in One Variable VI EquationsEquationsEquationsin inOne OneVariablein VariableOne VariableVI VI VI A zero of a f(x) is a z E R such that f(z) = 0. A zero of a functionzerof(x) is a number z ∈ such that f(z) = 0. = A zero of a functionA f(x)ofisa afunctionnumberf(x)z E Ris suchRa numberthat f(z)z E R= such0. that f(z) 0. Examples. Examples.Examples. Examples. • The number 1 is a zero of the function loge(x). That’s because • The number 1 is a zero of the function log (x). That’s because loge(1) = 0. • The number 1 is• Thea zeronumberof the 1functionis a zerologe(x).of thee functionThat’s becauseloge(x). That’s because log (1) = 0. = loge(1)e = 0. loge(1) 0. JOSe () JOSe () JOSe ()

• 6 is a zero of the function x — 6. That’s because [6] — 6 = 0.

• 6 is a zero of• the6 is functiona zero ofxthe− 6.function That’s becausex — 6. That’s [6] − 6 = 0. — = — because [6] 6 0. • 6 is a zero of the function x 6. That’s because [6] — 6 = 0.

// // // 4 4 4 / / / • To state a more general version of the previous example, a is a zero • To state a more• To generalstate a versionmore general of the previousversion of example,the a is a zero of x — a. • To state a more general version of the previous example,previousa is a zeroexample, a is a zero

of x − a. of x — a. of x — a.

x-a x-a x-a / 278 / / 312 278 278 • The function x2 − 9 has zeros 3 and −3. ..TheThefunctionfunctionx2x2——9 9hashaszeroszeros3 3andand—3.—3.

• ••TheTheThezeroszeros zerosofof ofaapolynomial apolynomial polynomialareare arecalledcalled calledroots.roots. roots. x • ••e’e’ehashashasnono nozeros.zeros. zeros. xx

• The set of zeros of sin(x) is {..., −2π, −π, 0, π, 2π, 3π, . . .}. • •TheThesetsetofofzeroszerosofofsin(x)sin(x)isis{...{..., —27r,, —27r,—7r,—7r,0,0,r,r,2r,2r,3ir,.3ir,.. . ThatThatThatis,is, is,thethe thesetset setofof ofzeroszeros zerosofof ofsin(x)sin(x) sin(xis)is isthethe thesetset set

{n7r{n7r{ nπTi|Tin ∈Z}Z}Z }

Sin()Sin()

313 279279 FindingFindingFindingFindingzeroszeros zeroszerosofof offunctionsfunctions functionsof functions If Iff(x)Iff(x)f(isxIf)isaf(x) isfunction,a afunction, function,is a function,thenthen thenthethethen thezeroszeros zerostheofofzerosf(x) off(x)f(areofx)aref(x) arethethe thesolutionsaresolutions solutionsthe solutionsof ofthe ofthe theequationofequation equationthe f(x)=O.f(x)=O.f(x)f(x)=O. = 0.

Examples.Examples.Examples.Examples.

. The.•TheThezeros.zerosThe zerosofzerosofthe ofthe thelinearoflinear linearthepolynomiallinearpolynomial polynomialpolynomial2x2x+ 2x+5+are52x 5are are+the5the thearesolutionssolutions solutionsthe solutionsof ofthe ofthe theof the equation 2x+5 = 0. Subtract 5: 2x = −5. Then divide by 2: x = −5. Thus, = = —. —. equationequationequation2x2x+ 5+ 52x0.+0.Subtract5 Subtract0. Subtract5: 5:2x2x 5:=—5.—5.2xThen=Then—5.divideThendividebydivideby2: 2:x xby=2:—.xThus,=2 Thus, Thus, −5 is the only zero of 2x + 5. — —is2isthe—theonlyisonlythezerozeroonlyof of2xzero2x+ +5.of 5.2x + 5. 2.52.52.5

2 2 2

• The zeros of log (x − 2) are the solutions of log (x − 2) = 0. This — — — = — • The• Thezeros•zerosTheofzerosoflog(xlog(xofe log(x—2) 2)arearethe2)thesolutionsaresolutionsthe solutionsof oflog(xlog(xofe log(x—2) 2) =0. 0.2)ThisThis= 0. This equation is equivalent to x − 2 = e0 = 1, and thus is equivalent to x = 3. — = — = equationequationequationis isequivalentequivalentis equivalenttotox x —2to2 x= 2 ==1, 1,andand= thus1,thusandis isequivalentthusequivalentis equivalenttotox x 3.to3.x 3. That is, 3 is the only zero of the function log (x − 2). — ThatThatis,is,That3 is3 istheis,the3onlyisonlythezerozeroonlyof ofthezerothefunctionoffunctionthe functionloge(xloge(xe loge(x—2).2). — 2). 1o1o(x-2)(x-2)1o (x-2) z(t3z(t3 z(t3

• The zeros of x2−4x+3 are the solutions of the equation x2−4x+3 = 0. • The• Thezeros•zerosTheofofx2—4x+3zerosx2—4x+3of x2—4x+3arearethethesolutionsaresolutionsthe solutionsof ofthetheequationofequationthe equationx2—4x+3x2—4x+3x2—4x+3= =0. 0. = 0. UsingUsingUsingtheUsingthe thequadraticquadratic quadraticthe quadraticformula,formula, formula,formula,wewe wecancan canwefindfind findcanthatthatfind thatthethethat thesolutionssolutions solutionsthe solutionsareare are1 1and 1areand and3.1 3.and 3. 3. Therefore, 1 and 3 are the zeros of x2 − 4x + 3. — Therefore,Therefore,Therefore,1 and1 and3 1are3 andarethethe3 zerosarezerostheofofzerosx2x2 —4xof4xx2+ +3.— 3.4x + 3. I I 3 I 3 3

314 280280 280 √ √ • The zeros of the function 2 + x − 3 are solutions of 2 + x − 3 = 0. . The zeros of the function 2+ /‘x —√3 are solutions of 2+ \/x — 3 = 0. This equation is equivalent to x − 3 = −2, and thus it has no solutions This equation is equivalent to /x — 3 = —2, and thus it has √no solutions since square-roots are never negative. Therefore, 2 + x − 3 has no zeros. since square-roots are never negative. Therefore, 2 + /x — 3 has no zeros.

3 An illustrationAn illustrationof when ofnot whento divide not to divide If asked toIffind askedthe tosolutions find the solutionsof the equation of the equation (x—1)(x—2)=3(x—1)(x − 1)(x − 2) = 3(x − 1) you might be tempted to divide both sides of the equation by (x − 1) leaving you might be tempted to divide both sides of the equation by (x — 1) leaving you with you(x —2) with= (3,x −a much2) = 3,simpler a muchequation. simpler equation.However, However,while (x —2) while= 3 (xis− 2) = 3 is certainly certainlya much simpler a muchequation simpler(the equationsolution (theto solutionit is 5) toit is itnot is 5)equivalent it is not equivalent to the originalto theequation original(x—1)(x—2) equation (x−=1)(3(x—1).x−2) =That’s 3(x−1).because That’sthe becausefunction the function (x − 1) has a zero (the zero is 1), and multiplying or dividing an equation by (x — 1) has a zero (the zero is 1), and multiplying or dividing an equation by a functionathat functionhas a thatzero hasin athe zerodomain in theof domainthe equation of theis equationnot a valid is notway a valid way of obtainingof obtainingan equivalent an equivalentequation. equation. The stepsThebelow stepsdescribe belowthe describegeneral theprocess generalfor processsolving forequations solving equationsof the of the form h(x)f(x)form h=(xk(x)g(x).)f(x) = hAfter(x)g(xlisting). Afterthe listinggeneral thesteps, generalwe’ll steps,return we’llto the return to the example of (x − 1)(x − 2) = 3(x − 1). example of (x — 1)(x —2) = 3(x — 1).

Steps forStepssolving for solvingh(x)f(x)h(=x)h(x)g(x)f(x) = h(x)g(x)

Step 1: FindStepthe 1: Finddomain theof domainthe equation. of the equation.Call this set CallD. this set D.

Step 2: FindStepthe 2: Findzeros theof h(x) zerosthat of hare(x)in thatD. areCall inthisD.set CallZ. thisThe setnumbersZ. The

in Z will inbe Zsolutionswill be solutionsof h(x)f(x) of h=(xh(x)g(x))f(x) = becauseh(x)g(x)if becauseh(z) = 0 if h(z) = 0 then h(z)f(z)then =h(zOf(z))f(z)= =0 0=f(Og(z)z) = 0= =h(z)g(z). 0g(z) = h(z)g(z).

Step 3: FindStepthe 3: Findsolutions the solutionsof the equation of theh(x)f(x) equation =h(xh(x)g(x))f(x) = withh(x)gthe(x) with the restricted domain D − Z. The function h(x) has no zeros in D − Z, restricted domain D — Z. The function h(x) has no zeros in D — Z, so you cansostart you canwith startdividing withby dividingh(x) to byobtainh(x) tothe obtainequivalent the equivalent equation f(x) = g(x). equation f(x) = g(x). Step 4: CollectStep 4:theCollectzeros thefrom zerosStep from2 and Stepthe 2solutions and the solutionsfrom Step from3. These Step 3. These are the solutionsare the solutionsof the original of theequation. original equation. 281 315 Examples.

• Let’s return to the equation (x−1)(x−2) = 3(x−1) and proceed through the four steps outlined above. = (2 (Step. Let’s 1):find (x −the1)(xsolutions− 2) = 3(ofxthe− 1)equation is a (x2 — 16) equation,1og(x) so the— domain16). is R(Step. Written1): The in thedomain notationof the of theequation generalis steps(0, oc), above,becauseD =weRcan. only take the(Steplogarithm 2): Theof functiona positive thatnumber. we’d like to divide both sides of the equation by 2 is (x(Step− 1).2): ItsThe zero iszeros 1, whichof (x2 will— be16) aare solutionthe solutions of (x − 1)(ofx − 2)— =16 3(=x −0, 1)or sinceequivalently, ([1] − 1)([1]the −solutions2) = 0(of−x21)= =16, 0 =which 3(0)are =4 3([1]and−—4.1).However, Writtenof inthese the = (2 notationtwo zeros, of theonly general4 is in steps(0, oc), above,the Zdomain= {1}.of (x2 — 16) — 16) log(x),

so(Stepit’s the 3):only We needzero toof find(x2 — the16) solutionsthat concerns of theus equationin this problem. (x − 1)(xWith− 2)the = 3(notationx − 1) withof the thegeneral restrictedsteps domainlisted ofabove,R − {1Z}.= On{4}. this domain, the function

(x −(Step1) has3): noWe zeros,need to sofind we canthe dividesolutions byof (x(x2− 1).— 16) Theloge(x) resulting= (x2 equivalent— 16) with equationthe restricted is (x −domain2) = 3.(0, Weoo) can— {4}. addThe 2 tofunction see that(x2 the— solution16) has no is xzeros= 5.in the

set(Step(0, 4):oc) The— {4}. zeroTherefore, of (x − 1)—thewe can numberdivide the 1—andequation the solution(x2 — 16) ofloge(x) = (x(x2− —1)(16)x −by2)(x2 = 3(— x16)− to1)obtain with domainthe equivalentR − {1}equation—the numberloge(x) 5—are= 1. theThe solutionsonly solution of (x −of 1)(thisx −equation2) = 3(isx −x =1).e1 To= repeat,e. the set of solutions of the = equation(Step (4):x −The1)(xsolutions− 2) = 3(ofx −(x21)— is4){loge(x)1, 5}. (x2 — 4) are 2 (from Step 2) and e (from Step 3).

(x-;)Cx-z

5

316

283 Examples. 2 2 • Let’s find the solutions of the equation (x − 16) loge(x) = (x − 16). (Step 1):• Let’s Thereturn domainto the of theequation equation(x — is1)(x (0, ∞—2)), because= 3(x —1) weand canproceed only takethrough the logarithmthe four ofsteps a positiveoutlined number.above.

(Step(Step 2): The1): (x zeros— 1)(x of—2) (x2 −= 16)3(x are— 1) theis a solutionspolynomial ofequation,x2 − 16so =the 0, ordomain equivalently,is R. Written the solutionsin the ofnotationx2 = 16,of whichthe general are 4 andsteps−above,4. However,D = JR. of these 2 2 two zeros,(Step only2): 4The is infunction (0, ∞),that the domainwe’d like ofto (dividex − 16)both =sides (x −of16)the logequatione(x), by 2 so it’sis the(x only— 1). zeroIts zero of (xis−1,16)which thatwill concernsbe a solution us in thisof (x problem.— 1)(x — With2) = the3(x — 1) notationsince of the([1] general— 1)([1] steps— 2) listed= O(—1) above,= 0Z == 3(0){4}. = 3([1] — 1). Written in the notation of the general steps above, Z =2 {1}. 2 (Step 3): We need to find the solutions of (x −16) loge(x) = (x −16) with 2 the restricted(Step domain3): We (0need, ∞)to−{find4}.the Thesolutions functionof (xthe−16)equation has no(x zeros— 1)(x in the— 2) = 2 set (0,3(x∞)—−1) {4with}. Therefore,the restricted we candomain divideof theJR — equation{1}. On this (x −domain,16) logthee(x)function = 2 2 (x − 16)(x — by1) (hasx −no16)zeros, to obtainso we thecan equivalentdivide by (x equation— 1). The logeresulting(x) = 1.equivalent The 1 only solutionequation ofis this(x — equation2) = 3. isWex =cane add= e.2 to see that the solution is x = 5. 2 2 (Step 4):(Step The4): solutionsThe zero ofof (x(x−—16)1)—the loge(xnumber) = (x −1and16) arethe 4 (fromsolution Stepof 2) and e (from(x — 1)(x Step— 3).2) = 3(x — 1) with domain JR — {1}—the number 5—are the

solutions of (x — 1)(x — 2) = 3(x — 1). To repeat, the set of solutions of the

equation (x — 1)(x —2) = 3(x — 1) is {1, 5}.

(a

282 317 x x • Let’s• findLet’s thefind solutionsthe solutions of xe =of exeX. = e’. (Step 1):(Step The1): impliedThe implied domaindomain is R. is R. (Step 2):(Stepex has2): noeX has zeros.no zeros. x x x (Step 3):(Step Dividing3): Dividingxe = xexe by= ecx yieldsby eX theyields equivalentthe equivalent equationequationx = 1.x = 1. Thus, 1Thus, is the1 onlyis the solution.only solution.

(Step 4):(Step There4): areThere noare zerosno fromzeros Stepfrom 2,Step so 12, isSO the1 onlyis the solution.only solution.

i

2x x • Let’s• findLet’s thefind solutionsthe solutions of (x +of 2)(xe +=2)e2x (x += 2)(xe .+ 2)ex. (Step 1):(Step The1): impliedThe implied domaindomain is R. is R. (Step 2):(Step The2): zeroThe ofzero (x +of 2)(x is+−2)2.is It’s—2. in theIt’s domainin the domain of the equation,of the equation, so so it’s a solution.it’s a solution. (Step 3):(Step Dividing3): Dividing (x + 2)e(x2x+=2)e2’ (x + 2)e(xx +by2)ex (x +by 2)(x gives+2) usgives theus equivalentthe equivalent 2x x x e2x equationequatione = ee2x. We= cx. canWe thencan dividethen bydividee toby getcx toex get= 1 which= 1 iswhich the sameis the same 2x−x x equationequation as e as=e2 1 or= more1 or simply,more simply,e = 1.cx Applying= 1. Applying the logarithmthe logarithm yields yields that x =that logxe(1)= log(1) = 0. = 0. (Step 4): −2 and 0 are the solutions of (x + 2)e2x =2)e2x (x + 2)ex. 2)ex. (Step 4): —2 and 0 are the solutions of (x + = (x + 1’

318 284 *************

Chapter Summary

The solutions of h(x)f(x) = h(x)g(x) are the solutions of f(x) = g(x) together with the solutions of h(x) = 0.

319 Exercises

For #1-6, find the zeros, if there are any, of the given functions.

1.) x − 2

2.) 3x + 4

3.) x2 − 81

4.) x2 + 7 √ 5.) 2 x − 7

6.) e3x−2

Find the solutions of the following equations in one variable. √ 7.) x3 x = 9x3

8.) (x − 2)(x3 − 2x)2 = −5(x − 2)

2 2 9.) (x − 4) loge(x + 2) = 2(x − 4)

x 10.) loge(x − 1)e = −2 loge(x − 1)

2 2 2 11.) −8(x − 25) = (x − 25)[loge(x) − 6 loge(x)] 12.) (2x + 1)5 = 9(2x + 1)3 √ 13.) (x2 − 4) x = −(x2 − 4)

x 1  x 14.) e x + x = 2e √ √ 15.) e3x x = ex−2 x

16.) (x + 3) loge(2x) = (x + 3) loge(x + 1) 17.) (x − 4)5 = (x − 4)6

18.) (x − 2)(x − 3) = (x − 2)(x + 1)

19.) x = x2

320