ONLY FOR PERSONAL USE. This digital version of the DictaatRekenvaardigheden - Algebraic Skills is for personal use because of copyright.

Transcription

1 ONLY FOR PERSONAL USE This digital version of the DictaatRekenvaardigheden - Algebraic Skills is for personal use because of copyright. c

2 Algebraic Skills Department of Mathematics and Computer Science June, 0

3 Preface Preface This tet aims at refreshing the algebraic skills required for following courses in basic mathematics at an academic level. Familiarity with many basic concepts is implicitly assumed throughout. The topics covered in Chapters 0 show which algebraic skills freshmen students are assumed to have mastered already. The topics covered in the remaining Chapters 4 are typically beyond what is assumed as prior knowledge upon entering university, but nevertheless basic and essential to any course in calculus. As the main purpose of this tet is to be used for self-study, answers to all the eercises are provided in the final chapter. Major deficiencies in algebraic skills and lack of familiarity with formulas can not be remedied within a few practice sessions. Therefore, to allow for etensive and repetitive practicing, many eercises of the same type are provided for each topic. To make sure that you master a certain topic the idea is that you do all the eercises and check that you can do these correctly and quickly! Note: the goal is that you can do the eercises without the use of a formula sheet or calculator. You should know the rules and formulas in boes like this by heart or be able to derive these quickly, without access to a formula sheet or calculator. Finally, according to a long and widespread tradition, parentheses around the argument to the standard functions sin, log, etc. are omitted if no confusion arises. So y sin sin()y sin y sin(y), sin cos sin() cos() sin( cos ) and sin + y sin()+ y sin( + y). Use of superfluous parentheses, especially when in doubt, is allowed. Note that these functions are typeset in a regular Roman font. ii

4 Preface ii Sets, Logic, Terms, and Factors. Sets and Logic Terms and Factors Powers Simplifying Epressions 5. Removing Parentheses Factoring Polynomials Factoring Polynomials, continued Fractions 9 5 Trigonometry 6 Trigonometric Identities 5 6. Basic Formulas Factoring Proofs Differentiation 9 7. Derivatives of Elementary Functions Differentiation Rules Antiderivatives 9 Graphing Functions 5 0 Equations and Inequalities 9 0. Polynomial Equations Polynomial Inequalities Equations Involving Fractions Inequalities Involving Fractions Eponential Equations Eponential Inequalities Logarithmic Equations Logarithmic Inequalities iii

5 Contents 0.9 Trigonometric Equations and Inequalities Equations Involving Square Roots Inequalities Involving Square Roots Rationalizing Denominators (etra) 5 Partial Fraction Decomposition A (etra) 55 Partial Fraction Decomposition B (etra) 57 4 Inverse Trigonometric Functions (etra) 59 5 Answers to the Eercises 6 iv

6 Chapter Sets, Logic, Terms, and Factors. Sets and Logic A set may be defined by enumerating its elements: V { a, b, c, d }. Eample: {, 4, 6, 7 } is the set containing the numbers, 4, 6, and 7. If is an element of the set V we write V. Common sets of interest are: the set of Natural numbers Integer numbers N {,,, 4,... } Z {...,,,, 0,,,, 4,... } Rational numbers Q all numbers that can be written in the form p with p Z and q N q Real numbers R all numbers that can be written as an (infinite) decimal epansion Comple numbers C all numbers that can be written in the form a + ib with a, b R If W is a subset of a set V, we write W V. Hence, N Z Q R C. If W is defined in terms of a given condition B, we write W { V B() }. We say: W consists of all elements V such that B(). Eample: the natural numbers which are a multiple of 5: The union of sets A and B is defined as { n N 5 n N}. A B { all elements that are in A or in B }. The intersection of sets A and B is defined as A B { all elements that are in A and in B }.

7 . Terms and Factors Subsets of R can be represented on the real line. A contiguous subset is called an interval. We use common notation for intervals. Eamples: U { R < 5 } (, 5] V { R } [, ) W { R } (, ] 5 Some intersections and unions of these intervals are U V { R > } (, ) U V { R 5} [, 5] V W { R } (, ] [, ) V W { R } (, ] [, ) denotes the empty set (the set without any elements). denotes the logical or: a b is true if a is true or b is true (or both). denotes the logical and: a b is true if a is true and b is true. Note the parallel between and, and between and.. Terms and Factors We distinguish terms and factors. A term is a part of a sum (or difference); a factor is a part of a product. For eample, a is a sum of the terms and a; a is a product of the factors and a. Eamples: a b c consists of two terms, namely a b and c. a b consists of three factors:, a, and b. c consists of two factors: and c. a(b cd) consists of three factors:, a, and (b cd). b cd is a sum of which the first term consists of three factors and b; the second term consists of, c, and d. It is important to realize whether you are dealing with terms or factors since the rules of arithmetic are different!

8 Chapter Powers The following rules hold, assuming that a > 0 and b > 0: a p a q a p+q, (ab) p a p b p, a a p n a n, a a p q, (a p ) q a pq, q an a n p, a q a q p, ab a b. For eample, using these rules we have: (a b) a 4 b a b a 4 b a 0 b 5 6. The epression is reduced to a product of numbers and powers of the form C a p b q c r If R then and, where denotes the absolute value of { if 0, if < 0. Note: c c but c c c. Powers (eponentials) and logarithms are inverses of each other. For instance: log y y For more properties of logarithms refer to Section 0.7, page 4.

9 . Powers Eercises Reduce the epressions below to the form C a n b m c o.... All variables are positive numbers.. (p 4 q ) (p q 5 ). (p q 4 ) (p q ) 4. ( a 5 b ) 4 (a b). ( a b ) 4 ( a 4 b). ( cd 4 ) (c d). ( c d 4 ) 4 (c d) 4. ( a b) 4. ( ab b) 5 5. ab a 5. ab a b 6. p q p q 4 6. p q 4 p q 7. (a b ) a 7 b 8. (a b) 4 (6a b ) 7. (a b ) a b 8. (a b) (6a b ) 4 9. a b a b 9. a 5 b a b 0. (a) 4 a 0. (a) a 4

10 Chapter Simplifying Epressions. Removing Parentheses We have: (a + b)(c + d) ac + ad + bc + bd. Special forms of this rule are: (a + b) a + ab + b, (a b) a ab + b, (a b)(a + b) a b. Important for factoring a quadratic polynomial: ( + a)( + b) + (a + b) + ab. Eample: ( + )( 7) 4, where 4 is obtained as + plus 7, and as + times 7. (a b) 9a a b + 4b, where a b is twice the product of a and b. We also need to reduce epressions involving square roots. Eamples: (factor 7 into a square times a number, 7 7 (multiply both the numerator and denominator with 7)

11 . Removing Parentheses Eercises Reduce the epressions below using the above rules. Make sure to remove all square roots from the denominators. Reduce square roots as much as possible.. (a b). ( a + a). ( 6 6 ) 4. (m n)(m + n) 5. (6 )( + ) 6. ( a b + a4 b) 7. ( a + )(a + ) 8. (a b + )(a + b 5). ( a + b). (a a). ( 5 ) 4. (m n)(m + n) 5. ( 5)( 5 + ) 6. ( a 4 b + 4 a b) 7. ( a + 0)(a + 0) 8. (a b + )(a + b 4) 9. ( ) 9. ( ) 0. (a )(a + )(9a + ) 0. (a )(a + )(4a + ) 6

12 . Factoring Polynomials. Factoring Polynomials When factoring a polynomial the goal is to write the polynomial as a product of as many factors as possible. To do so, first move as many common factors as possible outside the parentheses, and then try to factor the remaining part. Eamples: ( + )( 4). +7 and product 44. Look for two numbers with sum ( + ) ( + ) ( + )( ). numbers with sum + and product. Look for two 8 ( 8) ( + 4)( 7) So, first move outside the parentheses. 4 6 ( + 4)( 4) ( + 4)( + )( ) Here we have the form a b twice. Note: a + b cannot be factored any further! ( +) ( +) ( +)( ) Here you take ( +) as factor outside the parentheses. + ( + ) ( + )( ) parentheses. Here you first take outside the Eercises Factor the following polynomials (using factors with integer coefficients only) ( ) ( ) 8. ( ) + ( ) 9. ( ) ( + ) ( + ) + ( + ) 8. ( ) ( ) 9. (5 ) ( + 5)

13 . Factoring Polynomials, continued The eercises below are also about factoring polynomials. Eamples:. Factoring Polynomials, continued ( ) ( ) (6 )( ). Here, it turns out that if you take a factor outside parentheses for two terms at a time, these factors happen to coincide, and the polynomial can be split into two factors In cases like this, sometimes two number a and b can be found such that + 0 ( +a)( +b). Here, a times b must be equal to 0 and b+a must be equal to +. This is satisfied by a 5 and b, hence: + 0 ( +5)( ). Eercises Factor the following polynomials (using factors with integer coefficients only)

14 Chapter 4 Fractions numerator A fraction,, does not eist if the denominator is equal to 0, and it is indeterminate denominator if both the numerator and the denominator are equal to 0. Assume throughout this chapter that denominators are 0. We have the following rules: ab ac b c a b + c d ad bd + bc ad + bc bd bd a b c d ac bd a b : c d a b d c ad So, to divide by a fraction c bc d, multiply by its reciprocal d c. Simplify fractions by factoring its numerator and denominator. Eamples: a + ab a(a + b) ab + b b(a + b) a b a b (a b)(a + b) a b a + ab + b (a + b) a + b b a a b (a b) a b For the eercises below you need to write your answer as a single fraction. This is called finding a common denominator. Eamples: ( )( ) ( ) ( )( ) ( )( ) ( ) ( )( ) 9

15 4. Fractions ( ) ( ) Remarks on the above eample: ( ) ( ) ( ) ( 4 + 4) ( ) + + ( ) ( ) - Use the least common denominator, as this simplifies calculations considerably. Compare this to , where using 4 as denominator instead of 4 would also 4 complicate calculations. - Make sure to carefully multiply with all terms of between the parentheses. - Usually, parentheses do not need to be removed from the denominator. Sometimes, the resulting fraction can be reduced further by also factoring the numerator. Eercises ( + ) + ( ) ( ) + ( )

16 Chapter 5 Trigonometry The trigonometric functions sine, cosine, and tangent are introduced in terms of ratios of sides in right-angled triangles. This covers the case of angles between 0 and 90. We have: tangent() sine() / cosine(). A natural generalization to arbitrary angles is obtained by using the unit circle. We define: A point P on the unit circle has -coordinate cos(α) and y-coordinate sin(α), so P (cos(α), sin(α)), or in a slightly different notation P (cos α, sin α). y P : cos Α, sin Α O Α We then have in general: tan α sin α cos α and sin α + cos α. Note that the second identity corresponds to Pythagoras theorem. The angle α is commonly measured in radians. In this case α is equal to the length of the arc on the unit circle corresponding to the angle α. For instance, π rad corresponds to 60. Throughout this syllabus, we measure angles in radians.

17 5. Trigonometry Eact values of the sine, cosine, and tangent can be given for some special angles. These values can be found using the triangles depicted below. The table list these values for angles between 0 and π. It is strongly recommended that you memorize this table. π 6 π π 4 π π sin π Note that tan is undefined (und.) for π. cos tan 0 6 π 4 π π π 0 und. For angles eceeding π the corresponding trigonometric function values can be derived directly from the definitions. Eample: The coordinates of point Q which corresponds to an angle 7 π can be derived from the 6 coordinates of point P corresponding to an angle of 6 π. 7 Π 6 O Π 6 P Q Therefore sin 7 6 π sin 6 π cos 7 6 π cos 6 π tan 7 6 π sin 7 6 π cos 7 6 π We call this reducing an angle to the first quadrant.

18 5. Trigonometry Another type of question goes like this: Given: sin α. How large is α, under the restriction that 0 α < π? 0 Use the unit circle By definition, sin α is equal to the y-coordinate of a point on the unit circle. Since is negative, we know that the y-coordinate is negative. The y-coordinate is negative in the third and fourth quadrant, so that is where angle α is to be found. Recall that: sin π. From the figure and the above table one can readily see that sin α corresponds to angles π + π and π π. Thus the answer is: α 4 π α 5 π.

19 5. Trigonometry Eercises Find the values of the following trigonometric epressions. Reduce angles to the first quadrant and use the above table.. sin( 4 π). tan( 4 π). cos( 5 6 π) 4. sin( π) 5. tan( 85 4 π) 6. sin( π) 7. cos( 7 4 π) 8. tan( 5 6 π) 9. cos( π) 0. sin( 7 4 π). cos( 4 π). sin( 4 π). tan( 5 6 π) 4. cos( π) 5. sin( 85 4 π) 6. tan( π) 7. sin( 7 4 π) 8. cos( 5 6 π) 9. tan( π) 0. sin( 4 π) 4

20 Chapter 6 Trigonometric Identities The following identities can easily be found using the definition in terms of the unit circle: sin + cos, sin( ) sin, cos( ) cos, tan( ) tan sin(π ) sin, cos(π ) cos, tan(π ) tan sin cos( π ), cos sin( π ) The following formulas are given without proof. These formulas are strongly connected. For instance, if you start from one formula you can find the other ones by using the simple identities above: sin( + y) sin cos y + cos sin y sin( y) sin cos y cos sin y cos( + y) cos cos y sin sin y cos( y) cos cos y + sin sin y tan + tan y tan( + y) tan tan y tan tan y tan( y) + tan tan y sin sin cos cos cos sin cos sin tan tan tan 5

21 6. Basic Formulas 6. Basic Formulas Eercises. Derive the formula for sin( +y) from the formulas for sin( y), cos( +y), and cos( y).. Derive: tan( + y) tan + tan y tan tan y, and tan( y) tan tan y + tan tan y.. Derive from the formulas for sin( + y), cos( + y), and tan( + y): sin sin cos cos cos sin cos sin tan tan tan 4. For an angle [0, π] we have: cos. Find cos( 6 π). 5. For an angle [ π, π] we have: Find sin. cos. 6. For an angle [0, π] we have: cos 4. Find tan. 7. Simplify as much as possible sin cos + cos + sin Simplify as much as possible (cos sin ) tan(). 6

22 6. Factoring 6. Factoring Eercises Use the formulas at the start of this chapter:. Factor: sin + sin. Factor: sin( + y) + sin( y). Factor: cos + sin 4. Factor: + sin cos 5. Factor: cos 6. Factor: sin 5 sin Factor: sin + 5 cos Find the eact value of cos 8 π. 9. Simplify cos sin cos + sin. 0. f () cos sin can be written as f () a + b cos(c). Determine a, b en c.. Factor: sin sin. Factor: cos cos. Factor: sin sin 4. Factor: cos Factor: sin sin 6 6. Factor: cos + sin + 9 cos 7. Eliminate the square root: + cos 8. Find the eact value of sin 8 π. 9. Simplify cos 4 cos sin + sin f () cos sin can be written as f () a + b cos(c). Determine a, b en c. 7

23 6. Proofs 6. Proofs Eercises Use the formulas at the start of this chapter:. Show that cos + cos.. Show that cos 4 sin 4 cos.. Show that cos 4 + sin + sin Show that cos ( + tan ). 5. Show that tan 6. Show that 7. Show that 8. Show that 9. Show that tan + tan y tan tan y sin + cos. sin( + y) sin( y). tan sin tan sin. tan tan + sin cos. tan( + tan π + ) 4 tan. 0. Show that cos cos.. Show that sin cos.. Show that cos 4 ( tan 4 ) cos.. Show that 4 sin 4 sin 4 sin. 4. Show that cos 4 ( + tan 4 ) sin cos. 5. Show that tan 6. Show that + tan tan y tan tan y cos. sin cos( y) cos( + y). 7. Show that sin tan tan cos. 8. Show that 9. Show that sin + cos cos. sin tan( 4 π +)+tan( 4 π ) cos sin. 0. Show that cos sin cos + sin + cos + sin cos sin cos. 8

24 Chapter 7 Differentiation If a function f assigns the value f () to, we also call f () the mapping rule. The graph of f is the curve in the (, y)-plane given by the equation y f (). We will often use a function f, a mapping rule f (), and the equation y f () interchangeably, even though this is sloppy use of terminology. Under certain conditions we can take the derivative of a function f (). In terms of the graph of f, the derivative f () is equal to the slope of the tangent at to the graph y f (). The derivative of y f () with respect to is denoted as f () or d d f () or d f () d or y or dy d. 7. Derivatives of Elementary Functions d d n n n, d d e e, d d ln, d d a a ln a, d sin cos, d d cos sin, d d d tan cos + tan, 9

25 7. Differentiation Rules 7. Differentiation Rules We use the following rules to calculate derivatives: f () u() + v() f u + v (Sum Rule) f () cu() f cu (Rule for Constant Multiples) f () u()v() f u v + v u (Product Rule) f () t() n() f nt tn n (Quotient Rule) Given functions f (u) and u(), the derivative with respect to of the composite function y f (u()) is equal to the derivative of f with respect to u multiplied by the derivative of f with respect to u: dy d dy du du d (Chain Rule) Eamples: Calculate the derivative of the function y 6( ). Define: u(). Then we need to differentiate y 6u with respect to u and multiply the result by the derivative of u with respect to. We have: dy du u and du d 6. So dy d dy du du d u 6 ( ) 6. Hence: y 7( ) y ( ) 5 y 5( ) 4 ( ). Useful to memorize: y u n y nu n u y u y u u y u y u u y ln u y u u 0

26 7. Differentiation Rules Eamples: y 4( ) y 8( ) 6 48( ) y 6 y y y ( ) 6 8 ( ) y ln( ) y 5 y y 5 ln ln 5 y ( sin ) y ( sin ) 4 sin cos sin cos ( sin )

27 7. Differentiation Rules Eercises Find the derivatives of:. y ( ) 5. y. y 5 4. y ( ) 5. y 6 6. y e 7. y ln( + 4) ( ) 8. y ln + 9. y 0. y. y ln ln. y ( ) e. y tan 4. y sin cos sin + cos 5. y sin cos. y ln. y ln. y sin 4. y + 5. y y sin ( 6 π) 7. y ln sin 8. y + 9. y ln( ) 0. y e sin. y e + e. y ln 4. y sin cos 4. y sin cos 5. y cos cos

28 Chapter 8 Antiderivatives Antiderivatives play an essential role in integral calculus. Taking antiderivatives is the reverse operation of taking derivatives. Knowledge of differentiation is therefore required. Elementary formulas (with a 0, n, and c a constant of integration): a n d a n + n+ + c a d a ln + c e a d a ea +c sin(a) d a cos(a) + c cos(a) d a sin(a) + c (a + b) n d a a + b d ln a + b + c a cos d tan + c n + (a + b)n+ + c Always make sure to check the antiderivative by differentiating it.

29 8. Antiderivatives Eercises Find the antiderivatives of the following functions.. ( ). (5 ). ( + ). (8 ) ( + ) ( ) 5 5. sin ( 6 π) 5. cos ( π) sin + e 9. tan cos + e 9. tan

30 Chapter 9 Graphing Functions The goal of the eercises below is to sketch the graphs of the given functions without using any electronic tools. The scales used for the -ais and the y-ais do not have to be the same. Each time choose the domain such that the properties of the graph are clearly visible, such as points of intersection with the aes, asymptotes, periods, and so on. Also put values net to the points of intersection and asymptotes if these values can be calculated easily. From the graph of y f () one can obtain the graphs of related functions by scaling the aes and by horizontal or vertical shifts. the graph of: a f () is obtained from the graph of f () by multiplying the y-value by a f (/b) multiplying the -value by b f ( c) shift it c units to the right f () + d shift it d units upward Note the order in which these steps are applied: If we want to construct the graph of ( c ) y a f + d b from the graph of y f (), then we have apply this order: first scale, then shift. 5

31 9. Graphing Functions Elementary Functions 4 y y y y ep() y ln() y 0 y sin() y cos( ) π π.5π π.5π 0 0.5π π.5π π.5π 5 y / y tan() π 0.5π 0 0.5π π.5π π 6

32 9. Graphing Functions The graph of y tan() (see previous page) has the following properties: periodic with period π; zeros at kπ, k Z; asymptotes at π + kπ, k Z. Eample of scaling and shifting: The graph of y tan ( π) is obtained from the graph of y tan by shrinking the 4 -value by a factor of (the period changes from π to π) and shifting it by π to the right π π 4 π π 4 π 0 4 π 7

33 9. Graphing Functions Eercises. f () ( + ) 4. f () ( ) +. f () f () f () 6( + ) 5 Series C. f (). f () ( ) +. f () + 4. f () 5. f () e Series E. f () +. f () 4. f () f () 5. f () 6. f (). f () 4. f () + 4. f () 5. f () Series D 4 ( ). f () log. f () log ( + ). f () log 4. f () log 5. f () ln( e) Series F. f () sin. f () cos π. f () + 8 sin π( ) 4. f () sin 5. f () tan ( + 4 π) 8

34 Chapter 0 Equations and Inequalities 0. Polynomial Equations The following rules apply when solving equations: a + b 0 a b b a A B C... 0 A 0 B 0 C 0... A B A B A B A B A C A 0 B C Quadratic equation: a + b + c 0 b ± b 4ac a (quadratic formula) We do not need to use the quadratic formula in simple cases such as a + c 0, a + b 0, + b + b 0, or when the epression can easily be factored: +4 6 ( + ) ( + ) ( )( +) 0. Also completing the square may be a useful option: + 4 ( + + ) 5 ( + ) ± 5 ± 5. In general, polynomial equations can be written in the form (with n a positive integer): a n n + a n n + a n n a + a 0 0. Solutions may sometimes be found by factoring the polynomial. Note: the quadratic formula can be derived by completing the square: a + b + c a ( + a b + a c ) ( a + a b + ( a b ) ( b ) a + c ) [ ( a a + b ) ] a b 4ac (a) 0. 9

35 0. Polynomial Equations Eamples: Solve: ( 4) 7( 4) 0 ( 7)( 4) 0 ( 7)( )( + ) 0 Solve: ( 4)( + 4) 5( + 4): ( 4)( + 4) 5( + 4) 0 ( 4 5)( + 4) 0 ( + )( 7)( 4) Solve: ( 4)( ) ( + )( ): ( + 6)( + ) Eercises Solve the following equations: ( )( + 0) 7. ( ) ( 4)( + ) ( )(4 ) ( + )( + ) ( )( + ) 9. ( 4)( ) ( + )( ) 0. ( ) ( + ) ( + ) 0

36 0. Polynomial Inequalities 0. Polynomial Inequalities Basic rules (assume a > 0) A B A C B C a A ab a A ab (inequality is reversed) A a a A a A a A a A a Analogously for > and <. Eamples: + 8 < 4 < 4 > 4 as dividing or multiplying by a negative number reverses the inequality. Note: 0 < < y 0 < y < < y < 0 y < < 0 < 0 < y < 0 < y In all these cases we multiply by (y). In the first two cases this is positive, while in the last case it is negative hence the inequality is reversed. For a quadratic inequality the solution set may consist of one or two intervals: > 5 < 5 > 5 If the graph can be sketched quickly, one can also first solve the equation and then read out the solution to the inequality from the graph. Also take the domain into account. In more advanced cases of polynomial inequalities a sign chart is a useful tool. By means of + and signs on the real line, sign charts show where a function is positive or negative. On the real line, we also indicate where the function is equal to zero. Sign changes only take place at the zeros on the real line. Two eample sign charts: f () ( + )( )( ) has sign chart: Evaluate the value of f () at simple values for (other than the zeros). If the function value is positive (or negative) then the same holds everywhere between the neighboring zeros. At each zero on the real line, the sign changes.

37 0. Polynomial Inequalities g() ( + ) ( )( ) has sign chart: Eamples At on the real line 0 is written three times to represent the solution to ( + ) 0, that is ( + )( + )( + ) 0. We call this a triple zero (or root). This also means that the sign changes three times, since the sign changes once for each zero. So, if g() is positive for < then g() will be negative for >. At on the real line 0 is written twice as ( ) can be written as ( )( ). Therefore the sign changes twice, which amounts to no change in sign at. Hence, at a zero of multiplicity n there is no change in sign if n is even, and the sign changes if n is odd. Solve: First rewrite as: Net factor the left-hand side: ( 6 + 8) 0 ( )( 4) 0 Sign chart: Hence the solution is: 0 4 or, in terms of intervals: (, 0] [, 4] Solve ( + ) ( ) ( + ) ( ) ( + ) ( ) ( + ) ( ) 0 ( )( + ) ( ) 0 Sign chart: Hence the solution is: or, in terms of intervals: (, ] [, )

38 0. Polynomial Inequalities Eercises Solve the following inequalities:. 5. ( ). ( ) < + 5. ( + ) ( + ) > ( ) 6. 5 < + 7. ( 4)( 4) 0 8. ( + ) ( + ) > 5( ) 9. ( 7 + )( + 4) ( ) < + 5. ( ) ( + ) > 5( ) ( 9)( ) 0 8. ( 5) ( + )( 5) > 0 9. ( )( ) ( )

39 0. Equations Involving Fractions 0. Equations Involving Fractions The following rules apply when solving equations: A B 0 A 0 B 0. A B C D A D B C B 0 D 0 (cross-multiplication) Cross-multiplication and finding common denominators are important techniques when solving equations involving fractions. Make sure to check the solutions found, as the denominators should all be different from 0! Eamples: Cross-multiplication yields: ( 6)( + 5) 5( + 8) ( 5) + 4( 5) 0 ( + 4)( 5) (all solutions are valid) + We obtain: ( )( 6) 0 6 (solution is invalid) 6 4

40 0. Equations Involving Fractions We obtain: 6( + ) + 5( ) ( )( + ) ( 4) + ( 4) 0 ( 4)( + ) 0 4 (all solutions are valid) Eercises Solve the following equations: ( + )

41 0.4 Inequalities Involving Fractions 0.4 Inequalities Involving Fractions Fractions may change in sign not only at zeros of the numerator, but also at zeros of the denominator. Sign charts therefore indicate both the zeros of the numerator (0) and the zeros of the denominator ( ). The function is undefined for the zeros of the denominator. Eample 5 + ( + ) ( )( + ) 5( ) ( )( + ) ( )( + ) 0 ( + 7)( ) ( )( + ) 0 ( )( + ) ( )( + ) 0 Sign chart: X X 0 7 so 7 < <. Again: if a zero of the numerator (0) or the denominator ( ) occurs an even number of times, the sign of the function does not change. 6

42 0.4 Inequalities Involving Fractions Eercises Solve the following inequalities involving fractions < > > > ( )( + ) ( + )( ) ( ) ( ) < > > > < 7 6 7

43 0.5 Eponential Equations 0.5 Eponential Equations For the eercises below you are required to reduce the given equation to an eponential equation, that is, an equation of the form with a > 0. a epression in a number Since a is an increasing function (if a > ) or a decreasing function (if a < ), one can simply equate the eponents. Subsequently, solve the resulting equation. Use the rules for eponents, for eample 4 ( ) ( ). Eamples ( ) 6 e + e e + e ( 6) Suppose a, then a + 8 a 6 a 6a (a )(a 4) 0 a a 4 4 8

44 0.5 Eponential Equations Eercises Solve the following eponential equations e + ( e ) e e e e e 5 e log 4 8. ( ) ( 4 ) ( ) (0.4) 50 (0.8) 9

45 0.6 Eponential Inequalities 0.6 Eponential Inequalities For eponential inequalities you must be careful with bases smaller than. For instance, > 4 > 4 but ( ) > ( )4 < 4. You may reach the same conclusion as follows: ( ) > ( ) 4 > 4 > 4 < 4. Eamples Solve: > > > 6 5. Multiply both sides by 5 : > > > 5. Solve: + ( ) and + ( ) + ( ) Assume a and note that therefore a > 0! a + 7 a a + 7 a. This is valid because a > 0. Hence a a (a )(a 9) 0 a 9. 40

46 0.6 Eponential Inequalities Eercises Solve the following eponential inequalities... ( ) < ( 4)( 8) > ( 4 ) < > > > 0. ( ) > < < ( ) ( ) < 5 0. ( ) 4 ( ) 5 < 0 4

47 0.7 Logarithmic Equations 0.7 Logarithmic Equations The logarithm is defined as (with a > 0 and a ): a L L log a log 0 is commonly used in scientific tets, written as log. log e is called the natural logarithm and is usually written as ln (in some mathematical tets also as log!) Important rules: log a eists only if > 0 log a is an increasing as a function of if a > and decreasing if 0 < a < log a 0, log a a, log a a, (etc.) log a a a log a provided > 0 log a + log a y log a y log a log a y log a ( y ) log a r r log a log a log a p log p log a log p log p a ln ln a log log a Note: y n log a y log a n, but y log a n y n log a. Make sure to verify the validity of the solutions found, as the argument of a logarithm must be positive (larger than 0)! Eamples: Solve: log ( + ) + log ( ). Then we have: log ( + ) log 4 + log ( ) log 4( ) Under the condition that the arguments must be positive this is equivalent to: (check: the solution is valid). 7 Solve: ln ln. This is equivalent to (why?): ln ln 0. Suppose ln a a a 0 (a + )(a ) 0 a a ln ln e e 4

48 0.7 Logarithmic Equations Eercises Solve the following equations.. log 6 8. log ( ) 9. ln( 7 + 7) 0 4. log 5 log ( + 4) 5. log ( ) + log ( + ) 5 6. log( 0) 7. ln(7 ) ln( ) 8. ln( + ) ln 5 9. ln + 6 ln 5 0. ln + ln 0. log 7. 4 log 4 (8 ). ln( + e) ln 4. log (5 ) + log 5. + log ( ) 6. log + log( + ) ( ) 7. log ( + ) + log 5 8. ln( + ) ln( + ) 9. ln ln + 0 ( 0. ln( ) 8) ln 4

49 0.8 Logarithmic Inequalities 0.8 Logarithmic Inequalities An important step in solving logarithmic inequalities is determining the domain. Of course, solutions need to be in the domain. Again, be careful: log < log 4 0 < < 4 but log < log 4 > 4! Just as for eponentials, the logarithm function is decreasing if the base is less than, and we need to reverse the inequality sign. Eamples: Solve: log + log ( + ). Here: > 0 >, so D (0, ). hence log log 8 + log ( + ) log log ( + ) 8 8 ( + ) Taking into account the domain D, the solution is: (0, 7 ] Solve: log ( ) < log. We have: ( > ) ( > 0) D (, ). On this domain: log ( ) + log < log 7 log ( ) < log 7 7 < 0 ( 9)( + ) < 0 < < 9 Taking into account the domain D, the solution is: (, 9 ) 44

50 0.8 Logarithmic Inequalities Eercises Solve the following logarithmic inequalities.. log 5 ( + ). log 4 ( ) >. log ( 4 5) 4 4. ln + ln 0 5. log log ( 6) 6. log > 7. ln( e) > 8. ln ln > 0 9. log ( ) log ( + 7) 0. ln ln( ). log ( ). log 4 ( + 6). log ( 8 + 7) 4 4. ln + ln > 0 5. log ( ) < log 6. ln( ) ln 0 7. log ( + ) 8. log( + 4) + 4 log( + 4) 0 9. log ( 8) < 0. log( + ) log < 45

51 0.9 Trigonometric Equations and Inequalities 0.9 Trigonometric Equations and Inequalities A few rules: sin sin a a + kπ π a + kπ cos cos a a + kπ a + kπ tan tan a a + kπ Here k Z, so k 0, ±, ±,... Eamples: Solve: cos( π) sin 4 cos( 4 π) sin cos( 4 π) cos( π ) 4 π π + k π 4 π π + + k π 5 4 π + k π 4 π + k π 5 π + k π 4 π + k π Solve: sin cos sin cos ( cos ) cos cos + cos 0 Suppose cos a then Inequalities: a + a 0 (a + )(a ) 0 a a Note that cos has no solution, hence cos ± π + k π. Determine where equality holds.. Determine where the function is undefined.. This yields all points where the inequality possibly reverses. These boundary points define the intervals where the inequality, or its reverse, holds. 4. Draw a sketch of the function and write down the solution. Eample: Solve: tan( π). We have tan( π) als π 4 π + kπ, dus 8 π + kπ. The function is undefined if π π + kπ, so kπ. Consider the graph in Chapter 9 on page 7. Hence the solution is 8 π + kπ < π + kπ. 46

52 0.9 Trigonometric Equations and Inequalities Eercises Solve the following trigonometric equations and inequalities. Use R as domain... sin sin. tan + tan 0. sin( + π 4 ) + sin( π 4 ) 0 4. cos + cos 0 5. cos + sin 6. sin cos sin 0 7. tan tan 8. sin tan 9. cos sin 0 0. sin + sin. sin cos. cos + tan 0. cos cos. tan sin. cos( π) + sin( 6 π) 0 4. sin sin 0 5. cos + sin cos sin 6. sin cos 7. tan tan 8. sin tan 9. cos + cos sin 0. sin + 6 sin. 6 cos + sin 0. sin cos. cos sin. sin + cos + sin cos 4. (tan sin )(tan + sin ) cos 5. sin sin cos Series C.. sin >. tan. sin < cos 4. sin tan sin 5. sin + cos 0 Series D. cos <. tan 0. tan sin 47

53 0.0 Equations Involving Square Roots 0.0 Equations Involving Square Roots Equations involving square roots are often solved by squaring both sides of the equation. However, this introduces additional solutions compared to the solutions to the original equation. Phrased differently: y y, but, in general, y y. Always remember that the epression under the square root sign cannot be negative, and that the square root is a non-negative number too. Check the solution found by substituting it in the original equation. Eamples: Solve: ( 5) ( 4)( ) 0 4 Since 5 < 0 for, only 4 remains as solution. Solve: ( 4)( 44) The only solution is 4. 48

54 0.0 Equations Involving Square Roots Eercises Solve the following equations involving square roots p ( 4 p) p 4 p 8 0. For which values of p is the graph of f () + tangent to the graph of g() + p? ( + ) For which values of p is the graph of f () 5 tangent to the graph of g() p +? 49

55 0. Inequalities Involving Square Roots 0. Inequalities Involving Square Roots Sometimes it helps to draw two graphs and compute the points of intersection. But keep the domain in mind when determining the solution. For instance, for <, the solution is [, 5) because. Eamples: Solve: + < 0 + < < 0 with 0. First solve: The quadratic formula yields: 5 6. Only 5 satisfies the original 4 4 equation. If we now substitute, for eample, 4 (4 < 5 ) in the inequality, we get a value less 4 than 0. But if we substitute, for eample, 9, we get a value larger than 0. The solution is: [0, 5 4 ). Solve: + 5. The domain is D (5, ). Then the fraction is always positive. First solve: + 5. We have: ( 5) The quadratic formula yields: 8 9. Only 9 satisfies the original equation. 6 If we now substitute, for eample, 6 (6 < 9) in the inequality, we get a value less than. But if we substitute, for eample, 4, we get a value larger than. This means that we need to take values larger than 9. The solution is: [9, ). 50

56 0. Inequalities Involving Square Roots Eercises Solve the following inequalities involving square roots < 0 5. > ( ) > > <. < ( ) > < ( ) + 5 > < 9. 6 > 0. + < 5 0. < 9 5

57 5

58 Chapter Rationalizing Denominators (etra) In the eercises below you need to remove the square roots from the denominators of the fractions. This is called rationalizing denominators. In case of just a square root, this is done by multiplying top and bottom by this square root. Eample:. In more comple cases we can apply the rule: (a b)(a + b) a b. Eample: ( )

59 . Rationalizing Denominators (etra) Eercises Rationalizing the denominators: ( + ) 8. ( )

60 Chapter Partial Fraction Decomposition A (etra) In the fractions given below, the denominator is a product of two factors or can be written as such. The goal is to split the given fraction into a sum of two fractions of which the denominators are these two factors, respectively. This is called partial fraction decomposition. Eamples: We have: + 0 ( )( + ) +. This can be checked easily afterwards, but how do we actually find the numerators? + 0 We replace ( )( + ) by A + B where we need to compute A and B. + Write this epression as a single fraction: A( + ) + B( ) ( )( + ) (A + B) + (A B). ( )( + ) Then A + B and A B 0 must hold. If you solve this system of two equations, you get A and B. Consider: ( + )( + 4)? We put: ( + )( + 4) A + + B + 4 A( + 4) + B( + ) ( + )( + 4) (A + B) + (4A + B). ( + )( + 4) Then A + B 0 and 4A + B must hold. Solving this system yields A 6 5 and B 5. Therefore we can replace ( + )( + 4) by 6 5( + ) 5( + 4). 55

61 . Partial Fraction Decomposition A (etra) Because of the double zero in the denominator, the net fraction is slightly different: + ( )? We change the numerator into the following form containing : + ( ) + 7 ( ) ( ) + 7 ( ). The resulting partial fraction decomposition has as denominators all powers of ( ) up to the power ( ) occurring in the given fraction. Eercises Apply partial fraction decomposition to the following fractions:. 5 ( )( ). 5 ( + )( + ) ( )( ). 4 ( + )( ) ( ) 6. 6 ( ) ( ) 8. 4 ( ) 9. ( ) 9. 6 ( ) ( + ) ( + ) 56

62 Chapter Partial Fraction Decomposition B (etra) Below a few more eercises are included, where a given fraction is decomposed into several parts. Eamples Consider We may obtain this result in two ways: We also have: hence: 7/\ The is always possible if the numerator eceeds the denominator. The same applies to rational functions: Also: / + \ hence:

63 . Partial Fraction Decomposition B (etra) We have: / + \ hence: The method used in the last two eamples applies if the degree of the numerator is larger than or equal to the degree of the denominator. Eercises Apply partial fraction decomposition in the eercises below

64 Chapter 4 Inverse Trigonometric Functions (etra) In high school you have already seen inverse functions. For instance, log and 0 are inverses of each other and so are and, for 0. On a scientific calculator, these function often share the same key. So you can tell that also ln en e are each other inverses. A property of inverse functions is that the graphs of a function and its inverse are reflections of each other in the line y. The domain and range of a function and its inverse need not be identical; you can check this for the above functions. The functions sin, cos and tan have inverse functions. sin, cos and tan is often used on calculators; we will use arcsin, arccos and arctan. We will now check the domain and range of these functions by reflecting their graphs. In the figure on the left you may recognize the graph of y sin reflected in the line y. π 4 π 0 4 π π y arcsin() π 4 π π 4 π 0 y arccos() π 4 π 0 4 π π 5 4 y arctan() We note that we take only a small piece of the sin-graph, since otherwise the reflection of the graph would not be a function. If the domain for sin is equal to D [ π, π], the inverse is indeed a function. In choosing the domain we include the origin and ensure that all values for sin are reached (from to ). See the figure above. The range of y sin is R [, ]. The inverse function of y sin is called y arcsin. Its domain is D [, ], while its range is R [ π, π]. In this way we can also reflect the graphs of y cos and y tan in the line y.. Check that for y arccos the domain is D [, ] choosing as range R [0, π].. Check that for y arctan the domain is D (, ) choosing as range R ( π, π). 59

65 4. Inverse Trigonometric Functions (etra) Similar as before when we determined which satisfy sin, we can now determine a value for arcsin( ). The important difference is that we now get eactly one result, as a function may take on one value only. Therefore: arcsin( Or, in general: ) π. y arcsin sin y, but sin y y arcsin, and similarly for arccos and arctan. Eercises Give your answer in radians (in terms of π).. arcsin( ). arccos( ). arctan( ) 4. arccos( ) 5. arctan() 6. arcsin( ) 7. arccos( ) 8. arctan( ) 9. arcsin(0) 0. arccos( ). arccos( ). arcsin( ). arctan( ) 4. arcsin( ) 5. arccos() 6. arctan( ) 7. arcsin( ) 8. arccos( ) 9. arctan(0) 0. arcsin( ) 60

66 Chapter 5 Answers to the Eercises Powers. p 6 q 6. a b c d a b 5. ab 6. p 4 q 5 7. a b ab 4. p 8 q 6. b c d 4. a 5 b 5 5. ab 6. p q a b a b 6 9. a 8 b a 5 b a a 6 6

67 5. Answers to the Eercises. Removing Parentheses. 9a 6ab + b. 9a a + 4a m 5mn n a 4 b 6 a 6 b a8 b 7. a 8. 7b 7a + ab + 6a b a 4. 9a ab + 4b. 9a a 4 + 4a m 5mn 6n a4 b a 6 b 4 + 4a 8 b a 8. 4b 5a 5ab + 6a 6b a 4. Factoring Polynomials. ( )( + )(4 + 9). ( + )( 7). ( )( + )( + )( 4 + )( 8 + ) 4. ( )( + ) 5. ( )( 7) 6. ( + )( 7) 7. ( ) 8. ( )( + + ) 9. (5 + )( 5) 0. ( ). ( )( + )(9 + 4). ( )( 4). ( )( + )( 6 + 4) 4. ( + 4)( 5) 5. ( )( 9) 6. ( + )( 9) 7. ( + ) 8. ( )( + ) 9. 4(4 + )( 4) 0. ( 5 + 4) 6

68 . Factoring Polynomials, continued. ( )( 6). ( + )( + ). ( )( ) 4. ( + )(5 ) 5. ( )( + ) 6. ( )( 4)( + ) 7. ( )( + ) 8. ( + 5)( )( + ) 9. ( )( ) 0. ( )( + 5). ( 5)( ). ( + )( + ). ( 4)( ) 4. ( + )(7 ) 5. ( + )( )( + ) 6. ( )( 8)( + ) 7. ( 4)( + ) 8. ( + 5)( 4) 9. ( )( ) 0. ( + )( 6) 4 Fractions. ++ ( )(+). ( ) ( )(+) ( )(+) ( )(+) 4. ( ) ( )(+) (+)(+) ( ) +6 ( )(+) ( +) ( )(+) ( )(7+4) ( +)(+) (+) 7 ( )(+) 8. + (+) 8. + ( ) 9. + ( ) (+) 9. + ( )(+) 0. + (+) ( ) 6

69 5. Answers to the Eercises 5 Trigonometry undefined Trigonometric Identities: Basic Formulas 64. We have: sin( y) sin( + ( y)) sin cos( y) + cos sin( y) sin cos y cos sin y cos( + y) sin( π y) sin(( π ) y) sin( π ) cos y cos( π ) sin y cos cos y sin sin y cos( y) cos( + ( y)) cos cos( y) sin sin( y) cos cos y + sin sin y. We have: tan( + y) sin( + y) sin cos y + cos sin y cos( + y) cos cos y sin sin y tan( y) tan( + ( y)). We have: tan + tan y tan tan y tan + tan( y) tan tan y tan tan( y) + tan tan y. sin sin( + ) sin cos + cos sin sin cos cos cos( + ) cos cos sin sin cos sin ( sin ) sin sin cos ( cos ) cos tan + tan tan tan( + ) tan tan tan tan. 4. We have: cos( 6 π) cos cos 6 π + sin sin π. Since lies in the first quadrant also 6 sin. So cos cos 6 π + sin sin 6 π We have: cos sin sin sin sin ±. Since [ π, π] we find: sin.

70 6. sin cos , hence sin 7 6 because [0, π]. So tan 7, and tan 7 7. sin cos + sin 4 + cos sin (cos + sin ) + cos sin + cos. 8. (cos sin ) tan() cos() sin()/ cos() cos() sin() sin(4). 6. Trigonometric Identities: Factoring. ( cos + ) sin. sin cos y. ( sin )( + sin ) cos 4. sin (sin + cos ) 5. (cos )(cos + ) sin 6. (sin )(sin 4) 7. (cos + )(6 cos ) sin. sin (sin cos ). (cos )(cos + ) sin. (cos sin ) cos 4. (sin )(sin + ) 5. (sin + )(sin ) 6. (sin + )(5 sin ) 7. cos sin tan cos 0., en 0.,, 65

71 5. Answers to the Eercises 7. Differentiation.. 0( ) 4. ln + ln +. ( ). ln. 6( 5) 4. 7( ) ln 6. 4 e 7. ln( + 4) ( ln + ) (ln ). e ( ). (tan )(tan + ) 4. sin + 5. cos cos. cos (sin ) 4. ( + ) 5. 7( ) 6. sin(4 π) 7. ln sin ln cos 8. ( + ) 9. (6 ) sin 0. sin() e cos. e. 4 ln. sin ( cos sin ) cos + cos sin 66

72 8 Antiderivatives. ( )4 8. ( + )4. (5 ). (8 ) 6. 9 ( 4). ( ) 4. ( + ) 6 5. cos ( 6 π) ln 8. cos + e 9. tan 0. ln 4. ( ) sin ( π) ln sin + e 9. + tan 0. ln 67

73 5. Answers to the Eercises 9 Graphing Functions f f f f f O O O O f 4 f 8 f 6 9 f 6 5 f A A A A4 A5 f f f f f O O f 4 f O f O f O 4 4 f B B B B4 B5 f f f f f f O f O f O f O f O f C C C C4 C5 f f f f O O O O O f O f O f log f log f log f log f ln D D D D4 D5 f f 4 O 4 f 4 f O 4 f 4 4 f O f f O f 6 E E E E4 E5 Π f sin Π f O f cos Π f 0 4 O 7 f 8 sin Π f () f () sin F F F F4 F5 f () f () tan ( + 4 π) 68

74 0. Polynomial Equations. 4 or. or or 6 or 6. 0 or or or or or 6. 5 or 4 or 7. or or 8. or 9. or or or 0. or or or. or. 6 or 6. 0 or or 4. or 5. 0 or or 6. or 7. or 8. or or 4 9. or 0. or or 69

75 5. Answers to the Eercises 0. Polynomial Inequalities. [ 5, 5]. (, ] [4, ). [, 5 ] 4. (, ) (, 4) (5, ) (, 7) 7. {4} [, ] 8. (, 7 6 ) 9. {4} [ 6, ] 0. [, ]. [4, ) (, 4]. [ 4, 4 ]. (, 5 ] [ 7, ) 4. (, ) [, 5) 5. (, 6 ) 6. [, ) (, 5] 7. [, ] 8. (, 5 ) (, ) 9. {} (, ] [, ) 0. [, 0] [, ] 0. Equations Involving Fractions or. + or 4. or 5. or 6. or or or or. or.. or or 6. or or 8. 0 or or 70

76 0.4 Inequalities Involving Fractions. (, 5]. (, 4) (7, ). [, ] (, ) (, ) 4. (, ) (, ) 5. (, 5] (, ] 6. (0, ] (, ) (, ] 7. [, ) [, ) 8. (, ) (, 0) (, ) 9. (, ) (9, ) 0. (, ) (0, ). (, 7) (, 4). (, 0) (0, ). (, ) (0, ) (, ) 4. (, ) [5, ) [ 4, 5 ) 5. [, ) (, ] 6. ( 4, ] [0, ) 7. (4, 6) (, ) (, ) 8. (, ] (, 4 ] 9. (, ] [, ) 0. (4, 6) (, ) (, ) 0.5 Eponential Equations ln 4 or ln or or 0. ln or 5. ln 6. or

77 5. Answers to the Eercises 0.6 Eponential Inequalities. (, ). [, ). (, ) (, ) 4. [ 5, ) 5. (, ] [, ) 6. (, 0) (, ) 7. (, ) 8. (, ] [, ) 9. (, ) 0. (, ). [, ). (, ). [, ) 4. [ 5, ) 5. (, 0] (, ) 6. (, ) 7. (, ) 8. (, 0] 9. (, 0) (, ) 0. (0, ) 0.7 Logarithmic Equations.. 5. or or e 9. e or e 0. e or e... e /(e ) 4. or e e 9. e or e 0. 7

78 0.8 Logarithmic Inequalities. (, ]. (, ) (4, ). [, ) (5, 7] 4. (0, e ) [e, ) 5. (6, 9] 6. (, 0) (0, ) 7. (, e e) (e + e, ) 8. (0, ) (, ) 9. (, ] 0. (, 6] [, 6). [, 0) (, ]. [ 8, 6) (0, ]. [, ) (7, 9] 4. (e, e ) 5. (, 4) 6. (, e +] 7. (, ] 8. ( 4, ] 9. (, 4) 0. (0, ) (, ) 7

79 5. Answers to the Eercises 0.9 Trigonometric Equations and Inequalities. {kπ} { π + kπ}. { kπ}. { 5 kπ} { π + kπ} 4. { 5 π + 5 kπ} 5. { π + kπ} 6. {kπ} { 4 π + kπ} 7. {kπ} { 6 π + kπ} { 5 6 π + kπ} 8. { π + kπ} { 4 π + kπ} 9. { π + kπ} { π + kπ} 0. { 8 π + kπ}. { 4 π + kπ} { π + kπ}. { 7 6 π + kπ} { kπ + kπ} 6. {kπ} 4. { 4 π + kπ} 5. { π + kπ} { 4 π + kπ} Series C. ( π + 4kπ, 5 π + 4kπ). [ π + kπ, π + kπ). [ 4 π + kπ, 4 π + kπ]. {kπ} { 5 kπ}. {kπ} { 4 π + kπ}. { kπ} 4. {kπ} { 4 π + kπ} 5. { 8 π + kπ} 6. { π + kπ} { 4 π + kπ} 7. {kπ} 8. { 6 π + kπ} 9. { π + kπ} 0. { 6 π + kπ} { 5 6 π + kπ}. { 6 π + kπ} { 5 6 π + kπ}. { 4 π + kπ} { π + kπ}. { 4 π + kπ} { π + kπ} 4. { kπ} {kπ} 5. { 4 π + kπ} Series D. (0 + kπ, π + kπ). ( 4 π + kπ, π + kπ]. [0 + kπ, π + kπ) 74

80 0.0 Equations Involving Square Roots. 6.. or or 0 6. or or 7. or p 4 0. p or or 6 7. or p 0. Inequalities Involving Square Roots. [9, ). [ 6, 0). ( 5, 0) 4. [, 4] 5. (, 8 ) ( 0, ) 6. [, 5) 7. [4, ) 8. [, ] [, ) 9. (0, ) (4, ) 0. [, ) (8, ). [, 4) (4, ). [, 7). (, 4] 4. [4, 8] 5. (, ) (4, ) 6. (, ) 7. (, ) 8. [ 7, 64] 9. ( 5, ) (, 5) 0. [, 4) ( 4, ) 75

81 5. Answers to the Eercises Rationalizing Denominators (etra) Partial Fraction Decomposition A (etra). ( ) ( ). ( ) ( + ). ( ) ( ) 4. ( + ) + ( ) 5. ( ) 6. ( ) + 4( ) 7. ( ) + ( ) 8. ( ) + 9. ( ) + ( ) 0. ( + ) ( + ). ( + ) ( + ). 6 ( ) ( + ) 6. ( + ) + ( ) ( + ) + 9 ( ) 5 5. ( ) 6. 6( ) + 8( ) 7. ( ) + 9 ( ) 8. 4( ) ( ) + 8( ) 0. 6( + ) 76

82 Partial Fraction Decomposition B (etra) ( ) ( ) + ( + ) ( ) 8. + ( ) 9. + (7 0)/( 4) + ( ) + 6( + ) 0. + ( + ) ( ) ( + ) ( ) (5 )/( ) ( 4)/( + ) 4 Inverse Trigonometric Functions (etra). The graph of y sin can be reflected in the line y, where the resulting image is a graph of a function if we take [ π, π] as domain for. For this domain we have a graph which is increasing everywhere. It should be no problem to see that this also holds for a graph which is decreasing everywhere. So, for the domain [ π, π] the resulting image should also be the graph of a function. This can be checked using a graphing calculator, by entering on the TI-8: ysin() and then using DrawInv y on the Draw-menu to inspect the reflected graph. You can now check the same for y cos on [0, π].. As above. 77

83 5. Answers to the Eercises 4 Inverse Trigonometric Functions (etra) π 4 π π 6 π 4 π 6. π π 4 π 6 π. π. 4 π. π π 6. 4 π π 6 π 0. π 78