Loudoun Valley High School Calculus Summertime Fun Packet

Transcription

1 Loudoun Vlley High School Clculus Summertime Fun Pcket We HIGHLY recommend tht you go through this pcket nd mke sure tht you know how to do everything in it. Prctice the problems tht you do NOT remember! Solutions will be posted on our websites. Nme:

2 Comple Frctions When simplifying comple frctions, multiply by frction equl to 1 which hs numertor nd denomintor composed of the common denomintor of ll the denomintors in the comple frction. Emple: ( ) ( ) ( ) ( ) 5( )( ) 1( ) Simplify ech of the following

3 Functions To evlute function for given vlue, simply plug the vlue into the function for. Recll: f g( ) f( g( )) OR f[ g( )] red f of g of mens to plug the inside function (in this cse g ) in for in the outside function (in this cse, f). Emple: Given f ( ) 1 nd g( ) find fg. f( g( )) f( ) ( ) 1 ( 8 16) ( ( )) 16 f g Let f ( ) 1 nd g( ) 1. Find ech. 6. f () 7. g( ) 8. f( t1) 9. f g( ) 10. ( ) g f m 11. f ( h) f( ) h Let f ( ) sin Find ech ectly. 1. f 1. f Let f ( ), g( ) 5, nd h( ) 1. Find ech. 1. h f( ) 15. f g( 1) 16. g h ( )

4 f( h) f( ) Find for the given function f. h 17. f( ) f( ) 5 Intercepts nd Points of Intersection To find the intercepts, let y 0 in your eqution nd solve. To find the y intercepts, let 0 in your eqution nd solve. Emple: y int. ( Let y 0) 0 0 ( )( 1) 1or i ntercepts ( 1,0) nd (,0) yint. ( Let 0) y 0 (0) y y intercept (0, ) Find the nd y intercepts for ech. 19. y 5 0. y 1. y 16. y

5 Use substitution or elimintion method to solve the system of equtions. Emple: y169 0 y 90 Elimintion Method ( )( 5) 0 nd 5 Plug = nd 5 into one originl y y y 0 16 y y 0 y Points of Intersection (5,), (5, ) nd (,0) Substitution Method Solve one eqution for one vrible (1st eqution solved for y) y ( 169) 9 0 Plug wht y is equl to into second eqution. The rest is the sme s ( ) previous emple ( )( 5) 0 or 5 Find the point(s) of intersection of the grphs for the given equtions.. y 8 y 7. y 6 y 5. y y Intervl Nottion 6. Complete the tble with the pproprite nottion or grph. Solution Intervl Nottion Grph 1, 7 8 5

6 Solve ech eqution. Stte your nswer in BOTH intervl nottion nd grphiclly Domin nd Rnge Find the domin nd rnge of ech function. Write your nswer in INTERVAL nottion. 0. f 1. f( ). f( ) sin. ( ) 5 f( ) 1 Mth Mni Crossword 6

7 Inverses To find the inverse of function, simply switch the nd the y nd solve for the new y vlue. Emple: f( ) 1 Rewrite f() s y y= 1 Switchndy = y 1 Solve for your new y y 1 Cube both sides y1 Simplify y 1 Solve for y 1 f ( ) 1 Rewrite in inverse nottion Find the inverse for ech function.. f ( ) 1 5. f( ) if 0 Fourteen mtchsticks form two identicl equilterl tringles nd two identicl squres. Move five mtchsticks to form three identicl equilterl tringles nd three identicl squres. 7

8 Also, recll tht to PROVE one function is n inverse of nother function, you need to show tht: f( g( )) g( f( )) Emple: If: 9 f( ) nd g( ) 9 show f() nd g() re inverses of ech other f( g( )) 9 g( f( )) f( g( )) g( f( )) therefore they re inverses of ech other. Prove f nd g re inverses of ech other. 6. f( ) g( ) 7. f( ) 9, 0 g( ) 9 8 Euclid

9 Eqution of line Slope intercept form: y m b Verticl line: c (slope is undefined) Point slope form: y y1 m( 1) Horizontl line: yc (slope is 0) 8. Use slope intercept form to find the eqution of the line hving slope of nd y intercept of Determine the eqution of line pssing through the point 5, with n undefined slope. 0. Determine the eqution of line pssing through the point, with slope of Use point slope form to find the eqution of the line pssing through the point 0, 5 with slope of /.. Find the eqution of line pssing through the point, 8 nd prllel to the line y 1.. Find the eqution of line perpendiculr to the y is pssing through the point, 7.. Find the eqution of line pssing through the points, 6 nd 1,. 5. Find the eqution of line with n intercept, 0 nd y intercept 0,. 9

10 Rdin nd Degree Mesure Use degrees rdins to convert bck nd forth between degrees nd rdins Convert to degrees:. 5 6 b. 5 c..6 rdins 7. Convert to rdins:. 5 b. 17 c. 7 Angles in Stndrd Position 8. Sketch the ngle in stndrd position b. 0 c. 5 d. 1.8 rdins 10 Riddle Me This Why re mn hole covers round?

11 Reference Tringles 9. Sketch the ngle in stndrd position. Drw the reference tringle nd lbel the sides, if possible.. b. 5 c. d. 0 Qudrntls You cn determine the sine or cosine of qudrntl ngle by using the chrts. SIN COS TAN 1 0 ND Emple: sin 90 1 cos 0 0 ND 50..) sin180 b.) cos 70 c.) sin( 90 ) d.) tn e.) cos60 f.) cos( ) 11

12 Grphing Trig Functions f = sin f = cos ysin nd ycos hve period of π nd n mplitude of 1. Use the prent grphs bove to help you sketch grph of the functions below. For f AsinB C D, A mplitude, = period, B opposite of C phse shift nd D verticl shift. Grph one complete period of the function. 51. f ( ) 5sin 5. f ( ) sin 5. f( ) cos 5. f ( ) cos Trigonometric Equtions: Solve ech of the equtions for 0,. Isolte the vrible, sketch reference tringle, find ll the solutions within the given domin, 0,. Remember to double the domin when solving for double ngle. Use trig identities, if needed, to rewrite the trig functions. (See formul sheet t the end of the pcket.) sin 56. cos 1

13 57. 1 cos 58. sin sin 60. cos 1cos cos 0 6. sin cos cos 0 1

14 Inverse Trigonometric Functions: Recll: Inverse Trig Functions cn be written in one of wys: rcsin sin 1 Inverse trig functions re defined only in the qudrnts s indicted below due to their restricted domins. cos 0 sin 0 cos 0 tn 0 sin 0 tn 0 Emple: Epress the vlue of y in rdins. 1 y rctn Drw reference tringle. 1 This mens the reference ngle is 0 or. So, y = so tht it flls in the intervl from y 6 6 Answer: y For ech of the following, epress the vlue for y in rdins. 6. y rcsin 6. rccos 1 y 65. y rctn( 1) 1

15 Emple: Find the vlue without clcultor. 5 cosrctn 6 61 Drw the reference tringle in the correct qudrnt first. 5 θ Find the missing side using Pythgoren Theorem. 6 Find the rtio of the cosine of the reference tringle. cos 6 61 For ech of the following give the vlue without clcultor. 66. tn rccos secsin sin rctn sin sin

16 Circles nd Ellipses h y k r Minor Ais h yk b 1 b CENTER (h, k) Mjor Ais FOCUS (h - c, k) c FOCUS (h + c, k) For circle centered t the origin, the eqution is y r, where r is the rdius of the circle. y For n ellipse centered t the origin, the eqution is 1, where is the distnce from the center to the ellipse b long the is nd b is the distnce from the center to the ellipse long the y is. If the lrger number is under the y term, the ellipse is elongted long the y is. For our purposes in Clculus, you will not need to locte the foci. Grph the circles nd ellipses below: 70. y y y y

17 Logrithms There re TWO bses tht re used for logrithms Common Log (Bse 10) y log sme s 10 y Nturl Log (Bse e) y y ln sme s e y log rgument nswer bse Eponentil Form Logrithmic Form y y log Inverse Properties of Logrithmic Epnsion Eponentil Equivlent 1. log 1 0 nd ln1 0 comes from the fcts tht 0 1 nd 0 e 1. log 1 nd ln e 1 comes from the fcts tht 1 1 nd e e log. M ln M M nd e M comes from rewriting ech s logrithmic function log M log M nd ln M e ln M r. log r r nd ln e r comes from rewriting logrithm in eponentil form r r r r nd e e (remember the understood e in the ln) M N M N log MN log M log N comes from the eponent rule log M log M log N N comes from the eponent rule M N M N 7. r log M rlog M proved by using log M rewritten eponentilly, ech side rised to the r power then simplified 8. if log M log N then M=N comes from the eponent rule if M N then M=N 17

18 Epnd or condense: log 1 log 1 log log Solve: 76. log log log 77. log log 9 18

19 Formul Sheet Reciprocl Identities: 1 csc sin 1 sec cos 1 cot tn Quotient Identities: sin tn cos cos cot sin Pythgoren Identities: sin cos 1 tn 1 sec 1cot csc Double Angle Identities: sin sin cos tn tn 1 tn cos cos sin 1 sin cos 1 Logrithms: y log is equivlent to y Product property: log mn log m log n b b b m Quotient property: log log m log n b b b p Power property: log m plog m Property of equlity: If log m log b b b b n n, then m = n Chnge of bse formul: Derivtive of Function: log log n log b b n Slope of tngent line to curve or the derivtive: lim Slope intercept form: y m b Point slope form: y y1 m( 1) Stndrd form: A By C 0 19