คณิตศาสตร์พื้นฐาน1

50 Chapter2 Equations and Inequalities 2 Inequalities 2 1 Properties of inequalities For all real numbers, the expression that represents the relationships be- tween quantities with inequality signs, >, <, ≧ or ≦, is called an inequality. As for imaginary numbers, the relationships between quantities are not considered. For inequalities, the following properties are often used. Properties of inequalities (1) (1) If a > b, then a − b > 0. On the other hand, if a − b > 0, then a > b. (2) If a > 0 and b > 0, then a + b > 0 and ab > 0. Using the properties above, let’s look into the other properties of inequalities. I “ If a > b and b > c, then a > c.” a − b > 0, and b − c > 0 cb a a − c = (a − b) + (b − c) > 0 From this, a > c II “If a > b, then a + c > b + c.” ba For example, +4 +4 (a + 4) − (b + 4) = a − b > 0 b+4 a+4 Therefore, a + 4 > b + 4 III “When a > b, if c > 0, then ca > cb.” ba For example, ×2 ×2 2a − 2b = 2(a − b) > 0 Therefore, 2a > 2b 2b 2a “If c < 0, then ca < cb.” ×(−2) ba ×(−2) −2a −2b

§ 2 Inequalities 51 By summarizing the above, the following properties are obtained. Properties of inequalities (2) I If a > b and b > c then a > c II If a > b then a + c > b + c and a − c > b − c III When a > b, If c > 0 then ac > bc and a > b cc If c < 0 then ac < bc and a < b c c 2 2 Solving linear inequalities If the inequality for the letter x such as 3x − 12 > 0 is given, then the range of values for the real number x that satisfies the given inequality, is called the solutions of an inequality, and finding the solutions is called solving its inequality. If we simplify an inequality using transposition such that 0 is on the right side, then the inequality whose left side is represented in the form of a linear expression (for x) is called a linear inequality. Exercise1 Solve the following linear inequality. x+3 > 2x − 3 2 Solution Transform the inequality as follows. x+3 >2x − 3 Multiply both sides by 2(> 0) [ Properties III ] 2 x + 3 > 4x − 6 x + 3 − (4x − 6) > 0 Subtract 4x − 6 from both sides [ Properties II ] −3x + 9 > 0 Subtract 9 from both sides [ Properties II ] −3x > −9 Divide both sides by −3(< 0) [ Properties III ] x<3 Therefore, the solutions of the inequality is, x < 3. //

52 Chapter2 Equations and Inequalities Q. 1 Solve the following linear inequalities. (1) x − 10 < 26 − 3x (2) −x + 1 ≦ 7 (3) 8 ≧ 3x + 5 (4) 4x + 3 > 2x − 1 3 Q. 2 Suppose you want to buy eight of apples and persimmons all together for less than 700 yen. An apple costs 100 yen each and a persimmon costs 80 yen each. How many apples can you buy? 2 3 Various types of inequalities Simultaneous inequalities A set of two or more inequalities is called simultaneous inequalities. The range of real numbers that satisfies all the inequalities simultaneously is called the solutions of simultaneous inequalities. Exercise2 Solve the following simultaneous inequalities. (1) 3x + 10 > 4 (2) 2x + 3 ≦ 9 Solution The solutions of (1) are, x > −2. The solutions of (2) are, x ≦ 3. These solutions can be shown on a number line as below. (2) (1) −2 3x By finding the common range of x for the solutions of (1) and (2), −2 < x ≦ 3 // Note in the figure shows that the value (point) is not included in the solutions, and shows that the value (point) is included in the solutions. When solving simultaneous inequalities, as in the case with Exercise 2, we first find the solutions of each inequality and then find the common range of x for them.

§ 2 Inequalities 53 Q. 3  Solve each of the following sets of simultaneous inequalities.  x−3>1  3x − 8 < 4 (1)  x + 8 > 2(x + 1) (2)  x − 2 ≦ 3x + 4    6x − 1 > 2x + 4  3(x − 1) ≦ x + 5 (3)  3x + 5 ≦ 6x − 4 (4)  1 x 2x − 1 2 − 3 ≦ 6 Inequalities of second degree By simplifying an inequality using transposition such that 0 is on the right side, the inequality whose left side is represented in the form of a quadratic expression (for x) is called an inequality of second degree. When solving an inequality of second degree, we first determine how the sign of a quadratic expression on the left side changes depending on the value for x, and then find the range of x that satisfies the given inequality. Exercise3 Solve the following inequality of second degree. (x + 3)(x − 5) > 0 Solution First, determine how the sign of (x + 3)(x − 5) changes. The sign of x + 3 changes at the point where x + 3 equals 0, that is, it changes only if x is. Next, determine the sign of x − 5 in the same manner. From these two signs, the sign of (x + 3)(x − 5) can be shown in the table below. x · · · −3 · · · 5 · · · x+3 − 0 +++ x−5 −−− 0 + (x + 3)(x − 5) + 0 − 0 + ++ x −3 − 5 Therefore, the ranges of x for which (x + 3)(x − 5) > 0 are, x < 3, x > 5.// Note The solutions of (x + 3)(x − 5) < 0 are, −3 < x < 5.

54 Chapter2 Equations and Inequalities In general, the following can be said. Solutions of an inequality of second degree If α and β are real numbers with α < β, The solutions of (x − α)(x − β) > 0 are, x < α, x > β The solutions of (x − α)(x − β) < 0 are, α < x < β ++ x α−β The following can be also said for an inequality with an equal sign. The solutions of (x − α)(x − β) ≧ 0 are, x ≦ α, x ≧ β The solutions of (x − α)(x − β) ≦ 0 are, α ≦ x ≦ β Exercise4 Solve the following inequalities of second degree. (1) x2 − 5x + 6 > 0 (2) −6x2 − x + 2 ≧ 0 Solution (1) x2 − 5x + 6 > 0 Factor the left side. (x − 2)(x − 3) > 0 Therefore, x < 2, x > 3 (2) −6x2 − x + 2 ≧ 0 Multiply both sides by −1. 3 2 4 2 −1 −3 6x2 + x − 2 ≦ 0 6 −2 Factor the left side. 1 (3x + 2)(2x − 1) ≦ 0 // ( 2 )( 1 ) 0 6 x+ 3 x− 2 ≦ Therefore, − 2 ≦x≦ 1 3 2 Q. 4 Solve the following inequalities of second degree. (1) x2 − 6x + 8 < 0 (2) 2x2 − 3x + 1 > 0 (3) −x2 + 4 ≧ 0 (4) 3x2 − 11x + 6 ≧ 0

§ 2 Inequalities 55 Higher degree inequalities We can solve inequalities of third degree or more in the same manner as Exercise 3. Exercise5 Solve the following inequality. 2x3 − x2 − 7x + 6 < 0 Solution First, let P (x) = 2x3 − x2 − 7x + 6, and then find the range of x for which P (x) < 0. Since P (1) = 0, P (x) is divisible by x − 1. From this, P (x) = (x − 1)(2x2 + x − 6) = (x − 1)(x + 2)(2x − 3) By finding x for which P (x) = 0, x = −2 1 3 2 By dividing a number line into these three points and determining the signs of each factor and P (x), the following can be shown in the table below. x · · · −2 · · · 1 ··· 3 ··· x+2 −0+ ++ 2 ++ x−1 − − − 0 + + + 2x − 3 − − − − − 0 + P (x) − 0 + 0 − 0 + + + 3x − −2 1− 2 Therefore, the ranges of x for which P (x) < 0 are, x < −2 1<x< 3 // 2 Q. 5 Solve the following inequalities. (1) (x − 1)(x − 2)(x − 4) < 0 (2) x3 + 2x2 − 5x − 6 ≧ 0

56 Chapter2 Equations and Inequalities 2 4 Proof of inequalities The properties of inequalities described on page 50 and 51, are used for proving inequalities. Exercise6 If x ≧ 1 and y ≧ 1, prove that xy + 1 ≧ x + y holds. In addition, determine the cases where the equal sign holds. Proof The left side − The right side = xy + 1 − (x + y) = x(y − 1) − (y − 1) = (x − 1)(y − 1) From x ≧ 1 and y ≧ 1, x − 1 ≧ 0 and y − 1 ≧ 0. That is, (x − 1)(y − 1) ≧ 0 Therefore, xy + 1 ≧ x + y Here, by taking the equal sign as, x − 1 = 0 or y − 1 = 0, the equal sign holds only if x = 1 or y = 1. // Q. 6 If a > b and c > d, prove a + c > b + d. Q. 7 If a ≧ 1, prove that a2 ≧ 1 holds. In addition, determine the cases where the equal sign holds. The following property of real numbers is also often used. Properties of real numbers a2 ≧ 0 holds for any real numbers a. Ex1 a2 + b2 − 2ab = (a − b)2 ≧ 0 Therefore, a2 + b2 ≧ 2ab (The equal sign holds only if a = b.) Q. 8 If a, b, x and y are real numbers, prove the following inequality (called Schwarz’s inequality). In addition, determine the cases where the equal sign holds. (a2 + b2)(x2 + y2) ≧ (ax + by)2

§ 2 Inequalities 57 Inequality of arithmetical mean and geometrical mean If a and b are positive numbers, then a + b is called the arithmetical √2 mean and ab is called the geometrical mean. Relationship between arithmetical mean and geometrical mean If a and b are positive numbers, √ √ a + b ≧ 2 ab a+b ≧ ab where 2 The equal sign holds only if a = b. Proof a+b √ = √ √ 2 − ab a + b − 2 ab √ 2 √√ ( a)2 − 2 a b + ( b)2 = √ √2 = ( a − b)2 ≧ 0 2 √ Therefore, a+b ≧ ab. 2 √√ Here, the equal sign holds only if a − b = 0 where, a = b. Note When drawing the circle O with radius a and the circle O’ with radius b such that the two circles touch each other as shown in the figure, we have OO′ = a + b. From the Pythagorean a b O′ √ O b a B theorem, AB = 2 ab. We find that AB ≦ OO’ holds from the figure. All this shows the inequality of arithmeti- cal mean and geometrical mean. A √ 2 ab ( 1 ) ( 1 ) a+ a b b Exercise 7 If a>0 and b > 0, prove + ≧ 4. In addition, determine the cases where the equal sign holds. Proof From the relationship between arithmetical mean and geometrical mean, √ a+ 1 ≧2 a· 1 ∴ a+ 1 ≧2 (1) a a a

58 Chapter2 Equations and Inequalities √ b+ 1 ≧2 b· 1 ∴ b+ 1 ≧2 (2) b b b For (1) and (2), by multiplying the left sides of each and the right sides of each, ( 1 ) ( 1 ) a+ b b a + ≧ 4 The cases of equality are when the equal signs hold in (1) and (2). Therefore, a= 1 b= 1 . That is, the equal sign holds only if a=1, b = 1. // a b Q. 9 If a > 0 and b > 0, prove the following inequalities. In addition, determine the cases where the equal signs hold. ( ) ( ) a+ 1 b 1 b a (1) b + a ≧ 4 (2) a + b ≧2 Q. 10 If a, b, c and d are all positive numbers, prove the following in- equality. In addition, determine the cases where the equal sign holds. √ (a + b)(c + d) ≧ 4 abcd Inequalities with respect to second degree Exercise8 Prove the following inequality. In addition, determine the cases where the equal sign holds. x2 + xy + y2 ≧ 0 Proof The left side = x2 + xy + 1 y2 − 1 y2 + y2 44 ( )2 = x+ 1 y + 3 y2 24 Here, since ( 1 )2 ≧0 and y2 ≧ 0, x+ y 2 x2 + xy + y2 ≧ 0 x2 + xy + y2 = 0 holds for x + 1 y = 0 and y = 0. 2 By solving this, x = y = 0. Therefore, the equal sign holds only if x = y = 0. //

§ 2 Inequalities 59 Q. 11 Prove the following inequalities. (1) x2 + 8x + 16 ≧ 0 (2) x2 − 4x + 6 > 0 Q. 12 Prove the following inequality. In addition, determine the cases where the equal sign holds. x2 + y2 ≧ xy Q. 13 If x > y, prove the following inequalities. (1) x2 + xy + y2 > 0 (2) x3 > y3 2 5 Sets The collection of natural numbers less than or equal to 5, {1 2 3 4 5} (1) (2) or the collection of all positive even numbers, {2 4 6 8 } As the above, a collection of objects that satisfies a certain condition is called a set, and each object that makes up the set is called an element of the set. Although a set is represented by writing all the elements such as (1) and (2), it can also be written as follows using the condition C(x) which defines the elements of the set. {x | C(x)} Ex2 {1 3 5 7 9} can be written as, {x | x is a positive even number less than or equal to 10}. {1 2 3 4 5 } can be written as, {x | x is a natural number}. When a set is represented with the letters such as A and B, and if a is an element of set A, this is denoted as, either a ∈ A or A ∋ a, and we call this “a belongs to set A”. If not, this is denoted as, either a ∈\\ A or A ∋\\ a

60 Chapter2 Equations and Inequalities If all the elements of set A and all the elements of set B are the same, we say that “A and B equal to each other”, which is B denoted as A = B. Then, if all the elements of A are the elements of B, we say that “A is a subset of B”, which is denoted as, either A A ⊂ B or B ⊃ A. We describe this as, “A is contained in B” or “B contains A”, and the relationship between A and B is shown in the diagram. We call such a diagram that shows the relationships between sets a Venn diagram. Here we should keep in mind that A ⊂ B also contains the case of A = B. Ex3 If A = {x | x > 1} and B = {x | x > 0}, then A ⊂ B For sets A and B, the set of elements common to both A and B is called the intersection of A and B, which is denoted as A ∩ B. Then, the set of elements that belongs to either A or to B is called the union of A and B, which is denoted as A ∪ B. A ∩ B and A ∪ B are shown respectively in the shaded areas in the diagrams. A BA B A∩B A∪B Ex4 If A = x|x is a multiple of 3 and B = x|x is a multiple of 5, The elements of A ∩ B are the numbers that are divisible by both 3 and 5, where the common multiples of 3 and 5. Therefore, A ∩ B = {x | x is a multiple of 15} Then A ∪ B = {x | x is a multiple of 3 or a multiple of 5}

§ 2 Inequalities 61 When dealing with divisors and multi- U ples, we think them in the set of all natu- ral numbers. In this manner, when we first A A define set U of all objects considered in the theory and think of only the elements and the subsets, we call U a universal set. If universal set U and its subset A are given, then the set that does not belong to A is called the complement of A, and this is denoted as A. The inter- section A ∩ A of set A and the complement A, is a set that does not contain any elements. In this manner, the set that does not contain any elements is called an empty set, and this is represented by the symbol ϕ. The following holds for universal set U and its subset A. A=A A∪A=U A∩A=ϕ Ex5 Suppose the universal set is the set of all real numbers. The complement of A = {x | x ≧ 0} is, A = {x | x < 0}. Q. 14 Suppose the universal set is the set of all natural numbers. Find the complement B of B = {x | x2 > 16}. Q. 15 Suppose the universal set is U = {x|x is a natural number less than or equal to 10}. A = {1, 2, 4, 6} and B = {1, 3, 4, 5, 9}, Find each of the following sets, (1) A ∩ B (2) A ∪ B (3) A ∩ B (4) A ∪ B If U is the universal set and A and B are the subsets of U , for the comple- ments of A ∩ B and A ∪ B, the following de Morgan’s law is established. de Morgan’s law A ∪ B = A ∩ B, A ∩ B = A ∪ B

62 Chapter2 Equations and Inequalities A∩B U A∪B U AB AB A∩B A∪B A U A is part of and B is part of and B Q. 16 Prove the following equality using de Morgan’s law. (A ∪ B) ∩ C = (A ∩ B) ∪ C 2 6 Propositions A proposition is something that is referred to by sentences and expressions and can be either true or not true. If a proposition is correct, then we say the proposition is true, and if not, then we say the proposition is false. Exercise9 Determine whether the following propositions are true or false. (1) If a is a real number, then a2 ≧ 0. (2) If ab = ac, then b = c. Solution (1) Since a real number when squared always gives a number greater than 0, the proposition is true. (2) Although a = 0 gives ab = ac = 0, it does not always hold b = c. Therefore, the proposition is false. //

§ 2 Inequalities 63 In order to show the proposition is false, we need to provide examples showing the proposition cannot be established. We call such an example a counter-example. Q. 17 Determine whether the following propositions are true or false. Give the counter-example if it is false. (1) If x > 1, then x2 > 1. (2) If x < 1, then x2 < 1. When a value is substituted into x in the expression such as “x > 1”, the expression that can be either true or false is called a condition. The set that satisfies the condition is called a truth set of the condition. Let the truth set of the condition p be P and the truth set of the condition q be Q. In a proposition, for two conditions p and q, the most familiar form is If p, then q . Here, p is called an assumption and q is called a conclusion. If the proposition “If p, then q” is true, then U this is denoted as p ⇒ q. When p ⇒ q, and Q if the condition p is satisfied, then the condi- tion q is supposed to be always satisfied. From P this, the elements of truth set P are also the elements of truth set Q, and holds. If p ⇒ q, which means If p, then q is true, then we say p is a sufficient condition for q, and q is a necessary condition for p. If p ⇒ q and q ⇒ p, then this is denoted as p ⇔ q, which we say p is a necessary and sufficient condition for q. Here q is also a necessary and sufficient condition for p. We also say p and q are equivalent. If the proposition “If p, then q” is false, then this is denoted as p ⇒\\ q. Ex6 For a quadrilateral ABCD, since “Parallelogram ⇒ AB//DC”, being a parallelogram is a sufficient condition for being AB//DC. On the other hand, since “AB//DC \\⇒ Parallelogram”, being a parallelogram is not a necessary condition for being AB//DC.

64 Chapter2 Equations and Inequalities Q. 18 Fill in the blanks and complete the sentences choosing the words from - necessary, sufficient or necessary and sufficient. (1) ac = bc is a condition for a = b. (2) x = y is a condition for x + z = y + z. (3) For the integer n, “ n is a multiple of 6 ” is a condition for “ n is a multiple of 3 ”. condition for the diag- (4) For a quadrilateral to be a rhombus is a onals to intersect at right angles. For the condition p, the condition that says U “ Not p” is called the negation of p and this PP is denoted as p. If we let the universal set be U and the truth set of the condition p be P , then the truth set of the condition p is the complement P of P . Ex7 If we let “ The integer n is an odd number ” be p, then p is “ The integer n is an even number ”. Q. 19 Suppose the universal set is the set of all natural numbers less than or equal to 10. Write the negation of “ The integer n is a multiple of 3 ” and find the truth set P . If we let the universal set be U , the truth set of the condition p be P , and the truth set of the condition q be Q, then the truth set of the condition “p and q” is P ∩ Q and the truth set of the condition “p or q” is P ∪ Q. UU PQ PQ P ∩Q P ∪Q Condition “p and q” Condition “p or q”

§ 2 Inequalities 65 From de Morgan’s law, P ∩ Q = P ∪ Q, P ∪ Q = P ∩ Q Therefore, “ p and q ” is the same condition as “ p or q ”. “ p or q ” is the same condition as “ p and q ”. Ex8 The negation of “ a = 0 or b = 0 ” is, “ a =\\ 0 and b =\\ 0 ”. Q. 20 Write a negation for each of the following conditions. (1) x < 1 or x > 5 (2) The integer n can be divisible by both 3 and 5. Let the proposition “ If p, then q ” be denoted as “ p → q ”. Using this expression, if the proposition “ p → q ” is true, then it is “ p ⇒ q ”. For “ p q ”, let us find the converse, converse q→p p→q obverse and contraposition as follows. “ q → p ” : the converse of “ p → q ” obverse contraposition obverse “ p → q ” : the obverse of “ p → q ” “ q → p ” : the converse of “ p → q ” p→q converse q→p Ex9 The converse of “ x > 2 → x2 > 4 ” is, “ x2 > 4 → x > 2 ”. The obverse is “ x ≦ 2 → x2 ≦ 4 ”, and the contraposition is “ x2 ≦ 4 → x ≦ 2 ”. Q. 21 Write the converse, obverse and contraposition of “ x > 0 and y < 0 → xy < 0 ”. Let the universal set be U , the truth set of the condition p be P , and the truth set of the condition q be Q. If the proposition “ p → q ” is true, then it is p ⇒ q and P ⊂ Q holds. As shown in the figure on the next page, since Q ⊂ P holds at that time, it is q ⇒ p and its contraposition “ q → p ” is also true. Likewise, it follows that if the contraposition “ q → p ” is true, then its proposition “ p → q ” is also true.

66 Chapter2 Equations and Inequalities U U QQ PP Q P p ⇒ q ··· P ⊂ Q Q ⊂ P ··· q ⇒ p Relationship between a proposition and its contraposition The truth or falsity of a proposition is the same that of its contraposition. Note Even if P ⊂ Q, Q ⊂ P and P ⊂ Q do not hold generally. Therefore, even if “ p → q ” is true, its converse and obverse do not always be true. Exercise10 Using contraposition, prove that the following proposition is true. For the integer n, if n2 is an even number, then n is an even number. Proof Prove the contraposition “ If n is an odd number, then n2 is an odd number ”. By using the integer m, since the integer n is represented by n = 2m + 1, n2 = (2m + 1)2 = 2(2m2 + 2m) + 1 Therefore, n2 is also an odd number. // Q. 22 Using contraposition, prove that the following propositions are true. (1) If x + y > 2, then either x > 1 or y > 1. (2) If m and n are integers, and mn is an odd number, then m and n are also odd numbers.

§ 2 Inequalities 67 √ Here, let us prove that 2 is an irrational number. √ Assume that 2 is not an irrational number. By using two positive integers √ a and b, 2 is represented as follows. √ b (Note that both a and b are relatively prime.) 2= a By squaring both sides of the equation, 2a2 = b2 That is, b2 is an even number, where b is also an even number. Therefore, we have it as follows. b = 2c (where c is an integer.) Then, 2a2 = 4c2 where a2 = 2c2 From this, we find that a is also an even number. This contradicts the as- √ sumption that a and b are relatively prime. Therefore, 2 is an irrational number. In this manner, when proving a certain fact, “ by assuming that the fact is not correct, the result that contradicts our assumption will be derived, from which we will find our assumption is wrong and the fact is correct in conclusion ”. Such a method of proof is called the reduction to absurdity.

68 Chapter2 Equations and Inequalities Column Propositions In general mathematics, when we simply say propositions, they mean only the true propositions. Propositions are classified into a lemma, a theorem, a corollary and a proposition (in the narrow sense). A theorem is the most important proposition. A lemma is a minor proposi- tion which is necessary in proving a theorem, and a corollary is a proposition which is established almost immediately from a theorem. A proposition (in the narrow sense) is generally less important than a theorem and a relatively easy to be proved. These classifications do not have clear boundaries, which depend on people. When we describe a term used in a proposition and a proof as clearly as possible to avoid misunderstandings, then we say we define the term. Using the propositions and definitions we already know, the process which allows us to establish the truth of a certain proposition is called a proof of the proposition. There are the following methods for proving propositions. Other than the direct method for proving “ p ⇒ q ” directly, there are methods such as the proof using contraposition in which instead of using “ p ⇒ q ”, its contraposition “ q ⇒ p ” is proved, and also the reduction to √ absurdity which is used in proving 2 is an irrational number. As a starting point to the mathematical theory of general systems, some ba- sic rules and assumptions that are regarded as taken for granted without proof are called axioms or axiomatic systems. Starting with a set of axioms, the method for constructing a mathematical system with demonstrations is called deduction. On the other hand, as often seen in natural science, the method for deepening the knowledge by piling up observations and experiments is called induction. Around 300 BCE, in his work “ Elements ” (original ge- ometry), a Greek mathematician Euclid, constructed a mathematical system called plane geometry which dealt with flat shapes. He constructed the sys- tem by using deduction from seven axioms. Since then, for over 2,000 years, Euclidian geometry had been the base of mathematics.

§ 2 Inequalities 69 Exercise Problem 2-A 1. Solve the following inequalities. (2) 4 + 2 x ≧ 3x + 1 (1) 2x + 3 < 6x + 8 32 (3) 5x > 2x2 − 3 (4) x2 − x − 6 ≧ 0 2. Suppose you want to make a rectangle such that the sum of the length and width is 12 cm and the area is 20 cm2. If the width of the rectangle is xcm, find the range of x. 3. Solve the following inequalities. (1) x3 − 2x2 − x + 2 ≧ 0 (2) 2x3 + 3x2 − 9x − 10 < 0 4. Prove the following inequalities. In addition, determine the cases where the equal signs hold. Note that (2) is a > 0 and b > 0. (1) a2 + b2 ≧ ( a + b )2 (2) a3 + b3 ≧ a2b + ab2 22 5. Write the converse and contraposition for each of the following propositions, and determine the truth or falsity. (1) If x = 0, then xy = 0. (2) If x2 = 1, then either x = 1 or x = −1. 6. Fill in the blanks and complete the sentences choosing the words from - necessary, sufficient or necessary and sufficient. (1) x = 1 is a condition for (x − 1)(x + 3) = 0. (2) For the real number a, |a| = 2 is a condition for a2 = 4. (3) For a quadrilateral ABCD, AB = CD is a ccondition for a quadri- lateral ABCD to be a parallelogram.

70 Chapter2 Equations and Inequalities Exercise Problem 2-B 1. Solve each of the following sets of simultaneous inequalities. 3x + 3 ≧ 2x − 1 x2 − 2x − 3 ≦ 0 (1) 2x < 1 − x (2) x > 4x + 3 3x + 2 < 4x 2. Solve the following inequalities. (2) x−1 ≦3 (1) −x + 4 > 0 x−2 x−2 3. Using the relationship between arithmetical mean and geometrical mean, prove the following inequalities. In addition, determine the cases where the equal signs hold. Note that the letters are all positive numbers. ( 1 ) ( 1 ) ( 1)≧8 a+ b c (1) + + bca (2) √ 2 ab ≧ 1+1 ab 4. Suppose a is a real constant. Answer the following questions. (1) Compare magnitude of 2 + a2 and 2a − a2. (2) Solve the following inequality of second degree. x2 − 2 (a + 1) x + a (2 + a2) (2 − a) > 0 5. Determine the truth or falsity for each of the following propositions. In addi- tion, determine the truth or falsity of its converse and contraposition. (1) If a and b are complex numbers, and a2 + b2 = 0, then a = 0 and b = 0. (2) If x = 1 and y = 3, then x − y = −2. 6. Using contraposition, prove that the following propositions are true. (1) If x and y are real numbers, and x + y > 0, then either x > 0 or y > 0. (2) If n is an integer, and n2 is a multiple of 3, then n is also a multiple of 3.

71 3Chapter Functions and Graphs § 1 Quadratic functions 1 1 Functions and graphs Considering two variables x and y, when value of x is specified, a single value of y is also specified accordingly, which means y is a function of x. In this case, x is an independent variable and y is a dependent variable and expressed as y = f (x), y = g (x). For the function y = f (x), the value of y corresponds to x = a is expressed as f (a) and called the value of the function. Ex.1 Given f (x) = 5x − 2 f (−2) = 5 × (−2) − 2 = −12 f (a) = 5a − 2, f (a + h) = 5(a + h) − 2 = 5a + 5h − 2 Q. Given f (x) = 3x + 1, find f (−1) f (−a), f (a + 1). The possible range of variables is called variable range. For the function y = f (x), a variable range of x is called domain and a variable range of y that corresponds to x domain is called the range of the function.

72 Chapter 3 Functions and Graphs Ex.2 (1) For the function y = 2x + 1, the domain and range are all real numbers unless stated otherwise. (2) For the function y = 1 , the domain is all real numbers except for 0 ( x x =\\ 0), and the range is y =\\ 0. ) Fractions with 0 denominator ( 1 , etc. are not defined. Note 0 00 When considering a coordinate plane with two coordinate axes cross at the origin 0, the points on the plane can be y expressed with sets of real numbers (x, y). Second First Apply this to functions and discover their quadrant quadrant properties. A coordinate plane is divided O x into four areas. Each of them are called First quadrant, Second quadrant, Third Third Fourth quadrant quadrant quadrant and Fourth quadrant respec- tively as shown below. The points directly on the coordinate axes don’t belong to any of the quadrants. When considering a as the value within a given domain of the function y = f (x), the points (a, f (a)) on the coor- y y = f (x) dinate plane form a figure. The figure is f (a) called a graph of the function y = f (x), and the equation y = f (x) is called the equation of the graph. The graph gen- O ax erally creates curved lines. The curve is called curve y = f (x). When variable y is expressed with a linear y y = ax + b expression of variable x, it is called a linear function y = ax + b (a and b are constant, a x a =\\ 0). A graph of linear function has a b1 straight line with a slope a and intercept b. O

§ 1 Quadratic functions 73 A function represented by y = c (constant) is called a constant function. This means y remains constant c no matter what the value of x would be. A graph of constant function y = c has a straight line parallel to x-axis and passes through the points (0, c). x-axis is represented by y = 0. Note A function y = f (x) is not ap- y y=c plicable to a straight line parallel to y- c axis. When the straight line parallel to y- axis passes through points (d, 0), all the O x=d x points on x-axis remain d. Therefore, the d straight line can be represented by x = d, and y-axis by x = 0. When you think about the range of function with restricted domain, it is useful to apply a graph of function. y Ex.3 (1) For the function y = −x + 3 , 2 y = −x + 3 2 2 1 x If the given domain is 1 ≦ x ≦ 2, 2 −1 O 2 the range is y= 1 ≦y≦2 2 (2) A rectangle with a perimeter of 20 m has a width of x m and a length x m Perimeter of 20m of y m, thus y = 10 x. The given domain is 0 < x < 10, and the range ym is 0 < y < 10. Q. Find the ranges of the following functions corresponding to the do- mains in the bracket ( ). (2) y = −3x + 2 (1 ≦ x < 2) (1) y = 2x − 1 (0 ≦ x ≦ 1)

74 Chapter 3 Functions and Graphs 1 2 Graphs of quadratic function When variable y is expressed with a quadratic expression of variable x, the function is called a quadratic function. y = ax2 + bx + c (a, b, and c are constant, a =\\ 0) A graph of quadratic function y = ax2 (a =\\ 0) is a parabola with the origin as the vertex and y axis as the axis of symmetry. For a > 0 the parabola is downwards convex and for a < 0 upwards y −x O convex y = ax2 (a > 0) Let’s look into the graphs of quadratic func- downwards tions and their relations. convex y = x2 (1) xx y = (x − 2)2 (2) upwards convex y = (x − 2)2 + 3 (3) y = ax2 (a < 0) x −3 −2 −1 0 1 2 3 4 5 y = x2 9 4 1 0 1 4 9 ········· y = (x − 2)2 +2 9 4 1 0 1 4 9 y = (x − 2)2 + 3 · · · · · · · · · 12 7 4 3 4 7 12 +3 As you see in the table above, if the values y (1) (3)(2) of y = x2 shifted two to the right, they will be equal with the value of y = (x 2)2. There- 3 x fore, the graph (2) has the same parabola as the graph (1) but translated 2 units towards x- +3 axis. Shifting figures to a certain direction and units are called parallel translation. If 3 is O +2 2 added to the value of y = (x 2)2, the value will be y = (x 2)2 + 3.

§ 1 Quadratic functions 75 That means the graph (3) has the same parabola as the graph (2) but translated 3 units parallel towards y-axis. The graph of the function y = (x 2)2 + 3 has the same parabola as y = x2 but translated 2 units parallel towards x-axis and 3 units towards y-axis, and taking x = 2 as an axis of a straight line and the coordinates (2, 3) as the vertex. For the graph of quadratic function y = a(x p)2 + q, the following is established. The graph of quadratic function   The graph of quadratic function y y = a(x − p)2 + q y = a(x p)2+q has the same parabola as y = ax2 but translated p units paral- y = ax2 lel towards x-axis and q units towards y-axis, and taking x = p as an axis of a q straight line and the coordinates (p, q) O px as the vertex. When a > 0, downwards convex. When a < 0, upwards convex.   Ex.4 The graph of quadratic function y y = −3(x 1)2 +2 has the same parabola y = −3(x − 1)2 + 2 as y = −3x2 but translated 1 unit parallel 2 towards x-axis and 2 towards y-axis, and opening upwards. The straight line x = 1 O x is the axis and the coordinates (1, 2) are 1 the vertex. −1 y = −3x2 Q. Plot graphs of the following functions. Find the axis and coordinates of the vertex. (1) y = 2x2 + 1 (2) y = −2(x − 1)2 (3) y = −(x − 2)2 − 1 (4) y = 3(x + 1)2 + 2

76 Chapter 3 Functions and Graphs Q. Find an equation of parabola that has the same parabola as y = 2x2 but translated −1 parallel towards x-axis and 3 towards y-axis. Let’s look into a graph of quadratic function y = −2x2 + 8x − 5. The right side of the equation can be transformed as, y y = −2x2 + 8x − 5 y = −2(x2 − 4x) − 5 3 = −2 {(x − 2)2 − 4} − 5 = −2(x − 2)2 + 3 O x Therefore, the graph is a parabola with a 2 straight line x = 2 as the axis and coordinates (2, 3) as the vertex. The graph has the same parabola as y = 2x2 but translated 2 parallel −5 towards x-axis and 3 towards y axis. y = −2x2 Note A quadratic function can be transformed into y = a (x − p)2 + q form as mentioned above. This form of equation is called a standard form of quadratic function. Q. Transform the following quadratic functions into the standard forms. In addition, plot graphs and find the axis and the coordinates of the vertex. (1) y = x2 − 4x + 5 (2) y = −x2 − 6x − 7 (3) y = 4x2 + 4x (4) y = −3x2 + 9x − 6 Exercise Find the equation of parabola that has the vertex at the coordinates (−2, 1) and passes through the point (−1, 4). Solution Since the coordinates of the vertex are (−2, 1), the equation of parabola (quadratic function) can be expressed as y = a(x + 2)2 + 1. Since the parabola passes through the point (−1, 4), 4 = a(−1 + 2)2 + 1, where a = 3, Thus, y = 3(x + 2)2 + 1 //

§ 1 Quadratic functions 77 Q. Find the equations of parabolas that satisfies each of following con- ditions. (1) The coordinates of the vertex are (1, 2) and passes through the origin. (2) Having a straight line x = 1 as the axis and passes through two points (0, 1) and (3, 7). (3) Having the vertex on y-axis and passes through two points (1, 0) and (2, −3). Exercise Find the equation of parabola that passes through 3 points (−1, 4), (1, 4) and (3, −4). Find the axis and the vertex, and plot a graph. Solution The equation of the parabola can be express as y = ax2 + bx + c. Since the parabola passes through three points (−1, 4), (1, −4) and (3, −4), 4=a−b+c (1) y −4 = a + b + c (2) −4 = 9a + 3b + c (3) 2 x By solving the simultaneous equations (1), (2) and (3), O −1 a = 1, b = −4 and c = −1 Therefore, the equation of parabola is y = x2 − 4x − 1. −5 Transform this into a standard form. y = (x − 2)2 − 4 − 1= (x − 2)2 − 5 Thus, x = 2 is the axis and the coordinates (2, −5) are the vertex. The graph is shown above. Q. Find the equations of parabolas that satisfies each of the following conditions. Find the axis and the vertex, and plot graphs. (1) It passes through three points (−1, 7), (0, 1) and (1, −1). (2) It crosses x-axis at the points (1, 0) and (−2, 0) and crosses y-axis at the point (0, 2).

78 Chapter 3 Functions and Graphs 1 3 Maximum and minimum values of quadratic functions Let’s think about maximum and minimum values of quadratic functions. When transforming a given quadratic y a>0 y a<0 function into standard form, q p x y = a(x − p)2 + q, O Op x the graph is a parabola with the straight line x = p as its axis and q the point (p, q) as its vertex. From the graphs, the followings are explained with respect to the function. If a > 0, the quadratic function has a minimum value when x = p , and the minimum value is equal to q. If a < 0, the quadratic function has a maximum vale when x = p , and the maximum value is equal to q. Exercise Find the maximum and minimum values of the following quadratic functions. (1) y = x2 − 4x + 2 (2) y = −2x2 + 4x Solution y x (1) Transform into standard form, x 2 y = (x − 2)2 − 2 2 Thus, the minimum value is O −2 (when x = 2) −2 and there is no maximum value. (2) Transform into standard form, y 2 y = −2(x − 1)2 + 2 Thus, the maximum value is O1 2 (when x = 1) and there is no minimum value. //

§ 1 Quadratic functions 79 Q. Find the maximum or minimum values of the following quadratic functions. (1) y = −x2 + 4x − 3 (2) y = 10x2 + 5x − 1 Exercise Find the maximum and minimum values of the following quadratic functions. (1) y = −x2 + 2x + 2 (−1 ≦ x ≦ 2) (2) y = x2 − 2x − 1 (−1 ≦ x ≦ 1) Solution y x (1) Transform into standard form, 3 2 y = −(x − 1)2 + 3 Then, when x = −1, y = −1 −1 O1 2 when x = 2, y = 2 −1 y The graph is shown in solid line in the figure. 2 Thus, the maximum value is 3 (when x = 1) and the minimum value is −1 (when x = −1). (2) Transform into standard form, y = (x − 1)2 − 2 1 x −1 O Then, when x = −1, y = 2 when x = 1, y = −2 −1 The graph is shown in solid line in the figure. −2 Thus, the maximum value is 2 (when x = −1) and the minimum value is −2 (when x = 1). // Q. For each of the functions below, find the maximum and minimum values within a given domain in the bracket ( ). In addition, find the value of x. (1) y = x2 − 4x + 3 (0 ≦ x ≦ 3) (2) y = −x2 + 2x + 1 (−3 ≦ x ≦ 0) Q. When letting the length of a rectangle with the perimeter of 20 m be x m, find the value of x for x m Perimeter of 20m the rectangular maximum area and the maximum value.

80 Chapter 3 Functions and Graphs 1 4 Quadratic functions and equations Lets’look into the relationships between a parabola which is the graph of a quadratic function. y = ax2 + bx + c (1) ax2 + bx + c = 0 (2) For the point at which the parabola (1) intersects the x-axis (common point), since y-coordinate is 0, the x-coordinates are real solutions to (2). On the other hand, real solutions to (2) are x-coordinates of the common point at which the parabola (1) intersects the x-axis. From the relation between discriminant and the types of solutions, the number of real solutions are de- termined by the discriminant. D = b2 − 4ac Therefore, the positional relationships between the parabola (1) and the x-axis are explained as follows. D When D > 0, When D = 0, When D < 0, Relation x x x between parabora Intersect the x-axis Tangent the x-axis No common point and x-axis 1 point No point Common point 2 points ( I ) If D > 0, The equation (2) has two distinct real solutions and the parabola (1) has two common points on the x-axis. If D > 0, the graph intersects in two points on the x-axis.

§ 1 Quadratic functions 81 (II ) If D = 0, The equation (2) has one real solution (one double real solution) and the parabola (1) has only one common point on the x-axis. If D = 0, the graph is tangent to the x-axis and the common point is called a tangent point. (III) If D < 0, The equation (2) has no real solution and the parabola (1) has no common point with the x-axis. Exercise Find the common points at which the following graphs of quadratic functions cross the x-axis. In addition, find the x-coordinates when there is a common point. (1) y = x2 − 2x − 3 (2) y = 2x2 − 8x + 9 (3) y = −x2 + 4x − 4 Solution (1) D = (−2)2 − 4 × 1 × (−3) = 16 > 0 Therefore, the graph intersects in two points on the x-axis. By solving x2 − 2x − 3 = 0, the x-coordinates of the intersection points are, x = −1, 3 (2) D = (−8)2 − 4 × 2 × 9 = −8 < 0 Therefore, the graph has no common point with the x-axis. (3) D = 42 − 4 × (−1) × (−4) = 0 Therefore, the graph is tangent to the x-axis. By solving −x2 + 4x − 4 = 0, the x-coordinate of the tangent point is, x = 2 // (1) y (2) y (3) y −1 O 3x O2x O2x

82 Chapter 3 Functions and Graphs Q. Find the common points at which the following graphs of quadratic functions cross the x-axis. In addition, find the x-coordinates when there is a common point. (1) y = 9x2 − 6x + 1 (2) y = −2x2 + 8x − 10 (3) y = x2 + x − 1 Exercise For the parabola y = x2 − 4x + k , answer the following questions. (1) Find the range of the values for the constant k such that the parabola intersects in two points on the x-axis. (2) Find the value of the constant k when the parabola is tangent to the x-axis and also find the x-coordinate of the tangent point. Solution y (1) The discriminant of x2 − 4x + k = 0 D = (−4)2 − 4k = −4(k − 4) O 2 x Since the parabola intersects in two points on the x-axis when D > 0. ∴ k<4 y D = −4(k − 4) > 0 (2) Since the parabola is tangent to the x-axis when D = 0, ∴ k=4 O2 x D = −4(k − 4) = 0 // By solving x2 − 4x + 4 = 0, x = 2. Therefore, the x-coordinate of the tangent point is (2, 0) Q. Find the value of the constant k or the range of the values for the constant k such that the following graphs of quadratic functions satisfy each of the conditions in the brackets ( ). (1) y = x2 + 2x + k ( Intersect in two points on the x-axis.) (2) y = 6x2 − 2kx + 5 ( Tangent to the x-axis.) (3) y = 2x2 + 3x + k ( No common point with the x-axis.)

§ 1 Quadratic functions 83 1 5 Quadratic functions and inequalities In this section, we’ll learn how to solve quadratic inequalities ax2 + bx + c > 0, ax2 + bx + c < 0, using the graph of a quadratic function y = ax2 + bx + c. Note that, if a < 0, multiply both sides by (−1) to obtain a > 0. First, as in the method described in the previous section (1 4), find the common points how the graph crosses the x-axis. Then solve quadratic in- equalities by examining the sign of y of the quadratic function ( I ) If D > 0, The graph intersects in two points (α, 0), (β, 0) on the x-axis. Where,α < β. Since the solution to an inequality ax2 + bx + c > 0 is the range of x for which + + y is positive, α x x < α, x > β −β Also, the solution to ax2 + bx + c < 0 is α<x<β (II) If D = 0, The graph is tangent to the x-axis at the point (α, 0). The solution to ax2 + bx + c > 0 is x =\\ α While there is no solution to ax2 +bx+c < 0. (No solution) + + The solution to ax2 + bx + c ≧ 0 is all real numbers. αx The solution to ax2 + bx + c ≦ 0 is x = α. (III) If D < 0, The graph has no common point with the x-axis. + + + The solution to ax2 + bx + c > 0 is all real x numbers. There is no solution to ax2 + bx + c > 0. (No solution)

84 Chapter 3 Functions and Graphs Exercise Solve the following inequalities. (1) x2 + 2x − 1 > 0 (2) −x2 + 4x − 4 ≧ 0 (3) x2 + 2x + 3 < 0 Solution Let the discriminant of the corresponding quadratic equations be D. (1) D = 22 − 4 × (−1) = 8 > 0 y √ By solving x2 + 2x − 1 = 0, x = −1 ± 2 From the graph of y = x2 + 2x − 1 √ O √ −1 − 2 −1 + 2 The solution to x2 + 2x − 1 > 0 is √√ x < −1 − 2, x > −1 + 2 (2) By multiplying both sides by (−1), y x2 − 4x + 4 ≦ 0 Since D = 0, x2 − 4x + 4 = 0 has a real double solution x = 2 From the graph of y = x2 − 4x + 4, O2 x The solution to x2 − 4x + 4 ≦ 0 is x = 2 Therefore, the solution to −x2 + 4x − 4 ≧ 0 is x = 2 (3) Since D = −8 < 0, the graph of y y = x2 + 2x + 3 has no common point with the x-axis and the values of y are always positive. Therefore, there is no solution to O x x2 + 2x + 3 < 0 (No solution) // Q. Solve the following inequalities. (1) x2 + 2x − 3 ≧ 0 (2) −x2 + 6x − 9 < 0 (3) x2 − 6x + 7 ≦ 0 (4) 2x2 + x + 1 > 0

§ 1 Quadratic functions 85 Exercises 1–A 1. Find ranges of the following functions within given domains in the brackets ( ). (1) y = x2 (1 ≦ x < 2) (2) y = x2 (−1 ≦ x < 2) 2. Find each equation of the parabola that satisfies each of the following conditions and whose axis is parallel to the y-axis. (1) The coordinate of the vertex is (2, 1) and the parabola passes through the point (0, 3). (2) The parabola is tangent to the x-axis at the point (1, 0) and passes through the point (2, 3). (3) The parabola takes the straight line x = 2 as the axis and passes through two points (0, 0), (3, 6). 3. Find maximum and minimum values of the following functions. (1) y = x(x − 1) (2) y = −3x2 + 4x + 1 (3) y = − 1 x2 − 6x + 1 2 4. Find the maximum and minimum values of the following functions. Note that within ( ) are given domains. (1) y = x2 − 2x − 2 (−1 ≦ x ≦ 5) (2) y = −2x2−4x+1 (−3 ≦ x ≦ −1) (3) y = −x2 + 5x − 1 (1 ≦ x ≦ 5) 4 5. Solve the following quadratic inequalities. (1) −x2 − x − 1 ≧ 0 (2) 6x2 − 5x − 6 < 0 6. For the graph of a quadratic function y = x2 − 5x + 6, how many units the parabola needs to be translated parallel towards x-axis in order to pass through the origin. 7. Express how is the parabola y = x2 + 2x + 3 translated parallel to the parabola y = x2 − 4x + 4. 8. Find the values of the constants b and c when a parabola y = x2 + bx + c is translated (−1) unit parallel towards x-axis and 3 units towards y-axis and the vertex of the translated parabola is (0, 1). 9. Find the number of real solutions to a quadratic equation x2−2x+k+3 = 0 when k is a constant.

86 Chapter 3 Functions and Graphs Exercises 1–B 1. When the vertices of the following parabolas are the same, find the values of constants a and b. y = x2 − 2x + a y = bx2 − 3x + 1 2. For the function y = ax2 − 2ax + b (−2 ≦ x ≦ 1), find the values of constants a and b such that the maximum value of the function is 6 and the minimum value is 1. Note that a > 0. 3. Find the range of values for the constant a such that a quadratic inequality ax2 + 2x + a > 0 for all real values of x. y 4. In the area enclosed by a parabola y = 4 − x2 4 and the x-axis, a rectangle of ABCD is inscribed A D as shown in the figure. Find the length of BC such that the perimeter of the rectangle has its −2 B O x maximum. C2 5. When the function s is the minimum value of a function y = x2 − 2mx + 4m, express the function s for m. In addition, for what values of m does s has its maximum? y 6. Let the y-coordinate of point P on a parabola P y = x2 be y. By taking the fixed point on the A y-axis as A(0, a), find the value of y when the length of the line segment AP has its minimum. Ox In addition, find the minimum value. Note a > 0. 7. Find standard form of a quadratic function y = ax2 + bx + c. In addition, find the axis of the graph of the function and the coordinates of the vertex.

§ 2 Various types of functions 87 § 2 Various types of functions 2 1 Power function A power function is a function that can be represented in the form y = xn (where n is a positive integer), such as y = x2, y = x3, y = x4. If 0 ≦ a < b, an < bn is provided from the properties of inequality. Thus if x ≧ 0, it means that as x increases, so does xn. Now, let’s think about graphs of power functions by taking n = 3, 4 as an example. The below Table shows the values of y for the various values of x. x · · · −3 −2 −1 0 1 2 3 · · · y = x3 · · · −27 −8 −1 0 1 8 27 · · · y = x4 · · · 81 16 1 0 1 16 81 · · · If y = x3, since (a, a3) and (−a, −a3) are the points on the graph, the graph is symmetric with respect to the origin. Then if y = x4, since (a, a4) and (−a, a4) are the points on the graph, the graph is symmetric with respect to the y-axis. Therefore, the two graphs are as shown in the figure. y = x3 Q a4 y = x4 a3 P 1 P 1 −a −1 1a x −a −1 x 1a −1 Q −a3

88 Chapter 3 Functions and Graphs In general, when f (−x) = −f (x) is provided for any x in its domain, a function f (x) is called an odd function, and when f (−x) = f (x) is provided, a function f (x) is called an even function. As in y = x3 and y = x4, the graph of an odd function is symmetric with respect to the origin, while the graph of an even function is symmetric with respect to the y-axis (and vice versa). Even and Odd functions   f (x) odd function ⇐⇒ f (−x) = −f (x) ⇐⇒ Graph is symmetric with respect to the origin f (x) even function ⇐⇒ f (−x) = f (x) ⇐⇒ Graph is symmetric with respect to the y-axis.   Graph of an Odd function Graph of an Even function y y b b −a a x −a O ax O −b Ex.1 A power function f (x) = xn (where n is a positive integer) If n is an odd number, f (−x) = (−x)n = −xn = −f (x). Therefore, an odd function. If n is an even number, f (−x) = (−x)n = xn = f (x). Therefore, an even function. A constant function y = c is an even function. √ Ex.2 Given : f (x) = x2 + 4 √√ f (−x) = (−x)2 + 4 = x2 + 4 = f (x) Therefore, an even function. Q. Determine the following functions are even or odd. (1) y = x2 + 1 (2) y = −x5 + x (3) y = x − x2

§ 2 Various types of functions 89 As we learned in p.75, when the graph of y = ax2 is translated p parallel towards x-axis and q towards y-axis, the function representing the graph is y = a(x − p)2 + q. The same things are also applied to a general function y = f (x).  Parallel translation on graph  When the graph of a function y = f (x) is translated p parallel towards x-axis and q towards y-axis, the function representing the graph is, y − q = f (x − p) that is to say, y = f (x − p) + q   Proof When the graph of a function y y = f (x − p) + q y = f (x) is translated p parallel to- wards x-axis and q towards y-axis, let y = f (x) P(a, b) be the point on the moved graph. Since Q(a−p, b−q) is the point b P(a, b) on the graph of a function y = f (x), q b−q = f (a−p) is provided. Therefore, Q(a − p, b − q) P(a, b) is the point on the graph of a b−q p x O a−p a function y − q = f (x − p). // Ex.3 The graph of y = (x − 1)4 − 2 y = x4 y y = (x − 1)4 − 2 1x is obtained when the graph of y = x4 is translated 1 unit parallel towards O x-axis and −2 units towards y-axis. See the figure. −1 Q. Plot graphs of the following functions. (1) y = (x + 2)3 + 1 −2 (2) y = −(x − 1)3 − 3 (3) y = 2(x + 1)4 − 2

§ 2 Various types of functions 89 As we learned in p.75, when the graph of y = ax2 is translated p parallel towards x-axis and q towards y-axis, the function representing the graph is y = a(x − p)2 + q. The same things are also applied to a general function y = f (x).  Parallel translation on graph  When the graph of a function y = f (x) is translated p parallel towards x-axis and q towards y-axis, the function representing the graph is, y − q = f (x − p) that is to say, y = f (x − p) + q   Proof When the graph of a function y y = f (x − p) + q y = f (x) is translated p parallel to- wards x-axis and q towards y-axis, let y = f (x) P(a, b) be the point on the moved graph. Since Q(a−p, b−q) is the point b P(a, b) on the graph of a function y = f (x), q b−q = f (a−p) is provided. Therefore, Q(a − p, b − q) P(a, b) is the point on the graph of a b−q p x O a−p a function y − q = f (x − p). // Ex.3 The graph of y = (x − 1)4 − 2 y = x4 y y = (x − 1)4 − 2 1x is obtained when the graph of y = x4 is translated 1 unit parallel towards O x-axis and −2 units towards y-axis. See the figure. −1 Q. Plot graphs of the following functions. (1) y = (x + 2)3 + 1 −2 (2) y = −(x − 1)3 − 3 (3) y = 2(x + 1)4 − 2

§ 2 Various types of functions 91 When a point on the graph gets infinitely closer to a straight line as it gets infinitely far away from the origin, the line is called an asymptote of the graph. The asymptotes of a hyperbola y = a is the straight line x = 0 (y-axis) and x the straight line y = 0 (x-axis). Since the two asymptotes are perpendicular, the hyperbola is called a rectangular hyperbola. The graph of a fractional function y = x a p +q can be plotted by applying − the formula of parallel translation on graph(p.89). Exercise Find domains and ranges of the following functions and plot the graphs. In addition, find the asymptotes. (1) y= 1 +2 (2) y= 2 x−1 5 − 2x Solution (1) The graph of the function y is obtained when the graph of a function 3 2 y= 1 is translated 1 unit parallel towards 1 x O 12 x-axis and 2 units towards y-axis. y The domain is x =\\ 1 and the range is 2 5 y =\\ 2. The asymptotes are the straight O 53 x x line x = 1 and the straight line y = 2. 2 −2 (2) By transforming, 2 = − 1 5 . 5 − 2x x− 2 The graph is obtained when the graph of a function y = − 1 is translated 5 unit x 2 parallel towards x-axis. The domain is x =\\ 5 and the range is y =\\ 0. The asymptotes 2 are the straight line x = 5 and the straight 2 line y = 0. //

92 Chapter 3 Functions and Graphs Q. Find domains and ranges of the following functions and plot the graphs. In addition, find the asymptotes. (1) y = − 3 −1 (2) y= 4 −2 (3) y= 5 x x+1 2−x Let’s think about a fractional function y = cx + d (a =\\ 0). Let the quo- ax + b tient be q which is obtained by dividing a numerator cx + d by a denominator ax + b, and also let the reminder be r. cx + d =q+ r = R +q ) ax + b ax + b x−p ( Note that R= r, p = − b a a Therefore, the graphs can be plotted in the same way as Exercise 1. Exercise Find the domain and the range of a function y = 3x + 4 x+2 and plot the graph. In addition, find the asymptotes. Solution When 3x+4 is divided by y x + 2, the quotient is 3 and the re- mainder is −2. Hence, 5 y = 3x + 4 = −2 + 3 4 x+2 x+2 3 The graph is obtained when a parabola y = − 2 is translated −2 units paral- 2 x 1 −2 x lel towards x-axis and 3 units towards −4 −3 −1 O y-axis. The domain is x =\\ −2 and the range is y =\\ 3. The asymptotes are the straight line x = −2 and the straight line y = 3. The graph is shown in the figure. Q. Find domains and ranges of the following functions and plot the graphs. In addition, find the asymptotes. (1) y= 4x − 3 (2) y= −2x x−1 x+1

§ 2 Various types of functions 93 2 3 Irrational function An equation which contains the variable inside a root sign is called an irrational equation, and the function which can be represented in an irrational equation with the variable x is called an irrational function. The domain of √ an irrational function y = P (x) is, unless otherwise stated, the range of the values for x such as P (x) ≧ 0. Q. Find domains of the following functions. √ (2) y = √ 5 √ (1) y = 3 − x x−2 (3) y = x2 − x − 6 Let’s think about the graph of a func- x0 1 2 3 4 ··· 9 √ y0 1 √√ 2 ··· 3 tion y = x. The domain is x ≧ 0. 23 The values of y for various values of x are shown in the Table. The graph y plotted by using these values is shown in the figure. 3 Note The graph is tangent to the 2 1 y-axis with respect to the origin. O1 x 49 Exercise √ Find the domain and the range of a function y = x − 1 + 2 and plot the graph. Solution The graph of a function y √ √ y= x−1+2 4 y = x − 1 + 2 is obtained when the 3 √ √ 2 y= x graph of y = x is translated 1 unit parallel towards x-axis and 2 units to- wards y-axis. (See the figure.) From O 12 5 x // x − 1 ≧ 0, the domain is x ≧ 1. The range is y ≧ 2. Q. Find domains and ranges of the following functions and plot the graphs. √ √ √ (2) y = x + 1 (3) y = x − 2 − 2 (1) y = x + 1

94 Chapter 3 Functions and Graphs Let’s think about the graphs of functions y = −√x, y = √ y = √ −x, − −x. √√ Since the domain of y = − x is x ≧ 0 and − a which is the value of y for a is the sign of √ changed, the graph of y = −√x is symmetric with respect a √ to the graph of y = x and the x-axis. √ Next, since the domain of y = −x is x ≦ 0 and the value of y for −a is √√ √ a, the graph of y = −x is symmetric with respect to the graph of y = x and the y-axis. √ Then, the graph of y = − −x is symmetric with respect to the graph of √ y = −x and the x-axis. Therefore, the graph is symmetric with respect to √ the graph of y = x and the origin. √ y √ y = −x y= x (a, f (a)) (−a, f (a)) Ox √ (−a, −f (a)) (a, −f (a)) √ y = − −x y=− x The same things are also applied to a general function and the following relationships are provided. Symmetry translation on graph   Each of the graphs of y = −f (x), y = f (−x) and y = −f (−x) is sym- metric with respect to the graph of y = f (x) and the x-axis, symmetric with respect to the y-axis and symmetric with respect to the origin.   Q. For the function y = x2 − x, find the following functions. (1) A function whose graph is symmetric with respect to the x-axis. (2) A function whose graph is symmetric with respect to the y-axis. (3) A function whose graph is symmetric with respect to the origin.

§ 2 Various types of functions 95 Q. Find domains and range of the following functions and plot the graphs. √ √ √ (2) y = −x + 1 (3) y = − 2 − x + 1 (1) y = − x − 1 By letting C be a positive constant, y √ √ y=C x lets’think about the relationship between the Ca √ √ √ ×C y = x graph of y = x and each of the graphs of a √√ functions y = C x, y = Cx. √ Since the value for x = a of y = C x is the value multiplied by a factor C, the graph O ax is obtained by stretching or compressing the √ graph of y = x towards y-axis by a factor C. Also, since the value for x = a √√ of y = Cx is a , the graph is obtained C by stretching and compressing the graph of y √ y = Cx √ y = x towards x-axis by a factor 1 . √ C y= x √a The same things are also applied to a gen- × 1 eral function and the following relationships C O aa x are provided. C  Stretching and compression on graph  The graphs of y = Cf (x), y = f (Cx) are obtained by stretching and compressing the graph of y = f (x) in the following way. y = Cf (x) : towards y-axis by a factor C y = f (Cx) : towards x-axis by a factor 1 C   Q. How are the graphs of the following functions obtained by stretching and compressing the graphs of the functions in the brackets ( )? In addition, plot the graphs. √√ √√ (2) y = −2x (y = −x) (1) y = 3 x + 1 (y = x + 1)

96 Chapter 3 Functions and Graphs 2 4 Inverse function Consider a function y = x2 with a given do- y y = x2 main x ≧ 0. When the values of x correspond 9 (x ≧ 0) to various numbers such as x = 1, 2, 3, · · · , the obtained values of y will vary accordingly as y = 1, 4, 9, · · · . On the other hand, when find- 4 ing a value of x that satisfies y = x2 with respect to a single value of y within the range of y ≧ 0, 1 x x = ±√y O 123 Since x ≧ 0, Thus x = √y Therefore, the only possible value of x is derived. The corresponding equation is x = √y. Consider the domain of function y = f (x) as A and its range as B. When there is only one possible value of x within A satisfies y = f (x) with respect to the value of y within B, x is considered as a function of y. It is represented as x = g(y). Express an independent variable as x and a dependent variable as y, then a new function y = g(x) is obtained. The function y = g(x) is an inverse function of y = f (x). The inverse function y = g(x) can be obtained by switching x and y of the original function y = f (x) and express y with x. The following relationship is established. y = g(x) ⇐⇒ x = f (y) In addition, the domain of the original function is the range of the inverse function and the range of the original function is the domain of the inverse function.

§ 2 Various types of functions 97 Exercise Find the inverse function of the following function and plot a graph. y = x2 − 1 (x ≧ 0) Solution The domain and the range of this function is x ≧ 0 and y ≧ −1 respectively. Thus, the domain and the range of the reverse function are x ≧ −1 and y ≧ 0. y y = x2 − 1 y=x From the relationship above, the inverse function is x = y2 − 1 3 √ √ y= x+1 By solving y, y = ± x + 1 2 √ // −1 O 23 x Since y ≧ 0, y = x + 1 −1 and the graph is shown above. Q. Find the inverse functions of the following functions. In addition, find the domains and the ranges of the inverse functions. (1) y = −2x + 1 (1 ≦ x ≦ 3) (2) y= 2 +3 (1 ≦ x ≦ 2) (3) y = x2 + 1 (4 ≦ x ≦ 5) x You can see from the graph in exercise 4, the graphs of function y = f (x) and inverse function y = g(x) are symmetrical in relation to the straight line of y = x. Let’s prove this with general functions. y y=x Let a point on the graph of inverse B(b, a) function y = g(x) be A (a, b). Then, b = g(a) ⇐⇒ a = f (b) y = f (x) A(a, b) Therefore, the point B at the coordi- nates (b, a), is proved to be located on O x the function y = f (x). This can be y = g(x) said the other way around. The point A (a, b) and the point B (b, a) are sym- metrical in relation to the straight line of y = x, which means the two graphs are symmetrical in relation to the straight line y = x.

98 Chapter 3 Functions and Graphs [Column] function The original meaning of “function” is described as “the way in which some- thing works or operates”. Gottfried Wilhelm Leibnitz (1646 − 1716), Germen mathematician, was the first to employ the mathematical notion of a func- tion (taken from the Latin “functio”). A function is symbolized as y = f (x), whereas Leibnitz considered a function as follows - “When the value of a is placed into x, y is paired with the value of f (a)”. ax function y f (a) y = f (x) The function of pairing elements of the X Mapping f Y xy two sets is not limited to real numbers. For example, a map consists of pairing single area on a map with single point on a map. In this way, the function of pairing single element in one set with sin- gle element in another set is known as mapping. Let’s take an example. The y 2 mapping from the set of points (x, y) in −2 the plane to the set of points (u, v) in O the plane is defined as follows. −2 2x √ u = √21 x − 3 2 y v= 3 x + 1 y 2 2

§ 2 Various types of functions 99 Exercises 2–A 1. Determine whether the following functions are even or odd. (1) y= x (2) y = x4 − 5x2 x2 + 1 (4) y = |x| (3) y = x6 − 2x3 2. Plot graphs of the following functions. (1) y = (x + 1)3 − 2 (2) y = −(x − 1)4 + 2 (3) y = 1 − x (4) y = x √x + 1 2√− x (5) y = −3x + 6 + 1 (6) y = 2 x + 1 + 3 3. Find the range of the function y = x+2 in the domain −2 ≦ x ≦ 0. 2x − 1 4. When the graph of a function y = ax + b passes through the point (−1, 1) 2x + 1 and the one asymptote is y = 2, find the values of the constants a and b. √ 5. When the range of the function y = 2x − 6 (3 ≦ x ≦ a) is 0 ≦ y ≦ 2, find the value of the constant a. 6. Find the inverse of each of the following functions. (1) y = ax + b (a =\\ 0) (2) y = x2 − 2 (x ≦ 0) (3) y = a (a =\\ 0) (4) y = x − 1 x+b x+3 7. Find the inverse of a function y = (x − 1)2 + 3 (x ≧ 1). In addition, plot the graphs of the original function and the inverse function.