = ( ) Sme height Sme length, opposite signs y = y = y = y = y = y = log () y = ( (, ] (, ] ) [, ) (, ) { : < < } Sme [, ) height Section.6: Common { : Functions } (, ) nd Grphs { : < } (, ] { : y = Sme < } (, mirror length, ) { : } (y = ) { : opposite < < } { : > } signs { : y = } { : } { : < y = } { : } log () { : < } { : y = } { : > } { : } { : shdow < } y = y = shdow The domin is (, ); note tht this cks up wht I sid erlier out not eing le to tke logrithms of negtive numer g() = or of. The rnge is ll of, nd there s verticl symptote t =. The grphs of log (), f() = nd indeed log () for ny >, re very similr to this one. The log function is very importnt in clculus, so you should relly know how to drw the g() = f() = ove grph. We ll look t other properties of logrithms in Chpter 9. mirror g() (y = ). Trig functions: these re so importnt f tht the entire net chpter is f() = devoted to them. y = h() g() y = h 5. Functions involving solute vlues: let s = () f() = tke close look t the y = ( ) solute vlue function f given y f() =. Here s the mirror (y = ) definition of : { f () = if, y = h() = Sme height if < y =. h () y = ( ) Another wy of looking t is tht it is Sme the distnce length, etween nd on the numer line. More generlly, you should opposite lern this signs nice fct: Sme height y = y is the distnce etween nd y on the numer line. For emple, suppose tht you need to identify Sme y = the length, region on the numer line. You cn interpret the inequlity opposite s the signs distnce etween nd is less thn or equl to. Tht is, we re looking y = for ll the points tht re no more thn units wy from the numer y. = So, let s tke numer line nd mrk in the numer s follows: y = y = y = y = log () The points which re no more thn units wy etend y = to on the left nd on the right, so the region we wnt looks like this: y = y = y = log () units units So, the region cn lso e descried s [, ].

24 { : < < } { : } { : < f() = } { : < } Functions, Grphs, nd Lines g() { : = } It s lso true tht = f() { : = > }. To check mirror this, { suppose (y: = ) tht ; then =, no prolem. If insted <, then fit cn t () = e } { : < true } tht =, since the left-hnd side is positive ut the right-hnd side is negtive. The correct eqution is y = h() = ; now the right-hnd y = side h () is positive, since it s minus negtive numer. If you look ck t the definition of, you ll see tht we hve just proved tht = y = ( ). Even so, to shdow del with, it s much etter to use the piecewise definition thn to write it s Sme height. Finlly, let s tke look t some grphs. If you know wht the grph of function looks like, you cn get the grph of the solute vlue of tht function y reflecting everything elow the -is up to Sme ove length, the -is, using the -is s your mirror. For emple, here s theopposite grph of signs y =, which comes from reflecting the ottom portion of y = in theg() y -is: = = f() = y = g() y = = f() = mirror (y = ) f () = y = h() y y = mirror = h y = (-is) () y = ( ) y = log () Sme height units Sme length, How out the grph of y = log ()? Using opposite the reflection signs of the grph of y = log () ove, this is wht the solute vlue version y = looks like: y = y = log () y = y = mirror y = (-is) y = log () units y = Anywy, tht s ll I hve to sy out functions, prt from trig functions which re the suject of the net chpter. Hopefully you ve seen lot of the stuff in this chpter efore. Most of the mteril in this chpter is used over nd over gin in clculus, so mke sure you relly get on top of it ll s soon s you cn!