e.g. f(x) = x domain x 0 (cannot find the square root of negative values)

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1 CORE SUMMARY NOTES Functions A function is rule which genertes ectl ONE OUTPUT for EVERY INPUT. To be defined full the function hs RULE tells ou how to clculte the output from the input DOMAIN the set of vlues which will be used s inputs e.g. f() = domin 0 (cnnot find the squre root of negtive vlues) ALTERNATIVE NOTATION f : mens function f such tht mps to input is converted to output ÎR cn be n rel number There re different tpes of functions MANY-ONE = ONE-ONE = = + different inputs give the sme output for ech output there is onl one possible input The RANGE of function is the complete set of ll of the OUTPUTS An INVERSE function is denoted b f -. ONLY ONE-ONE FUNCTIONS HAVE INVERSES The DOMAIN of n inverse function is the RANGE of the function e.g. The function f is defined b f() = - find f - () Step : Write the rule in terms of nd Step : Rerrnge to mke the subject Step : Replce the s with s = - = + f - () = +

f() = domin 0 (cnnot find the squre root of negtive vlues) ALTERNATIVE NOTATION f : mens function f such tht mps to input is converted to output ÎR cn be n rel number There re different tpes of

2 Using the sme scle on the nd is, the grphs of function nd it s inverse hve reflection smmetr in the line = COMPOSITE FUNCTIONS The function gf is clled composite function nd tells ou to do f first then gf() e.g. f() = + g() = + gf() = (+) + fg() = ( + ) + gf() = + + fg() = + 7 The Modulus function is the modulus of or the bsolute vlue The modulus of rel number cn be thought of s its distnce from 0 nd it is lws positive. = - = The grph of = f() is 8 6 = trnsltion éù 8 ê ú ë0 û To sketch the grph of = (f) first sketch the grph of = f() Tke n prt of the grph tht is below the -is nd reflect it in the -is. SOLVING EQUATIONS Alws sketch the grph before ou strt to determine the number of solutions A function is defined b f() = + - Solve the inequlit f() < The grphs shows solutions (+) - = = - ( + ) - = - - = 5 = or = -

uk gf() = (+) + fg() = ( + ) + gf() = + + fg() = + 7 The Modulus function is the modulus of or the bsolute vlue The modulus of rel number cn be thought of s its distnce from 0 nd it is lws positive.

3 Trnsforming Grphs éù TRANSLATION - to find the eqution of grph fter trnsltion of ê ú ou ëbû replce b (-) nd b ( - b) - b = f(-) or = f(-) +b e.g. The grph of = - is trnslted through the grph formed. ( + ) = (-) = é ù ê ú. Write down the eqution of ë- û = - = REFLECTING 5 = -f() Reflection in the -is, replce with Reflection in the -is, replce with = f(-) STRETCHING Stretch of fctor k in the direction replce b k Stretch of fctor k in the direction replce b k = f( k ) = kf() COMBINING TRANSFORMATIONS When ppling trnsformtions the order does not mtter if one involves replcing nd the other replcing. If both trnsformtions involve replcing (or ) then the order could mtter e.g. The grph of = is first trnslted b Find the eqution of the finl imge. Trnsltion = ( - ) Reflection = (- ) = ( + ) éù ê ú nd then reflected in the -is ë0û TRIGONOMETRY INVERSE FUNCTIONS = sin - rcsin or sin domin p p rnge

Write down the eqution of ë- û = - = 6 + 6 5 5 REFLECTING 5 = -f() Reflection in the -is, replce with Reflection in the -is, replce with = f(-) STRETCHING Stretch of fctor k in the direction replce b

4 = cos - rccos or cos domin rnge 0 p = tn - rctn or tn domin Î R rnge p p sec is defined s co s = sec hs domin Î ÎR ¹ ± p, ± p, ±5p nd rnge nd cosec c is defined s si n = cosec hs domin ÎR ¹ 0, ± p, ± p, ± p nd rnge nd 5 Nturl Logrithms nd e e is n irrtionl; its vlue is correct to 9 deciml plces Nturl Logrithms use e s bse nd we write log e s ln e = Þ = ln

s = sec hs domin Î ÎR ¹ ± p, ± p, ±5p nd rnge nd cosec c is defined s si n = cosec hs domin ÎR

5 e.g. Solve the eqution e 5 = 0 e 5 = 5 = ln = 5 ln = 0 97 If the question sks for n ect nswer do not chnge into decimls e = Þ = ln so e nd ln re inverse functions e is positive for ll so ln is defined onl for positive vlues of 6 = e = ln 6 e.g The function g is defined b g() = e -5 + for ll rel. Find nd epression for g - () nd stte it s domin nd rnge. = e -5 + = e -5 ½( - )= e -5 ln(½( - )) = 5 ln(½( - )) + 5 = g - ()= ln(½( - )) + 5 The rnge of g is g() > so the domin of g - () is > The domin of g is ll rel vlues of so the rnge g - () is ll rel vlues Trnsformtion of grphs e.g. Describe the sequence of geometricl trnsformtions needed to obtin the grph of = e - from the grph of = e. Reflection in the -is gives = e - Stretch fctor of in the direction gives =e -

Solve the eqution e 5 = 0 e 5 = 5 = ln = 5 ln = 0 97 If the question sks for n ect nswer do not chnge into decimls e = Þ = ln so e nd ln re inverse functions e is positive for ll so ln is defined

6 6 Differentition Ke points from C nd C The derivtive of n = nn - If f () > 0, f is incresing t =. If f () < 0, f is decresing t = The points where f () = 0 re clled sttionr points If f () > 0 then = is locl minimum If f () < 0 then = is locl mimum The derivtive of e is e The derivtive of ln is e.g. If f() = e + ln( ) find f () f() = e + ln + ln f ()= e + Product Rule If = uv then d dv = u + d d du v d Quotient Rule du dv If = u v - u v then d = d d d v Chin Rule Find d given tht ln ( + ) d d du d = d du d Let u = + so = ln u du = d d = du u d = d + Differentiting sin, cos nd tn The derivtive of sin is cos. The derivtive of cos is sin The derivtive of tn is sec The derivtive of f() is f () The derivtive of f(+b) is f ( + b) e.g. Find f () given tht f() =sin cos Let u=sin nd v=cos du = d cos dv = -sin d d = (sin)( - sin) + (cos )( cos) d d = coscos - sin sin d

g. If f() = e + ln( ) find f () f() = e + ln + ln www.mthsbo.org.

7 7 Integrtion Ke points from C nd C ò n d = n + n + + c ò f() d b Gives the re under the grph of =f() between = nd =b Ares below the -is re negtive Ke integrls TO LEARN ò e d = e + c ò d = ln çç + c ò + b d = ln ç + bç + c ò sin d = cos + c ò cos d = sin + c Integrtion b SUBSTITUTION e.g. Use the substitution u = - to find ò First find du in terms of d du = so du = d d Rewrite the function in terms of u nd du ò d = ò ( )d = ò u du = ò u du Crr out the integrtion in terms of u Rewrite the result in terms of = æ è ç u ö ø + c = u + c ( ) + c Integrtion b prts ò u dv d = uv d ò v du d + c d

8 e.g. Find ò e5 d Let u = nd dv = e 5 d du = v = e 5 d 5 ò e5 = e 5 5 ò 5 e5 d = 5 e5 5 e5 + c Integrting f' ( ) f ( ) (the numertor is multiple of the derivtive of the denomintor) ò f '() = ln çf()ç + c f () e. g. Find ò + d The derivtive of the denomintor, + is, so think of the numertor s ¼ ( ) ò d = + ò + d = lnç + ç + c ò ò + d = tn æ è ç ö ø d = sin æ ç ö è ø + c + c STANDARD INTEGRALS TO LEARN 8 Solids of Revolution Revolution bout the -is The volume of solid of revolution bout the -is between = nd = b is given b ò b p d

9 Revolution bout the -is The volume of solid of revolution bout the -is between = nd = b is given b ò b p d e.g. The region shown is rotted through p rdins bout the is. Find the volume of the solid generted. First epress in terms of = 8 = 8 Þ 8 = Þ = Þ = When = 0 = 0 when = = Volume = ò p d = 0 ò 0 d = p é ë ê 5 ù 5 û ú = 5 p 9 Numericl Methods Chnge of sign For n eqution f() =0, if f( ) nd f( ) hve opposite signs nd f() is continuous between nd, then root (solution) of the eqution lies between nd Stircse nd Cobweb Digrms If n itertive formul (recurrence reltion) of the form n+ = f( n ) converges to limit, the vlue of the limit is the -coordinte of the point of intersection of the grphs =f() nd = The limit is therefore the solution of the eqution f() = A stircse or cobweb digrm bsed on the grphs of = f() nd = illustrtes the convergence e.g Solve the eqution + =0 First we will write it in the form = f() + = Þ + = Plotting the grphs = + nd = the solution is the point of intersection of the two grphs

First epress in terms of = 8 = 8 Þ 8 = Þ = Þ = When = 0 = 0 when = = Volume = ò p d = 0 ò 0 d = p é ë ê 5 ù 5 û ú = 5 p 9 Numericl Methods Chnge of sign For n eqution f() =0, if f( ) nd f( ) hve

10 We cn confirm tht there is point of intersection between = nd = b chnge of sign the vlues re substituted. Substituting = into = + gives =.66 (shown on the digrm) Substituting =.66. =.8 Repeting this the vlues converge to.57 The solution of + =0 is =.57 The mid-ordinte Rule (Numericl Integrtion) Gives nd pproimtion to the re under grph. The re is divided into strips of equl width. The vlue of the function hlfw cross ech strip (the mid-ordinte) is clculted Totl re = width of strip sum of mid-ordintes Y.5 ln.5.5 ln.5.5 ln ln 5.5 =ln Are = (ln.5 +ln.5 + ln.5 + ln 5.5) = ln ( ) = ln = Simpson s Rule Gives more ccurte pproimtion to the re under grph. An even number of strips of equl width re used. The ordintes 0,,,. re the vlues of the function on the verticl edges of the strips. The re is given b h( sum of end ordintes + sum of odd ordintes + sum of remining even ordintes)

The vlue of the function hlfw cross ech strip (the mid-ordinte) is clculted Totl re = width of strip sum of mid-ordintes Y.5 ln.5.5 ln.5.5 ln.5 5.5 ln 5.5 =ln Are = (ln.5 +ln.5 + ln.5 + ln 5.

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials

Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic