MA Lecture Notes Clculus Sections.5 nd.6 (nd other teril) Algebr o Functions Su, Dierence, Product, nd Quotient o Functions Let nd g be two unctions with overlpping doins. Then or ll x coon to both doins, the su, dierence, product, nd quotient o nd g re deined s ollows, Su ( g)( ( g( Dierence ( g)( ( g( Product ( g)( ( g( Quotient ( / g)( ( / g(, g ( 0 Exple: Find the su o the unctions. ( x x 5 nd g ( 9x 4x 8 Deinition o the Coposition o Two Functions The coposition o the unctions nd g is given by ( g)( ( g( ). The doin o g is the set o ll x in the doin o g such tht g( is in the doin o. Exple: Given the unctions ( x 5nd g x g x nd ( g )(. (, ind Given the unctions ( nd ( x 6 x g, ind gx nd ( g )(. Identiying coposite unction Soeties we re interested in breking down coposition o unctions into the originl unctions. For exple, i we know tht g(, we could identiy tht ( nd x x g ( x.
Exple: Identiy the unctions ( nd g(. Note tht ultiple solutions re possible. g( x 5 g( x Dierence Quotient In clculus we will hve to evlute the dierence quotient. There re ny dierent ors o the dierence quotient. Here is one or ( x h) ( D. Q. h Evlute the dierence quotient or the ollowing unctions: ( x ( x 5 ( 7
Inverse Functions Since unction cn be represented by the collection o ordered pirs stisying the eqution, we cn reverse the ordered pirs to crete nother reltion clled the inverse reltion. Only when tht inverse reltion is unction do we sy the unction hs n inverse unction. We y rrive t this by restricting the doin o the inverse reltion to crete unction. We use the ollowing nottion, or unction, we sy the inverse unction is. It is true then tht i we pply unction ollowed by its inverse unction, we rrive bck to the vlue we strted with. We cn think o n inverse unction s un-doing wht the unction did. Thus ( ( x. Soe coon exples o inverse unctions would be the ollowing: i ( x 5 then ( x 5 since the wy to undo dding ive is by subtrcting ive. Another exple would be ( x nd ( x but only i we restrict the doin on to be the set o nonnegtive rel nubers. Deinition o Inverse Functions Let nd g be two unctions such tht ( g( ) x or every x in the doin o g nd g( ( ) x or every x in the doin o. Under these conditions, the unction g is the inverse unction o the unction. The unction g is denoted by ( x nd ( x the rnge o ust be equl to the doin o (red inverse ). So. The doin o ust be equl to the rnge o. Note: ens the inverse unction, it does not en the reciprocl (like negtive exponents en in other prts o lgebr.) Finding Inverse Functions. In the eqution (, replce ( by y.. Interchnge the roles o x nd y.. Solve the new eqution or y. I the new eqution does not represent y s unction o x, the unction does not hve n inverse unction. I the new eqution does represent y s unction o x, continue to step 4. 4. Replce y by ( x ) in the new eqution. 5. Veriy tht nd equl to the rnge o re inverse unctions o ech other by sowing tht the doin o is ( ( ) x ( ( )., the rnge o is equl to the doin o, nd, nd
Exples: Find 5 ( x 7. ( x 5x ( x
Trigonoetry Whether we use specil right tringles, chrt, or the unit circle we need to be very coortble with the vlues o trigonoetric unctions or specil ngles. The ollowing re exples o ech. 0 sin 0 cos tn 0 6 4 0 Undeined
We cn express the reltionship between the length o sides o tringle with trigonoetric reltionships deined below. For given right tringle: The bsic reltionships re: o sin h h csc o sin cos h h sec cos o tn sin cos cot o cos tn sin
Exponentil nd Rdicl Rules Things to reeber 0 n n n n / n n Coon Powers 4 5 64 9 8 / 7 8 / ( 64) 4 0 5 00 5 Rdicls Things to reeber / b b b b Be creul! b b Exponent nd Rdicl Rules Exercise: Sipliy the ollowing, express with positive exponents. ) x b) c) x 4 y d) x x e) 5 y ) y 4 y x y 4 x y
Logriths Inverse unctions undo unction. A rith is siply the inverse o n exponentil unction. I n then n. We cn think o this s solving or the exponent. For exple, we know 8 or the bse rised to the power/exponent o gives us 8. Suppose we know the bse is rised to soe power nd the result is 8, we cn sk wht power it ws rised to, or ore precisely 8?. This is red bse o 8. The rule to reeber is tht the bse doesn t chnge but the other roles re reversed (we re given the vlue nd sked to ind the power.) Things to reeber n n Exercises: I we wnt to solve questions, we cn sy or results in vlue o? Fill in the blnks. ) 8 b) c) 7 d) 7 7 e) 4 64 ) 8 ; bse rised to wht power g) 9 7 h) 4 i) j) 4 k) 00 l) 5
Things to reeber Soeties we wnt to sipliy or cobine s. Here re soe things to reeber. r 0 n n r ( n) n It is understood tht. 0 Reeber tht exponentil unctions nd unctions re inverse unctions o ech other. Things to reeber n n nd We cn use s nd their properties to solve equtions. Since s re unctions we cn pply the to both sides o n eqution. Reeber, we cnnot tke the o negtive nuber or o the nuber zero. Not ll questions will result in siple nswers. We note tht nuber such s 5 is n exct vlue but tht clcultor could give us decil pproxition o the vlue. This is done with the chnge o bse orul i the bse is not 0 or e (s is vilble on ost clcultors.) Chnge o bse k k orul Solving Equtions First we wnt to look or coon bses. I we see coon bse we should rewrite the expression. Then we pply with the se bse to both sides nd use our rules to solve. Exples: x 8 Note 8 cn be expressed s bse.
x Now we pply to both sides x ( x ) x. Exercises Solve or x (write nswer s rtionl nuber). 5 x 5. x. 8 6 4. 9 x 8 5. x 64 6. 7 x 4 Solve or x 7. x 0 8. 5 8 9. 0 0. 5 0 Hoework Section. # 9,, 6, 7, 44 Section.5 # 5, 6 Section.6 #,,, 5, 47, 48