1105_LW5solutions_F11

1105_LW5solutions_F11 - 5 5;? )(Ai’i 0 V1 g MACllOS...

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Unformatted text preview: 5 5;? )(Ai’i 0 V1 g MACllOS Lecture Worksheet 5 Sections 5.1, 5.2 Fall, 2011 I. Let f(x) = J x ——l and g(x) = x2. Find a formula for each function below and state its domain. a) (f—g)(x)=m w x)“ Domain: {xixzq or Em) T Mws‘i“ make x4 :2 0 X3 i b) Domain; {Xi XE. \} affirm”) , Noie’. X1 i” Ae’mMJW‘A—mfl :2) X $0) 5" \0u,+ "Hag, 'M‘ieraX [\J +00) ' $\re,ao\7 exalt/wig; O . c) (go.r)(x)= 3&1») Domain Mag, ,3 W um) eywhm'ie by gubgiiilwi’l‘hj )0 in / fir Maia/g, Variable x I‘M/C “77;. J 2: _, ' LL?!er I H: _ 28‘? I) 1 z (m) WNW} '3: X‘Wl -2. Function f is given by the graph below. Function g is defined only for the x-vaiues in the table below. Evaluate each expression. Some expressions may be “undefined”. Graph of f a) (f+g)(—5)=R“5”)+ jiw-S):?>+\ air , 6 33+] b) (f-g)(0)= 4(0)'j(0)=2'£'“'2 c) £3),me unde‘gl‘neci (3 64—64-4344 123 45 678 d) (gof)(4)=3(rcaa—u)xj(o):é :: anw’: -4 e) f(g(0))= m) a “deem! gallons 3. A gallon of paint covers a 200 square foot area. a) Write an equation that expresses our coverage area, A, as a function of u, the number of gallons of paint we have. It may be helpful to complete the table first to see the pattern. A : Zoom b) Solve your equation from part a) for n to produce an equation that expresses the number of gallons of paint as a function of coverage area A. A 2" «’me loo A z ‘ A W = L00 200 c) These functions are inverses of each other. If we denote the function in part a) as A = f(n) , then the inverse function found in part b) is denoted n = f ‘1 (A) . Find f “1 (850) and write a sentence that interpets the answer in the context of coverage area and gallons of paint. «I n =2 t (Ale £7, 1% aka mega/w offset” PYQE’O): 8%: 5’325: 47,. 5.29%.,» 5550 szmmfa {seat (of area , C nut ODS be 0 ' ‘ f; e 'S “(i—{1'0 4 h {ll'l‘l‘e’ [3595 ex ofleh‘l Powehl u a) “1;; be, anemia—«owe / hams» Kchtiu as @3331“ Pme’img-éfl A :fi‘erew'f- yew: has. «A: warm” WWW , Raga-w; @151) 5. In #1, function g is one to one, so g”1(x) is a function. Use the table to find 3‘1 (HS) . g’] ("5): “‘3’ beam-5 a. 5,613) if“; in jemfiwwe “if Gl/ég) 5 lffggfi J ( lo) 691) S‘ou’l”;5’1€fi5 g6, . fig 6. A portion of the graph of f(x) = «J4 —x is shown in #2. Find its inverse function, f A' Sketch f and f ’1 on the axes below, laebaing the intercepts as ordered pairs. State the domain and range of each function. inhaling yflfijkw “I J solve glam “710 3d Kfiio’) - - - a . . . . a . . . . . . . _ y 23:: Q “X xsswyz... In {’ergkama ea vamriaioififi . , . . . fiffljeflg‘ y§¥(x)‘ . . . . . . .. ymawxthhamwAaegmfi : ::::::: Answers: +0 mg I), ‘5: Zone» NWX$Q::: ::::: i iiiiiii fl)=fi:§ . . _ _ _ . . _'.::§ :::: x H8 f 2x3$lwaq3 XEQ Domain off: Kg: (#3 Range off: \ {is} or- (" “:Jr‘fl EON”) f"(x)m In!th >< VES'inci’eci View X30, (Wife c: domahfl 97D Domainoff—1: bdxzmcq Rangeoff—'; { 4.3 0V“ 6 t” E0.) “19> (*3 mow) ...
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