Pre-Calculus Review Preemptive Strike

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1 Pre-Calculus Review Preemptive Strike Attaced are some notes and one assignment wit tree parts. Tese are due on te day tat we start te pre-calculus review. I strongly suggest reading troug te notes torougly - and not just skipping to te problems. Tis is material tat you sould already know. Te topics for te tree assignments are as follows: Part A: Eponents and radicals, absolute value, and piecewise-defined functions; solving non-linear inequalities by metod of sign analysis Part B: Basic graps; equations of perpendicular bisector lines and circles; Domain of functions Part C: Simplifying algebraic epressions often encountered in calculus; syntetic division and rational root teorem; solving trig equations Do te work for te assignment on your own paper. All assignments anded in during tis class will only be accepted if tey are college-level. Headings are epected on all pages wit your full name, te assignment, and te date. No fringed edges on note paper. If you don t follow tese instructions for acceptable-looking work, don t be surprised if your work isn t even scored. Sow work to justify all answers and eplain your solution metods. Answers in tis class submitted witout work are considered not complete. All material in tis packet even toug you are doing it on your own will be tested. Tis is wort 0 points and will be spot-cecked. You will not get full-credit if te answers tat are cecked contain errors. Notes for Part A: Eponents and Radicals: For basics on eponents and radicals, consult * Here are a few eamples: E. : Simplify: [Note: write answer in simplified radical form, wic indicates lowest possible inde, and factoring out as muc of te epression from under te radical symbol as possible.] Solution: = / / / / 9 / / / E. : Simplify write answer in simplified radical form: 8 Solution: 8 / 8 / 8 / E. : Find te solution set of te equation: / / 0

2 Solution: Let u= /, and substitute, to get te following quadratic equation: u 0 u u 0 u / u or u= / / Now, to solve for, we recall tat u= and substitute for u into tese equations and solve: / / /8 or / 7. You must ceck your solutions, for some equation solving tecniques suc as squaring bot sides of an equation sometimes produce etraneous solutions wic do not satisfy te original equation. Cecking tese two apparent solutions in te original equation, we find tat tey are solutions. So, te solution set is: /8, 7 Absolute Value and Piecewise Defined Functions: Definition of Absolute Value: Also Recall: 0 9, not and 0 E. : If a > 0 and b < 0, simplify ab. Simplify in te case of absolute values, means remove te absolute value sign Solution: It is ONLY possible to simplify tis epression because we ave been given information about te signs of a and b. Because te product of a positive and a negative is a negative, we know tat ab < 0. You need to realize tat for any < 0,. Terefore, ab ab We are only able to do tis problem because we know te sign of a and b from te given information. E. : Rewrite te following functions as a piecewise defined function witout absolute values: f Solution: To remove absolute values, we need to know wen te epression inside is positive and wen it is negative. We know tat = 0 wen = ½. Note tat wen > ½, ten > 0 and wen < ½ ten < 0. So, using te definition of Absolute Value and noting were is positive and were it is negative, we can remove absolute value signs and write: f wic simplifies to f For eamples on solving inequalities by te Metod of Analysis of Signs, see

3 Notes for Part B: Basic Graps: You are epected to be able to easily and quickly grap from memory were k, m, b, and r are constants: y, y, y, y, y k, k, y m b, y tan, and y = arctan,. y r, y r, y sin, y cos, Tese sould take no time at all. Drawing tem sould be automatic. You sould also be able to switc te & y in any of te above equations and draw tose graps easily. You sould be able to draw sifted and transformed graps i.e. if you know te grap of y f, ten you can easily grap y f, y a f, y f, y f k, y f, etc.. For furter notes on graping, see *. No graping eamples provided. Perpendicular Bisector Equation and Equations of Circles: E. 6: Given te coordinates of A and B, find te equation of te perpendicular bisector of segment AB. A, B, Solution: Step : Find te slope of segment AB. yb ya 6 m AB 7 B A A line perpendicular to AB as a slope tat is te negative reciprocal of perpendicular bisector is 6 7. m AB, so te slope of te Step : If M is te midpoint, let M M, ym. A B ya yb Ten M and y M. Tus M,. Step : Te perpendicular bisector is te like troug M, wit slope y y m y y 6 6 So te perpendicular bisector is 7 y or y m. 6 E.7: Find te center and radius of te circle y 6 y 0. Solution: 6 y y Completing te square, you get, 6 9 y y 9 y 9

4 y, wic is of te standard form r k y. So, te center is -, and te radius is. Domain of Functions: Wen defining a function, te domain of te function must be given eiter implicitly or eplicitly. For instance if f is defined by f it is implied tat can be any real number. However, if f is defined by f 0 Te domain of f consists of all real numbers inclusively between and 0. Similarly, if g is defined by te equation g it is implied tat because te quotient is undefined for ; ence te domain of g is te set of all real numbers ecept -. If 9 it is implied tat is in te closed interval, because 9 is not a real number for > or < -. Tus te domain of is, and te range is, 0. For furter notes of Domain of Functions, see *. Observe tat you will need te metods of analysis of signs on many problems. Notes for Part C: Algebraic Epressions tat occur in Calculus: E. 8: Simplify: 0 wic is an epression occurring in calculus. One Solution: Te LCD in te numerator is = Or you may prefer Anoter Solution =

5 E. 9: Simplify: Solution: Multiply numerator and denominator by to clear comple fractions. = Solution : You know ow to factor someting in tis form: 8. In tis factoring, you looked at all addends, and you pulled out teir common factor. In tis case, and. Remember tat wat you did was E. 0: Rationalize te numerator were, 0, 0 0 Solution: We multiply te numerator and denominator by te conjugate of te numerator.

6 For additional eamples on simplifying similar algebraic epressions, see te internet. You do not ave to rationalize denominators in tis class. Rational Root Teorem and Syntetic Division: No notes provided on tis topic. You sould go troug your pre-calc notes or turn to te net if you need additional elp Solving trig equations: You must ave a toroug trig background for calculus. You are epected to know all te trig values for te standard reference angles i.e. 0,,,, and be able to use tem to 6 find values for angles suc as 7,,, etc Know te Trig Fundamental Identities and te Double 6 Angle Formulas! Some eamples of solving trig equations follow. E. : Solve te equation sin t cost 0 for all t on0,. Solution: Since sin t cos t we substitute into te original equation, transforming it to: cos t cost 0 cos t cost 0 cos t cost 0. Tis is a quadratic form equation wic factors: cost cost 0 cost orcost Remember tat tere are two values of t on [ 0, for wic in te fourt quadrant, namely solution set is,,. t and E. : Solve te equation sec tan for in [ 0,. cost, one in te first quadrant and one t. Te value of t for wic cost ist. Tus te Solution to E. : Using te identitysec tan, we substitute to get: tan tan tan tan 0 tan tan 0 For tan 0 we get 0 or. For tan we get or. Tus te solution set is 0,,,. E. : Find all values of in [ 0, for wic tan.

7 Solution: Let and solve tan for all in, wic satisfies tis equation. Since tangent as a period of, we get n were n set of integers. Now substituting back, we get tat in, te solutions to our given equation must satisfy te condition n n were n. Now try values of n in a systematic fasion, until you ave found all te solutions for in te interval [ 0,. Usually, start wit n 0, and ten try n, n and so on, until you find a value of n for wic you are no longer on te interval [ 0,. Ten try negative values as well, n, n and so on, until you reac values of n for wic is no longer in te interval [ 0,. For tis equation, te values of n wic produce solutions were is in te interval [ 0, are n,,,, and. Tus, our solution set is 7 7,,,,, E. : Solve: sin cos sin for all in [ 0,. Solution: CAUTION! Often students will attempt to solve tis equation by dividing bot sides of te equation bysin, giving tem cos, from wic tey conclude te only solution is 0. NEVER divide bot sides of your equation by an epression containing a variable. Tis will usually cause you to lose solutions to your equation, wic is wat appened in tis case. Here is te correct tecnique for solving tis equation: Get zero on one side of te equation and factor sin cos sin sin cos 0 So our solutions are te values of for wicsin, wic means 0or. Also, consider te solutions forcos, wic means 0a repeat solution. So, te solution set to te original equation is0,. All angles tat we use in tis class will be in radians - to convert degrees to radians, use te formula radians 80 y deg rees It is assumed tat you know cos θ is an even function and sin θ is an odd function wic implies tat you remember tat cos θ = cos -θ and tat - sin θ = sin -θ sin A sinb sinc Law of sines a b c Law of cosines c a b ab cosc Te equation for finding arc lengt is s = r θ Te equation for finding sector area is r THIS IS THE END OF THE NOTES SECTIONS. THE ASSIGNMENT FOLLOWS.

8 Part A: Eponents and Radicals; Absolute Value and Piecewise Functions; Solving Non-linear Inequalities by Sign Analysis Metod 6 For #-, simplify.. y y.. d. 6 a. For #6 & 7, Solve: 6. / / t t 0 7. z / z / 0 For #8-, simplify, writing witout absolute values were possible. If not possible, state so, simplifying as muc as possible. 8. a 9. if a < 0, a a 0. if a < 0, b < 0, ab. y For #-, rewrite as piecewise defined functions witout absolute values.. f. f. f For #-, solve. Write solutions in interval notation werever possible < > 0... < 0 Part B: Basic Graps, Equations of Perpendicular Bisectors and Circles; Domain of Functions For #-, grap quickly and accurately. Note tat tese instructions are NOT to sketc tese. y. y. sin / For #-, find te domain of te functions: y. y cos. f 6. f 7. f / 8. f 6 9. f 0. f. f. Find te equation of te perpendicular bisector line of segment AB if A-, and B-, 9.. Find te equation of te perpendicular bisector of line segment CD if C 6,0 and D-,.. Find te equation of te circle wit center, - and radius.. Find te center and radius of te circle y y. Use te completing te square metod. 6. Find te center and radius of te circle y y 0 0

9 Part C: Algebraic Equations for Calculus; Synt. Div. And Rational Root Teorem Solving Trig Equations For #-, eliminate comple fractions and reduce to simplest terms. Make sure tat you do not simply epand. Epanded form is not simplified form For #s 6 and 7, rationalize te numerators and reduce. On #8 combine all tecniques For #9-, find all te rational zeros of eac polynomial, and if possible, any irrational or comple zeros. Suggested metod: Rational Root Teorem & Syntetic Division do not use calculators! 9. f 0. f 7 6. f. f 8 For #-7, solve for all on 0,, wic satisfy te equation. Use a calculator ONLY on #8, 9, 0, and. Give EXACT answers werever possible. Approimate answers sould be in radians. [TIP: Solving a trig equation suc as sin cos sin is difficult because some of te trig functions ave argument, wereas oters ave argument. It is easiest to solve tese equations if ALL functions ave te same argument. Use te double angles formulas for eample in tis case te formula sin sin cos, to switc te equation to sin cos sin. Now all te arguments of trig functions are te same. Net you would get te zero on one side and factor out sin to get sin cos 0, so eiter sin 0 or cos. Remember to not simply divide te sine out of te equation. Ten our solution set is 7 0,,,,,.]. sin cos 0. cos cot 0. sin sin 0 6. cos cos 0 7. sec tan 0 8. sin cos 0 9. sin sin 0 0. cos cos 0. tan cot. sin cos. cot csc. tan. cos 6. cos sin

10 Additional material from pre-calculus tat we will not cover in class but you are epected to know: You sould know ow to find te and y intercepts. You sould know te equation for slope. If two lines are parallel, tey ave te same slope If two lines are perpendicular, one slope is te negative reciprocal of te oter. Te standard equation of a orizontal line is y=k Te standard equation of a vertical line is = You sould know te distance formula You sould know tat you can never ave a decimal answer wen te question asks for an eact value