{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

1105_LW5solutions_F11

# 1105_LW5solutions_F11 - 5 5(Ai’i 0 V1 g MACllOS Lecture...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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) ...
View Full Document

{[ snackBarMessage ]}