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# CalculusI.pdf

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This 7 page One Day of Notes was uploaded by Rebecca Ramirez on Friday September 12, 2014. The One Day of Notes belongs to a course at a university taught by a professor in Fall. Since its upload, it has received 70 views.

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Date Created: 09/12/14

Calculus I Section 11 Domain and Range of a Function Review Interval Notation Interval Nctaticn Descripticn cf Interval Graph includes all real numbers r such that J r lE I 39quot i39 r1sgeaterthanE39I Irgtb 339 includes all real numbers 139 such that E I i39 r is geater than cur equal tc b r E bl 3 includes all real numbers r such that J F H w J 1SlESS39Ll391EIt3911 Irltl 3953 includes all real numbers J such that I l39I I W 39 r1slessthancreu1ualtc Iria E3 quot33939 ml includes all real nutnb ers J Q 2 includes all real numbers 139 such that E H 3 5 EJ H ss 39a39 I J1sbet1t39een1 andb I1ltJltbl I13 5339 includes all real nunibers r such that 1Ej 15 greaterthan nren1ualtc andris I 3 E i lessthanb IE151 l339lJ includes all real numbers J such that rEJ r is greater than1 andris less than cr 5 B 39 E E equaltcb I1iJ395E39 El includes all real numbers J such that air it is between andincluding andb E E E i caiiribj Finding the Domain of a Function Ifthe dcinain cf a functicn has nct been esplicitlgr stated then by ccntrenticn the dcinain is the set cf all real numbers fer which the enpressicn is defined as a real number EX Find the 1niain cat the f11I1I21ZitI I139 3939f 392x 1 Per quot2x 1tc he areal nuniher 2a 1 eannet he negative Selire the inequality 2t 1339C ftI3I393939f1IC find the cieniain eff 2x 1EU 2x 11EU1 221 31 E I2 DJ t rc 1 The EliI11EI1I11I11I11II3Iquotifi l netatien is EDj EX 5 ss s39 Find the values cat I where the lensininater is El hgr sehiring the equati en 3t 2 E Find the c1einain efthe funeti enf aj 3x 2U 3x 22U2 322 3 The funetien is defined far all real numbers a eseept these values sf 3 where the czieneniinate139 is El Thus the darn ain is I i 2 2 The cieinain in interval netatien is e LJ Finding the Range of a Function To rleterrnine the range of a funetieny P find the set of all real numbers i 39 that are ehtainerl by the equation as I sari es throughout the cleniain eff Find the rleniain and range ef the funetieny vJquot3 Fer J3 3 ta he a real number It eannet he negative Thus the rlernain in interval notation is 3 3933 Te find the range if the funeti an we must rcleterinine the set sf all values efy that are shtainerl la the equation fl 3 far nsnnegatiire values at 2 Per each x 2 Ilwe lmew that Jr 3 U Suhtraeting 3 frern hath sides sf this inequality we see that J 3 E 3 Thus the range ef the funetian in interval notation is 3 3933 Inverse of a Function Definition of a OnetoOne Function 4393i funeti snf is said to he ene te ene previlec1 that the following holds for all 31 and 32 in the domain eff ItquotfIJ1I zfifxgj then I1 I 2 EX Let quot3 3xTquot Use the czlefinitie11 te shew that f is ei1e te eiie Te shew that f is eheteene we hegin 1233 assuming thatj3931I Hquot xgjl We must then shew that 31 32 3393issi1methatfx1 f33 Then 3 fl 332 sa13 3sa23 3 339i3913F39i392 3I13I2 3 3 3939i3913939i392 This shears 123 clefinitien that f is ei1e te she EX LetfI3jI 32 Shaw that f is net ei1e te eiie Te shew that f is net ei1e te ei1ewe nee1 te give a speeifie eaample te shew that the eenditien fx1 fI2 fails te imply that 31 32 Using 31 1a11r1 2 1 we have 3 1w 1t1 an13w1u1t1 Thus f 1 hut 1 1 This shews thatf is net eheteehe Horizontal Line Test A funetien is ene te ene if np herisental line intersects its graph I39 1IJI39E than enee Te see why the herispntal line test is irali 1 the figure helpw shews the graph ef the funeti enflix 22 Fren1 the eirarnple aheire we 1meW thatf is net ene te ene Hete that the herisental line shewn en the graph intersects the graph in twp peints 11 ar1r111 This shewsthatf 1 a even theugh 1 1 439 33 23 E li ti 1 rquotquotiquotquot 1 EX Determine whether the funetien a r2 2 is ene te ene Begin with the graph efgifxjl 32 sh eW11 heleW Te graph the funetienf x 22 2 shift the graph efg 2 units rclewnward 21 n39 all II 1 3 I L 3 Sl the graph J 1 E f39g 2 111its If f H 5 11Itrat1i quot quotI 395kl J II 39I 1 II39 III39 I lquot393939 I 39T quot quot1 I III I 39 I Tm quotPquot39 39 I I 2 a 1 1 2 2 ah 393939I H 1 HH39hH PquotHI 2er The graph ef the given funeti an is shewn heleW 27 f a23 2 2 2 New use the IIeriaental Line Test 3 2 239quot I I 3 2 39 39 2 Since we ear1 find a heriaental line that interseets the graph mere than enee h the Herhsental Line Test f is net er1e te er1e The Inverse of a OnetoOne Function If f is a ene te ene funetien with r1ernainr39E1 anrlrange 5 then there is a ene te ene funetien g the inverse eff with c1en1ain 5 and range 1 such that f g xjj It fer eaeh r in 5 and gfr 53939 fer eaeh 3939fiI391rfi The funetiensf and g are eallerlinverses ef eaeh ether J5 are inverse funetiens Verify that the funeti ens 9 2v 5 and g EX Te verifv that the given funeti ens are inverses ef eaeh ether we must shew that fiat sit Jr aI1s1aff39rlUI eaeurquotquot5 sICII5 a a 2 395j s fa aa s I5 5 339i39 Efquotf39E3I 5I39 ffquotf393I 5 E f5 a E a E 2

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