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1. Ambo University Department of Mathematics Lecture Note Applied Mathematics I November 11, 2016

2. Contents 1 Vectors and Vector Spaces 4 1.1 Scalars and Vectors in R2 and R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Addition and scalar multiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.3.1 Angle between two vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.3.2 Orthogonal projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.3.3 Direction angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 1.4 Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.5 Lines and planes in R3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1.6 Vector space; Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 1.7 Linear Dependence and independence; Basis of a vector space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 2 Matrices and determinants 18 2.1 Definition of matrix and basic operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 2.2 Product of matrices and some algebraic properties; Transpose of a matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 2.3 Elementary operations and its properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.4 Inverse of a matrix and its properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.5 Determinant of a matrix and its properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2.6 Solving system of linear equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.6.1 Cramer’s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.6.2 Gaussian method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.6.3 Inverse matrix method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.7 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 1

3. 3 Limit and continuity 31 3.1 Definition of limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 3.2 Basic limit theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 3.3 One sided limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 3.4 Infinite limits,limit at infinity and asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.4.1 Limits at infinity (negative infinity) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.4.2 Infinite limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 3.4.3 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.4.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 3.5 Intermediate value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 4 Derivatives and application of derivatives 39 4.1 Definition of derivatives; basic rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 4.2 Derivatives of inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.2.1 Inverse trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.2.2 Hyperbolic and inverse hyperbolic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 4.3 Higher order derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.4 Implicit differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 4.5 Application of derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.5.1 Extrema of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.5.2 Mean value theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 4.6 First and second derivative tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 4.6.1 Concavity and inflection point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 4.6.2 Curve sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 4.7 Velocity, acceleration and rate of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 4.8 Indeterminate Form(L’Hopital’s Rule) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 5 Integration 49 5.1 Antiderivatives, indefinite integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 5.2 Techniques of Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 5.2.1 Integration by substitution, by parts and by partial fraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 5.2.2 Trigonometric integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 5.2.3 Integration by trigonometric substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 5.3 Definite integral, Fundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 5.4 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Ambo University DEPARTMENT OF MATHEMATICS 2

4. 6 Application of integrals 66 6.1 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 6.2 Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 6.2.1 Volumes by slicing; Disks and Washers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 6.3 Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 6.4 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Ambo University DEPARTMENT OF MATHEMATICS 3

5. Chapter 1 Vectors and Vector Spaces Introduction In Engineering applications we use two kinds of quantities; scalars and vectors. • A scalar is a quantity that is determined by its magnitude. Example: length, area, temperature, mass,...etc • A vector is a quantity that is determined by both its magnitude and direction. Example: Velocity, acceleration, force,...etc The concept of a vector is basic for the study of functions of several variables. It provides geometric motivation for everything that follows. 1.1 Scalars and Vectors in R2 and R3 Definition of points in space We know that a number x can be used to represent point on a line. A pair of numbers (x, y) can be used to represent a point in the plane. A triple numbers (x, y, z) can be used to represent a point in space. Definition 1.1.1. If n is positive integer, then an ordered n-tuples is a sequence of n real numbers (x1, x2, ..., xn). The set of all ordered n-tuples is called n-space and is denoted by Rn . Vectors in R2 and R3 Every pair of distinct points A and B in R2 and R3 determines a directed line segment with initial point at A and terminal point at B. We call such a directed line segment is a vector and denoted by − − → AB. A B initial point terminal point − − → AB The length of the line segment is the magnitude of the vector and the arrow indicates its direction. We denote a vector by printing a letter in boldface (v) or by putting an arrow above the letter (− → v ). Definition 1.1.2. A position vector is a vector whose initial point is at the origin. Definition 1.1.3. Two vectors u and v in R2 or R3 are said to be equal if they have the same magnitude and direction, and denoted by u = v. 4

6. (v1, v2) v v = (v1, v2) − → u − → v − → u = − → v Definition 1.1.4 (Component form). If v is a two-dimensional vector in the plane with initial point P(x1, y1) and terminal point Q(x2, y2) then the component form of v is v = → PQ = (x2 − x1, y2 − y1). If v is a three-dimensional vector in space with initial point P(x1, y1, z1) and terminal point Q(x2, y2, z2) then the component form of v is v = → PQ = (x2 − x1, y2 − y1, z2 − z1). The zero vector is the vector 0 = (0, 0) or 0 = (0, 0, 0) in R2 and R3 respectively. Example 1.1.1. Find the component form of a vector with initial point P(−3, −2, −1) and terminal point Q(1, −2, 3). Solution. The component form of − − → PQ is − − → PQ = (1 − (−3), −2 − (−2), 3 − (−1)) = (4, 0, 4) J 1.2 Addition and scalar multiplication Definition 1.2.1. Let u = (u1, u2) and v = (v1, v2) be vectors in R2 and let c be a real number. Then the sum of u and v is defined as the vector u + v = (u1 + v1, u2 + v2) and the the scalar multiple of u by c is defined as the vector cu = (cu1, cu2) Definition 1.2.2 (Parallel Vectors). Two non-zero vectors u and v are said to be parallel if they are scalar multiples of one another. In other word, the two vectors u and v are said to be parallel, denoted by u k v if there exists scalar c such that u = cv. − → u − → v 2− → u −− → v Triangle law(The head to tail rule): Given vectors u and v in R2 , translate v so that its tail coincides with the head of u. The sum u + v of u and v is the vector from the tail of u to the head of v. Parallelogram rule: The sum of two vectors in the plane can be repre- sented by the diagonal of a parallelogram having u and v as its adjacent sides, as shown in Figure below. Ambo University DEPARTMENT OF MATHEMATICS 5

7. The head-to-tail rule C u v w Parallelogram rule u w = u + v v Example 1.2.1. Find the sum of the vectors. a. u = (4, 3), v = (2, −1) b. u = (3, −2), v = (2, −3) Theorem 1.2.1 (Properties of Vector Addition and Scalar Multiplication in the Plane). Let u, v and w be vectors in R2 , and let c and d be scalars. 1. u + v ∈ R2 . Closure under addition 2. u + v=v + u. Commutative property of addition 3. u + (v + w) = (u + v) + w. Associative property of addition 4. u + 0 = u. Additive identity property 5. u + (−u) = 0. Additive inverse property 6. cu ∈ R2 . Closure under scalar multiplication 7. c(u + v) = cu + cv.Distributive property 8. (c + d)u = cu + du. Distributive property 9. c(du) = (cd)u. Associative property of multiplication 10. 1(u) = u. Multiplicative identity property The zero vector 0 in R2 is called the additive identity in R2 . Similarly, the vector −v is called the additive inverse of v. Theorem 1.2.2 (Properties of Additive Identity and Additive Inverse). Let v be vector in R2 , and let c be a scalar. Then the following properties are true. 1. The additive identity is unique. That is, if u + v=v, then u = 0. 2. The additive inverse of v is unique.That is,if u + v = 0, then u = −v. 3. 0v = 0. 4. c0 = 0. 5. If cv = 0, then c = 0 or v = 0. 6. −(−v) = v. Standard basis vectors Let i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1). The vectors i, j, k are called standard basis vectors. They have length 1 and points in the direction of the positive x, y, and z-axes respectively. Similarly in two dimensions we define i = (1, 0) and j = (0, 1). If a = (a1, a2, a3), then we can write a = (a1, a2, a3) = (a1, 0, 0) + (0, a2, 0) + (0, 0, a3) = a1(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1) = a1i + a2j + a3k Thus, any vector in space can be expressed in terms of i, j, and k. Example 1.2.2. If a = i+2j −3k and b = 4i+7k, express the vector 2a+3b in terms of i, j and k Ambo University DEPARTMENT OF MATHEMATICS 6

8. i = (1, 0) j = (0, 1) i j k x y z Solution. 2a + 3b = 2(i + 2j − 3k) + 3(4i + 7k) = 2i + 4j − 6k + 12i + 21k = 14i + 4j + 15k J 1.3 Scalar product Definition 1.3.1. Let u = (u1, u2, u3) be a vector in R3 . Then the magnitude (norm) of u, denoted by kuk is defined by: kuk = q u2 1 + u2 2 + u2 3 Similarly, for a vector v = (v1, v2) ∈ R2 , its norm is given by kvk = q v2 1 + v2 2 Example 1.3.1. a. If v = (−1, 4, 3), then find kvk. b. If kuk = 6, find x such that u = (−1, x, 5). Theorem 1.3.1. If v is a vector in R2 or R3 , then: i. kvk ≥ 0 ii. kvk = 0 if and only if v = 0 iii. kvk = k−vk iv. If c ∈ R, then kcvk = |c|kvk. Definition 1.3.2 (Unit vector). Any vector v satisfying kvk = 1 is called a unit vector. Example 1.3.2. The vectors (0, 1),(−1, 0),( 1 √ 2 , −1 √ 2 ),(1, 0, 0) are examples of unit vectors. Theorem 1.3.2. For any non-zero vector v the unit vector u corresponding to v in the direction of v can be obtained as: u = v kvk . Example 1.3.3. Let a = (1, 1, 1). Then find the corresponding unit vector. Solution. The unit vector u in the direction of a is u = a kak = (1, 1, 1) √ 3 = ( 1 √ 3 , 1 √ 3 . 1 √ 3 ) J Exercise 1.3.3. Find the unit vector in the direction of a. −3i + 7j b. 8i − j + 4k Definition 1.3.3. If u and v are points in R2 or R3 , then we denote the distance between u and v by d(u, v) and defined it to be d(u, v) = ku − vk The dot product Definition 1.3.4. If a = (a1, a2, a3) and b = (b1, b2, b3), then the dot product of a and b is the number a · b given by a · b = a1b1 + a2b2 + a3b3 The dot product of two dimensional vectors is defined in a similar fashion: (a1, a2) · (b1, b2) = a1b1 + a2b2 Ambo University DEPARTMENT OF MATHEMATICS 7

9. Theorem 1.3.4. The dot product of two non-zero vectors a and b is the number a · b = kakkbkcosθ where θ is the angle between a and b, 0 ≤ θ ≤ π. If either a or b is 0, we define a · b = 0 Remark 1.3.1. The dot product of two vectors is a scalar quantity, and its value is maximum when θ = 0 and minimum if θ = π. Example 1.3.4. If u = (1, −2, 3) and v = (0, 1, −5), then find u · v, u · u and (u + v) · v. Solution. u·v = (1, −2, 3)·(0, 1, −5) = 1(0)+(−2)(1)+3(−5) = −17, u·u = u2 = |u|2 = 14 J Properties of Dot Product If u, v and w are vectors with the same dimensions and c ∈ R, then 1. u · u = kuk2 2. u · v = v · u 3. u · (v + w) = u · v + u · w 4. (cu) · v = c(u · v) = u · (cv) 5. 0 · u = 0 1.3.1 Angle between two vectors If θ is the angle between two non-zero vectors u and v, then the angle between the two vectors can be obtained by: cos θ = u · v kukkvk ⇒ θ = cos−1 u · v kukkvk Theorem 1.3.5 (Cauchy-schwarz inequality). If u and v are vectors in R2 or R3 , then |u · v| ≤ kukkvk Proof. Exercise Theorem 1.3.6 (Triangle inequality). If u ,v and w are vectors in R2 or R3 , then: i. ku + vk ≤ kuk + kvk ii. d(u, v) ≤ d(u, w) + d(w, v) Proof. Exercise Theorem 1.3.7 (Parallelogram Equation for vectors). If u and v are vectors in R2 or R3 , then ku + vk2 + ku − vk2 = 2(kuk2 + kvk2 ) Proof. Exercise Theorem 1.3.8. If u and v are vectors in R2 or R3 with Euclidean inner product, then |u · v| = 1 4 ku + vk2 − 1 4 ku − vk2 Proof. Exercise Definition 1.3.5. Two non-zero vectors u and v are said to be orthogonal (perpendicular) denoted by u ⊥ v if and only if u · v = 0, i.e, if θ = π 2 . Example 1.3.5. show that 2i + 2j − k is perpendicular to 5i − 4j + 2k Solution. Since 2i + 2j − k · 5i − 4j + 2k = 10 − 8 − 2 = 0, we see that the vectors are perpendicular. J Example 1.3.6. Find the angle between a and b, where a = (1, 1, −1) and b = (1, 0, 0) Solution. cos θ = a·b |a||b| = 1 √ 3(1) = √ 3 3 ⇒ θ = cos−1 ( √ 3 3 ) = π 3 J Theorem 1.3.9 (Pythagoras theorem). If a and b are orthogonal vectors, then k a + b k2 = kak2 + kbk2 . Ambo University DEPARTMENT OF MATHEMATICS 8

10. Proof. Suppose a ⊥ b, then ka + bk2 = (a + b) · (a + b) = a · a + a · b + b · a + b · b = kak2 + 2a · b + kbk2 = kak2 + kbk2 , since a · b Theorem 1.3.10. Given two vectors u and v in space, ku + vk = ku − vk if and only if u and v are orthogonal vectors. 1.3.2 Orthogonal projection Definition 1.3.6. Let a = → PQ and b = → PR be two vectors. If S is the foot of the perpendicular from R to the line containing P and Q, then the vector from P to S is called the vector projection of b onto a and is denoted by projb a . P θ Q R S Projb a a b B C The scalar projection of b onto a (also called the component of b along a) is defined to be the length of projb a , which is equal to kbk cos θ and is denoted by compb a . Thus, compb a = a·b kak and projb a = a·b kak2 a Example 1.3.7. Find the scalar projection and Vector projection of b = (1, 1, 2) onto a = (−2, 3, 1). Solution. Since kak = p (−2)2 + 32 + 12 = √ 14, the scalar projection of b onto a is compb a = a · b kak = −2 + 3 + 2 √ 14 = 3 √ 14 The vector projection is this scalar projection times the unit vector in the direction of a. projb a = 3 √ 14 a kak = 3 14 (−2, 3, 1) = ( −3 14 , 9 14 , 3 14 ) J Remark 1.3.2. The vector projection of b onto a is the scalar projection times the unit vector in the direction of a Exercise 1.3.11. Let a = (−1, 3, 1) and b = (2, 4, 3). Then find projb a , proja b and compb a 1.3.3 Direction angles Let a = a1i + a2j + a3k be a vector positioned at the origin in R3 , making an angle of α, β and γ with the positive x, y and z axes respectively.Then the angles α, β and γ are called the directional angles of a and the quantities cos α, cos β and cos γ are called directional cosines of a, which can be computed as follows: cos α = a1 kak , α ∈ [0, π] cos β = a2 kak , β ∈ [0, π] cos γ = a3 kak , γ ∈ [0, π] Remark 1.3.3. cos2 α + cos2 β + cos2 γ = 1 Exercise 1.3.12. Let a = (−1, 2, 2). Then find the directional cosines of a. Ambo University DEPARTMENT OF MATHEMATICS 9

11. α β γ x y z a B C 1.4 Cross product Definition 1.4.1. Suppose that a = (a1, a2, a3) = a1i + a2j + a3k and b = (b1, b2, b3) = b1i + b2j + b3k be two vectors on R3 . Then the cross product a × b of the two vectors is defined as: a × b = (a2b3 − a3b2)i + (a3b1 − a1b3)j + (a1b2 − a2b1)k =

17. i j k a1 a2 a3 b1 b2 b3

23. = i

27. a2 a3 b2 b3

31. − j

35. a1 a3 b1 b3

39. + k

43. a1 a2 b1 b2

47. Example 1.4.1. Let a = 4i − 3j + 2k and b = 2i − 5j − k. Then find a. a × b b. b × a Solution. a × b =

53. i j k 4 −3 2 2 −5 −1

59. = i(3 + 10) − j(−4 − 4) + k(−20 + 6) = 13i + 8j − 14k b × a =

65. i j k 2 −5 −1 4 −3 2

71. = i(−10 − 3) − j(4 + 4) + k(−6 + 20) = −13i − 8j + 14k J Remark 1.4.1. For two non-zero vectors a and b, i. a × b is a vector which is orthogonal to both a and b ii. a × b is not defined for a, b ∈ R2 iii. i × j = −j × i = k j × k = −(k × j) = i k × i = −(i × k) = j Theorem 1.4.1 (Properties of Cross Product). Let a, b and c be vectors in R3 and α be any scalar. Then 1. a × 0 = 0 × a = 0, where 0 = (0, 0, 0) 2. a × b = −b × a 3. a × (b × c) 6= (a × b) × c 4. (αa) × b = a × (αb) = α(a × b) 5. a × (b + c) = a × b + a × c 6. a · (a × b) = b · (a × b) = 0 7. If a and b are parallel, then a × b = 0 8. ka × bk = kakkbk sin θ, θ ∈ [0, π] 9. ka × bk2 = kak2 kbk2 − (a · b)2 Example 1.4.2. If kak = 2, kbk = 4 and θ = π 4 for two vectors a and b then find ka × bk Solution. ka × bk = kakkbk sin θ = 2 × 4 × sin π 4 = 4 √ 2 J The angle θ between a and b can be obtained by sin θ = ka×bk kakkbk , for two non-zero vectors a and b. Definition 1.4.2 (Scalar Triple Product). Let a, b and c be vectors in R3 . Their scalar triple product is given by a · (b × c), which is a scalar. Ambo University DEPARTMENT OF MATHEMATICS 10

72. Applications of cross product 1. Area: The area of a parallelogram whose adjacent sides coincides with the vectors a and b is given by ka × bk = kakkbk sin θ θ |b| sin θ = h a b B C Note 1. The area of the triangle formed by a and b as its adjacent sides is given by area=1 2 ka × bk 2. Volume: The volume V of a parallelepiped with the three vectors a, b and c in R3 as three of its adjacent edges is given by: V = |a · (b × c)| =

78. det   a1 a2 a3 b1 b2 b3 c1 c2 c3  

84. Example 1.4.3. Find the area of a triangle whose vertices are A(1, −1, 0), B(2, 1, −1) and C(−1, 1, 2). Solution. The vectors on the sides of the triangle ∆ABC are − − → AB = (1, 2, −1) and − → AC = (−2, 2, 2) Then the area of the triangle ABC is A = 1 2 k − − → AB × − → ACk = 3 √ 2 square units J Example 1.4.4. Find the volume of the parallelepiped with edges u = i + k, v = 2i + j + 4k, and w = j + k. Solution. V = |u · (v × w)| = 1 J If the volume of the parallelepiped determined by a, b and c is 0, then the vectors must lie in the same plane; that is, they are Coplanar. Example 1.4.5. Use the scalar triple product to show that the vectors a = (1, 4, −7), b = (2, −1, 4) and c = (0, −9, 18) are coplanar. Solution. The scalar triple product V = |a · (b × c)| =

90. det   1 4 −7 2 −1 4 0 −9 18  

96. = 0 ∴ a, b and c are coplanar. J 1.5 Lines and planes in R3 Lines We know that in a plane(R2 ) a line is determined by a point on a line and slope of the line. i.e. y − y0 = m(x − x0) But in three dimensional space (R3 ) a line L is determined by a point P0(x0, y0, z0) and a vector v giving the direction of the line. Let P(x, y, z) be an arbitrary point on L and let r0 and r be the position vectors of P0 and P respectively. If a = → P0P, then r = r0 + a. But since a and v are parallel vectors, there is a scalar t such that a = tv. Thus r = r0 + tv——vector equation of L. If v = (a, b, c), then tv = (ta, tb, tc). Let r = (x, y, z) and r0 = (x0, y0, z0). Then the vector equation becomes (x, y, z) = (x0, y0, z0) + (ta, tb, tc) = (x0 + ta, y0 + tb, z0 + tc), t ∈ R2 This leads to x = x0+ta, y = y0+tb, z = z0+tc——–this is called parametric equation of a line Example 1.5.1. a. Find the vector equation and parametric equation for the line that passes through the point (5, 1, 3) and is parallel to the vector i + 4j − 2k Ambo University DEPARTMENT OF MATHEMATICS 11

97. x y z P0(x0, y0, z0) P(x, y, z) r0 r r − r0 v b. Find two other points on the line Solution. a. Here r0 = (5, 1, 3) = 5i + j + 3k and v = i + 4j − 2k. The vector equation is r = r0 + tv = 5i + j + 3k + t(i + 4j − 2k) = (5 + t)i + (1 + 4t)j + (3 − 2t)k Parametric equations are x = 5 + t, y = 1 + 4t, z = 3 − 2t b. Choosing t = 1 and t = 2 gives the two points (6, 5, 1) and 7, 9, −1 J Symmetric Equation The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. If a vector v = (a, b, c) is used to describe the direction of a line L, then a, b and c are called direction numbers of L. If none of a, b or c is 0, then we can solve each of the three parametric equations for t, equate the results and obtain the symmetric equations. x − x0 a = y − y0 b = z − z0 c Example 1.5.2. a. Find parametric equations and symmetric equations of the line that passes through the points A(2, 4, −3) and B(3, −1, 1). b. At what point does this line intersect the xy-plane? Solution. a. The vector − − → AB = (1, −5, 4) is parallel to the line L. If P0 = (2, 4, −3), then the parametric equations are x = 2 + t, y = 4 − 5t, z = −3 + 4t The symmetric equations are x − 2 1 = y − 4 −5 = z + 3 4 b. z = 0 ⇒ x−2 1 = y−4 −5 = 3 4 ⇒ x = 11 4 and y = 1 4 J Example 1.5.3. Show that the lines L1 and L2 with parametric equations x = 1 + t, y = −2 + 3t, z = 4 − t and x = 2s, y = 3 + s, z = −3 + 4s are skew lines. that is, the do not intersect and are not parallel. Solution. The lines are not parallel because their direction vectors are not parallel. If L1 and L2 had a point of intersection, then there would be values of t and s such that      1 + t = 2s −2 + 3t = 3 + s 4 − t = −3 + 4s. ( 1 + t = 2s −2 + 3t = 3 + s ⇒ t = 11 5 and s = 8 5 , but these values do not satisfy the third equation. J Ambo University DEPARTMENT OF MATHEMATICS 12

98. we know that the vector equation of a line through the vector r0 in the direction of a vector v is r = r0 + tv. If the line also passes through r1, then we can take v = r1 − r0 and so its vector equation is r = r0 + t(r1 − r0) = (1 − t)r0 + tr1 The line segment from r0 to r1 is given by the vector equation r(t) = (1 − t)r0 + tr1, 0 ≤ t ≤ 1 Planes A plane in space is determined by a point P0(x0, y0, z0) in the plane and a vector n that is orthogonal (perpendicular) to the plane. This orthogonal vector n is called normal vector. Let P(x, y, z) be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 and P. The normal vector n is orthogonal to every x y z P(x, y, z) P0(x0, y0, z0) r r0 r − r0 n vector in the given plane. In particular, n is orthogonal to r − r0. Thus n · (r − r0) = 0 or n · r = n · r0. This equation is called a vector equation of the plane. If n = (a, b, c), r = (x, y, z), and r0 = (x0, y0, z0), then (a, b, c) · (x − x0, y − y0, z − z0) = 0 ⇒a(x − x0) + b(y − y0) + c(z − z0) = 0 ⇒ax + by + cz + d = 0, where d = −(ax0 + by0 + cz0) This equation is called the equation of the plane through the point (x0, y0, z0) with normal vector n = (a, b, c) Example 1.5.4. Find an equation of the plane through the point (2, 4, −1) with normal vector n = (2, 3, 4). Find the intercepts and sketch the plane. Solution. Put P0 = (2, 4, −1) and n = (2, 3, 4) Hence an equation of the plane is 2(x − 2) + 3(y − 4) + 4(z + 1) = 0 =⇒ 2x + 3y + 4z = 12 the x-intercept y = z = 0 ⇒ 2x = 12 ⇒ x = 6 the y-intercept is y = 4. the z-intercept is z = 3 J Example 1.5.5. Find an equation of the plane that passes through the points P(1, 3, 2), Q(3, −1, 6) and R(5, 2, 0). Solution. Let a = − − → PQ = (2, −4, 4) and b = − → PR = (4, −1, −2). since both a and b lie in the plane, their cross product a × b is orthogonal to the plane can be taken as the normal vector. Thus n = a × b =

104. i j k 2 −4 4 4 −1 −2

110. = 12i + 20j + 14k The equation is 12(x − 1) + 20(y − 3) + 14(z − 2) = 0 ⇒ 12x + 20y + 14z = 100 ⇒ 6x + 10y + 7z = 50 J Ambo University DEPARTMENT OF MATHEMATICS 13

111. x y z (0, 4, 0) (0, 0, 3) (6, 0, 0) Angle between Planes Two plane are parallel if their normal vectors are parallel. If two planes are not parallel, then they intersect in a straight line and the angle between the planes is defined as the acute angle between their normal vectors. Example 1.5.6. Find the angle of intersection between the planes x+y+z = 1 and x − 2y + 3z = 1. Find symmetric equations for the line of intersection of the plane. Solution. The normal vectors are n1 = (1, 1, 1), n2 = (1, −2, 3) Thus if θ is the angle between the planes, then cos θ = n1 · n2 kn1kkn2k = 1 − 2 + 3 √ 3 √ 14 = 2 √ 42 ∴ θ = cos−1 ( 2 √ 42 ) = 72◦ To find a point on L take z = 0. Then ( x + y = 1 x − 2y = 1 ⇒ y = 0 and x = 1 So the point (1, 0, 0) lies on L. Since L lies on both planes, it is perpendicular to both of the normal vectors. Thus a vector v parallel to L is given by v = n1 × n2 =

117. i j k 1 1 1 1 −2 3

123. = 5i − 2j − 3k So the symmetric equation of L can be written x − 1 5 = y −2 = z −3 J Distance in Space a. Distance from a point to a line: The distance D from a point P1 (not on L) to a line L in space is given by D = kv × → P0P1k kvk , where v is the directional vector of L and P0 is any point on L Proof. execise b. Distance from a point to a plane: The perpendicular distance D of a point P1(x1, y1, z1) in space to the plane with the equation ax+by +cz + d = 0 is given by: D = |ax1 + by1 + cz1 + d| √ a2 + b2 + c2 = |n · → OP| knk where O is the foot of n within the plane. Ambo University DEPARTMENT OF MATHEMATICS 14

124. Proof. Exercise Exercise 1.5.1. 1. Find the distance of the point P1(−1, 3, 0) from the line with parametric equation L : x = 1, y = 1 + 3t, z = −1 + 2t 2. Find the distance from the point P(1, 2, 3) to the plane π : 3x + 5y − 4z + 37 = 0 3. Find the distance between the planes π1 : x + 2y − 2z = 3 and π2 : 2x + 4y + −4z = 7 c. Distance between two parallel planes: Given two parallel planes π1 : ax+by+cz = d1 and π2 : ax+by+cz = d2. Then the distance between π1 and π2 is the same as the distance from any arbitrary point P(x0, y0, z0) that has been taken from π1 to the plane π2 and is given by D = |ax0 + by0 + cz0 − d2| √ a2 + b2 + c2 = |d1 − d2| √ a2 + b2 + c2 Example 1.5.7. Find the distance between the parallel planes π1 : 10x + 2y − 2z = 5 and π2 : 5x + y − z = 1 Solution. The planes are parallel because their normal vectors (10, 2, −2) and (5, 1, −1) are parallel. The distance between π1 and π2 is the distance between any point in π1 and the plane π2. Let y = z = 0, then x = 1 2 . Hence (1 2 , 0, 0) lies in π1. Now the distance between the point (1 2 , 0, 0) and the plane π2 is D = |5(1 2 ) + 1(0) − 1(0) − 1| p 52 + 12 + (−1)2 = √ 3 6 ∴ the distance between the planes is √ 3 6 . J 1.6 Vector space; Subspaces Definition 1.6.1. Let V be a set on which two operations, called addition and scalar multiplication, have been defined. If u and v are in V , the sum of u and v is denoted by u + v, and if c is a scalar, the scalar multiple of u by c is denoted by cu. If the following axioms hold for all u, v and w in V and for all scalars c and d, then V is called a vector space and its elements are called vectors. 1. u + v is in V . Closure under addition 2. u + v=v + u. Commutativity 3. u + (v + w) = (u + v) + w. Associativity 4. There exists an element 0 in V , called a zero vector, such that u+0 = u. 5. For each u in V , there is an element −u in V such that u + (−u) = 0. 6. cu is in V . Closure under scalar multiplication 7. c(u + v) = cu + cv.Distributivity 8. (c + d)u = cu + du. Distributivity 9. c(du) = (cd)u. 10. 1(u) = u. Example 1.6.1. For any n ≥ 1, Rn is vector space over R Example 1.6.2. The set defined by S = {(x, y, z) : x, y, z ∈ Q} is not a vector space over R because if we take c = √ 2 ∈ R and u = (1, 3, 0) ∈ S, then we can see that cu is not in S. Subspace Definition 1.6.2. A non-empty subset W of a vector space V is called a subspace of V if W is itself a vector space with the same scalars, addition and scalar multiplication as V . Theorem 1.6.1. Let V be a vector space and let W be a non-empty subset of V . Then W is a subspace of V if and only if the following conditions hold: a. If u and v are in W, then u + v is in W. b. If u is in W and c is scalar, then cu is in W. Ambo University DEPARTMENT OF MATHEMATICS 15

125. Example 1.6.3. V and {0} are the trivial subspaces of any vector space V . 1. V and {0} are the trivial subspaces of any vector space V 2. For the vector space V = R3 = {(x, y, z); x, y, z ∈ R3 } over R. Then the set W = {(x, y, 0); x, y ∈ R} is a subspace of V . (verify!) 3. The set of all lines passing through the origin, L = {ax+by = 0, a, b ∈ R} is a subspace of the vector space V = R2 . Exercise 1.6.2. Is the set W = {x − 4y = 1} a subspace of V = R2 ? Justify. 1.7 Linear Dependence and independence; Ba- sis of a vector space Definition 1.7.1. A vector u in a vector space V is called a linear combination of the vectors v1, v2, . . . , vn in V when u can be written in the form u = α1v1 + α2v2 + · · · + αnvn where, α1, α2, . . . , αn are scalars. Definition 1.7.2. A set of vectors {v1, v2, . . . , vn} in a vector space V is linearly dependant(LD) if there are α1, α3, . . . , αn, atleast one of which is not zero, such that α1v1 + α2v2 + · · · + αnvn = 0 A set of vectors that is not linearly dependant is said to be linearly independent(LI). A set of vectors {v1, v2, . . . , vn} in vector space V is linearly independent if α1v1 + α2v2 + · · · + αnvn = 0 implies α1 = 0, α2 = 0, . . . , αn = 0 Example 1.7.1. Determine weather the following set of vectors in Vector space V = R3 are linearly dependant or independent i. {(1, 0, 0), (0, 1, 0), (0, 0, 3)} ii. {(2, 6, 0), (2, 4, 1), (1, 1, 1)} Theorem 1.7.1. Let S = {v1, v2, . . . , vr} be a set of vectors in Rn . If r n, then S is linearly dependent. Definition 1.7.3. If f1 = f1(x), f2 = f2(x), . . . , fn = fn(x) are functions that are n − 1 times differentiable on the interval (−∞, ∞) , then the determinant W(x) =

134. f1 f2 · · · fn f0 1 f0 2 · · · f0 n . . . . . . . . . fn−1 1 fn−1 2 · · · fn−1 n

143. is called the Wronskian of f1, f2, . . . , fn. Theorem 1.7.2. If the functions f1, f2, . . . , fn have n − 1 continuous derivatives on the interval (−∞, ∞) and if the Wronskian of these functions is not identically zero on (−∞, ∞) , then these functions form a linearly independent set of vectors in C(n−1) (−∞, ∞). Example 1.7.2. Let V be the vector space of all real valued functions of the variable t. Then which of the following set of functions are LD/LI? Justify! a. {t, t2 , sint} b. {cos2 t, sin2 t, 1} Theorem 1.7.3. A set of vectors {v1, v2, . . . , vn} in a vector space V is linearly dependent if and only if atleast one of the vectors can be expressed as a linear combination of the others. Proof. Exercise Basis of a vector space Definition 1.7.4. If S = {v1, v2, . . . , vn} is a set of vectors in a vector space V , then the set of all linear combinations of v1, v2, . . . , vn is called the span of v1, v2, . . . , vn and is denoted by span(v1, v2, . . . , vn) or span(S). If V = span(S), then S is called a spanning set for V and V is said to be spanned by S. Definition 1.7.5. A subset B of a vector space V is a basis for V if 1. B spans V and 2. B is linearly independent. Ambo University DEPARTMENT OF MATHEMATICS 16

144. Definition 1.7.6. A vector space V is called finite dimensional if it has a basis consisting of finitely many vectors. The dimension of V , denoted by dimV , is the number of vectors in a basis for V . The dimension of the zero vector space {0} is defined to be zero. A vector space that has no finite basis is called infinite dimensional. Example 1.7.3. 1. Show that the set S = {(1, 0, 0), (0, 1, 0), (0, 0, 5)} form a basis of the vector space R3 . 2. Determine whether the set S = {(0, 1), (1, 0), (2, 5)} is a basis of R2 . Theorem 1.7.4. If a vector space V has one basis with n vectors, then every basis for V has n vectors. Theorem 1.7.5. Let B = v1, v2, . . . , vn be a basis for a vector space V . a. Any set of more than n vectors in V must be linearly dependent. b. Any set of fewer than n vectors in V cannot span V . Theorem 1.7.6. Let V be a vector space of dimension n. 1. If S = {v1, v2, . . . , vn} is a linearly independent set of vectors in V , then S is a basis for V . 2. If S = {v1, v2, . . . , vn} spans V , then S is a basis for V . Ambo University DEPARTMENT OF MATHEMATICS 17

145. Chapter 2 Matrices and determinants 2.1 Definition of matrix and basic operations Definition 2.1.1. An m×n matrix A is a rectangular array of numbers, real or complex, with m rows and n columns. The following are all examples of matrices: 2 3 1 4 , √ 5 −2 1 π 3 i , [2], 1 1 1 1 ,   2 1 3   The size of a matrix is a description of the numbers of rows and columns it has. A matrix is called m × n (pronounced m by n) if it has m rows and n columns. A 1×m matrix is called a row matrix (or row vector), and an n×1 matrix is called a column matrix (or column vector). A general m × n matrix A has the form A =      a11 a12 . . . a1n a21 a22 . . . a2n . . . . . . ... . . . am1 am2 . . . amn      The diagonal entries of A are a11, a22, a33, . . . , and if m = n (that is, if A has the same number of rows as columns), then A is called a square matrix. A square matrix whose nondiagonal entries are all zero is called a diagonal matrix. A diagonal matrix all of whose diagonal entries are the same is called a scalar matrix. If the scalar on the diagonal is 1, the scalar matrix is called an identity matrix. For example let A = 2 4 5 0 3 4 , B = 3 1 4 5 , C =   2 0 0 0 6 0 0 0 4   , D =   1 0 0 0 1 0 0 0 1   The diagonal entries of A are 2 and 3, but A is not square; B is a square matrix of size 2×2 with diagonal entries 3 and 5; C is a diagonal matrix; D is a 3×3 identity matrix. The n×n identity matrix is denoted by In (or simply I if its size is understood). Remark 2.1.1. Two matrices are equal if they have the same size and if their corresponding entries are equal. Thus, if A = [aij]m×n and B = [bij]r×s, then A = B if and only if m = r and n = s and aij = bij for all i and j. Matrix Addition and scalar Multiplication If A = [aij] and B = [bij] are m × n matrices, their sum A + B is the m × n matrix obtained A + B = aij + bij Example 2.1.1. Let A = 1 4 0 −2 6 5 , B = 3 1 −1 3 0 2 and C = 4 3 2 1 18

146. Then A + B = −2 5 −1 1 6 7 but neither A + C nor B + C is defined. If A is an m × n matrix and c is a scalar, then the scalar multiple cA is the m × n matrix obtained by multiplying each entry of A by c. cA = c[aij] = [caij] Example 2.1.2. For matrix A in Example 2.1.1 2A = 2 8 0 −4 12 10 , 1 2 A = 1 2 2 0 −1 3 5 2 , (−1)A = −1 −4 0 2 −6 −5 The matrix (−1)A is written as −A and called the negative of A. As with vectors, we can use this fact to define the difference of two matrices: If A and B are the same size, then A − B = A + (−B) 2.2 Product of matrices and some algebraic properties; Transpose of a matrix Matrix Multiplication Definition 2.2.1. If A is an m × n matrix and B is an n × r matrix, then the product C = AB is an m × r matrix. The (i, j) entry of the product is computed as follows: ai1 ai2 . . . air b1j b2j . . . brj cij A (n × r) B (r × m) C = AB (n × m) row i column j cij = ai1b1j + ai2b2j + · · · + ainbnj Note 2. For AB to exist, the number of columns of A must equal the number of rows of B. A B m × n n × r same size of AB = AB m × r Remark 2.2.1. If A is an m × n matrix and B is an n × r matrix, then AB will be an m × r matrix. Example 2.2.1. Let A = 1 3 2 0 and B = 5 0 1 3 −2 6 . Determine AB and BA, if the product exists. Solution. A has two columns and B has two rows; thus AB exists. Interpret A in terms of its rows and B in terms of its columns and multiply the rows Ambo University DEPARTMENT OF MATHEMATICS 19

147. by the columns. We find that AB = 1 3 2 0 5 0 1 3 −2 6 = 14 −6 19 10 0 2 BA does not exist because B has three columns and A has two rows. We see that the order in which two matrices are multiplied is important. Un- like multiplication of real numbers, matrix multiplication is not commutative. In general, for two matrices A and B, AB 6= BA. J Matrix Multiplication in Terms of Columns Consider the product AB where A is an m × n matrix and B is an n × r matrix(so that AB exists). Let the columns of B be the matrices B1, B2, . . . , Br. Write B as B1 B2 . . . Br . Thus AB = A B1 B2 . . . Br Matrix multiplication implies that the columns of the product are AB1, AB2, . . . , ABr. We can write AB = AB1 AB2 . . . ABr For example, suppose A = 2 0 1 5 and B = 4 1 3 0 2 −1 . Then AB = 2 0 1 5 4 0 2 0 1 5 1 2 2 0 1 5 3 −1 = 8 2 6 4 11 −2 Theorem 2.2.1. Let A, B and C be matrices and r and s be scalars. Assume that the sizes of the matrices are such that the operations can be performed. Properties of Matrix Addition and Scalar Multiplication 1. A + B = B + A 2. A + (B + C) = (A + B) + C 3. A + 0 = 0 + A = A (where 0 is the appropriate zero matrix) 4. r(A + B) = rA + rB 5. (r + s)C = rC + sC 6. r(sC) = (rs)C Properties of Matrix Multiplication 1. A(BC) = (AB)C 2. A(B + C) = AB + AC 3. (A + B)C = AC + BC 4. AI = IA = A (where I is the appropriate identity matrix) 5. r(AB) = (rA)B = A(rB) Note 3. AB 6= BA in general. Multiplication of matrices is not commutative. Example 2.2.2. Compute the product ABC of the following three matrices. A = 1 2 3 −1 , B = 0 1 3 −1 0 −2 , C =   4 −1 0   Solution. Let us check to see if the product ABC exists before we start spending time multiplying matrices. We get The product exists and will be a A 2×2 B 2×3 C 3×1 ABC 2×1 = size of product is 2 × 1 match match 2 × 1 matrix. Since matrix multiplication is associative, the matrices in the product ABC can be grouped together in any manner for multiplying, as long as the order is maintained. Let us use the grouping (AB)C. This is probably the most natural. We get AB = 1 2 3 −1 0 1 3 −1 0 −2 = −2 1 −1 1 3 11 and (AB)C = −2 1 −1 1 3 11   4 −1 0   = −9 1 J Ambo University DEPARTMENT OF MATHEMATICS 20

148. Exercise 2.2.2. Compute each of the following expressions for A = 2 0 −1 5 , B = −1 1 2 4 , C = 3 4 0 2 1. A − 3B2 2. A2 B + 2C3 The Transpose of a Matrix Definition 2.2.2. The transpose of an m × n matrix A is the n × m matrix AT obtained by interchanging the rows and columns of A. That is, the ith column of AT is the ith row of A for all i. Example 2.2.3. Let A = 1 3 2 5 0 1 , B = a b c d , C = 5 −1 2 Then their transposes are AT =   1 5 3 0 2 1   , BT = a c b d , CT =   5 −1 2   Definition 2.2.3. A square matrix A is symmetric if AT = A—that is, if A is equal to its own transpose. Example 2.2.4. Let A =   1 3 2 3 5 0 2 0 4   and B = 1 2 −1 3 Then A is symmetric, since AT = A; but B is not symmetric, since BT = 1 −1 2 3 6= B. 2.3 Elementary operations and its properties Definition 2.3.1. A matrix is in row echelon form if it satisfies the following properties: 1. Any rows consisting entirely of zeros are at the bottom. 2. In each non-zero row, the first non-zero entry (called the leading entry) is in a column to the left of any leading entries below it. Example 2.3.1. The following matrices are in row echelon form:   2 4 1 0 −1 2 0 0 0   ,   1 0 1 0 1 5 0 0 4   ,   1 1 2 1 0 0 1 3 0 0 0 0   Elementary Row Operations Definition 2.3.2. The following elementary row operations can be performed on a matrix: 1. Interchange two rows. 2. Multiply a row by a non-zero constant. 3. Add a multiple of a row to another row. We will use the following shorthand notation for the three elementary row operations: 1. Ri ←→ Rj means interchange rows i and j. 2. kRi means multiply row i by k. 3. Ri + kRj means add k times row j to row i (and replace row i with the result). The process of applying elementary row operations to bring a matrix into row echelon form, called row reduction, is used to reduce a matrix to echelon form. Ambo University DEPARTMENT OF MATHEMATICS 21

149. Example 2.3.2. Reduce the following matrix to echelon form:     1 2 −4 −4 5 2 4 0 0 2 2 3 2 1 5 −1 1 3 6 5     Solution.     1 2 −4 −4 5 2 4 0 0 2 2 3 2 1 5 −1 1 3 6 5     R2 − 2R1 R3 − 2R1 R4 + R1 −→     1 2 −4 −4 5 0 0 8 8 −8 0 −1 10 9 −5 0 3 −1 2 10     R2←→R3 −→     1 2 −4 −4 5 0 −1 10 9 −5 0 0 8 8 −8 0 3 −1 2 10     R4+3R2 −→     1 2 −4 −4 5 0 −1 10 9 −5 0 0 8 8 −8 0 0 29 29 −5     1 8 −→     1 2 −4 −4 5 0 −1 10 9 −5 0 0 1 1 −1 0 0 29 29 −5     R4−29R3 −→     1 2 −4 −4 5 0 −1 10 9 −5 0 0 1 1 −1 0 0 0 0 24     With this final step, we have reduced our matrix to echelon form. J Definition 2.3.3. Matrices A and B are row equivalent if there is a sequence of elementary row operations that converts A into B. The matrices in example 2.3.2     1 2 −4 −4 5 2 4 0 0 2 2 3 2 1 5 −1 1 3 6 5     and     1 2 −4 −4 5 0 −1 10 9 −5 0 0 1 1 −1 0 0 0 0 24     are row equivalent. Definition 2.3.4. The rank of a matrix is the number of non-zero rows in its row echelon form. Definition 2.3.5. A matrix is in reduced row echelon form if it satisfies the following properties: 1. It is in row echelon form. 2. The leading entry in each non-zero row is a 1 (called a leading 1 ). 3. Each column containing a leading 1 has zeros everywhere else. For 2 × 2 matrices, the possible reduced row echelon forms are 1 0 0 1 , 1 ∗ 0 0 , 0 1 0 0 , 0 0 0 0 where ∗ can be any number. Exercise 2.3.1. Determine whether the given matrix is in row echelon form. If it is, state whether it is also in reduced row echelon form a.   1 0 1 0 0 3 0 1 0   b.   7 0 1 0 0 1 −1 4 0 0 0 0   c. 0 1 3 0 0 0 0 1 d.   0 0 0 0 0 0 0 0 0   Ambo University DEPARTMENT OF MATHEMATICS 22

150. e.   1 0 3 −4 0 0 0 0 0 0 0 1 5 0 1   f.   0 0 1 0 1 0 1 0 0   g.     1 2 3 1 0 0 0 1 1 0 0 1     2.4 Inverse of a matrix and its properties Definition 2.4.1. An n × n matrix A is invertible if there exists an n × n matrix B such that AB = In . Example 2.4.1. Prove that the matrix A = 1 2 3 4 has an inverse B = −2 1 3 2 −1 2 Solution. We have that AB = 1 2 3 4 −2 1 3 2 −1 2 = 1 0 0 1 = I2 and BA = −2 1 3 2 −1 2 1 2 3 4 = 1 0 0 1 = I2 Thus AB = BA = I2, proving that the matrix A has an inverse B . J Theorem 2.4.1. If A is an invertible matrix, then its inverse is unique. Proof. Let B and C be inverses of A. Thus AB = BA = In and AC = CA = In. Multiply both sides of the equation AB = In by C and use the algebraic properties of matrices. C(AB) = CIn (CA)B = C InB = C B = C Thus an invertible matrix has only one inverse. Definition 2.4.2. If an n × n matrix A is invertible, then A−1 is called the inverse of A and denotes the unique n×n matrix such that AA−1 = A−1 A = In . Determining the Inverse of a Matrix We now derive a method for finding the inverse of a matrix. The method is based on the Gauss-Jordan algorithm. Let A be an invertible matrix. Then AA−1 = InÂů Let the columns of A−1 be X1, X2, . . . , Xm and the columns of In be e1, e2, . . . , en. Express A−1 and In in terms of their columns, A−1 = X1 X2 . . . Xn and In = e1 e2 . . . en We shall find A−1 by finding X1, X2, . . . , Xn. Write the equation AA−1 = In in the form A X1 X2 . . . Xn = e1 e2 . . . en Using the column form of matrix multiplication, AX1 AX2 . . . AXn = e1 e2 . . . en Thus AX1 = e1, AX2 = e2, . . . , AXn = en Therefore X1, X2, . . . , Xn are solutions to the system AX1 = e1, AX2 = e2, . . . , AXn = en, all of which have the same matrix of coefficients A. Solve these systems by using Gauss Jordan elimination on the large augmented matrix A : e1 e2 . . . en . Since the solutions X1, X2, . . . , Xn are unique(they are the columns of A−1 ), A : e1 e2 . . . en ≈ · · · ≈ In : X1 X2 . . . Xn Thus, when A−1 exists, [A : In] ≈ · · · ≈ [In : B] where B = A−1 On the other hand, if the reduced echelon form of [A : In] is computed and the first part is not of the form In then A has no inverse. Ambo University DEPARTMENT OF MATHEMATICS 23

151. Example 2.4.2. Find the inverse of   1 2 −1 2 2 4 1 3 −3   if it exists. Solution. Gauss-Jordan elimination produces [A|I] =   1 2 −1 1 0 0 2 2 4 0 1 0 1 3 −3 0 0 1   R2 − 2R1 R3 − R1 −→   1 2 −1 1 0 0 0 −2 6 −2 1 0 0 1 −2 −1 0 1   −1 2 R2 −→   1 2 −1 1 0 0 0 1 −3 1 −1 2 0 0 1 −2 −1 0 1   R3−R2 −→   1 2 −1 1 0 0 0 1 −3 1 −1 2 0 0 0 1 −2 1 2 1   R1 + R3 R2 + 3R3 −→   1 2 0 −1 1 2 1 0 1 0 −5 1 3 0 0 1 −2 1 2 1   R1−2R2 −→   1 0 0 9 −3 2 −5 0 1 0 −5 1 3 0 0 1 −2 1 2 1   Therefore, A−1 =   9 −3 2 −5 −5 1 3 −2 1 2 1   (You should always check that AA−1 = I by direct multiplication.) J Example 2.4.3. Determine the inverse of the matrix   1 −1 −2 2 −3 −5 −1 3 5   Exercise 2.4.2. Find the inverse of   2 1 −4 −4 −1 6 −2 2 −2   if it exists. 2.5 Determinant of a matrix and its properties Definition 2.5.1. The determinant of a 2 × 2 matrix A is denoted kAk and is given by

155. a11 a12 a21 a22

159. = a11a22 − a12a21 Observe that the determinant of a 2 × 2 matrix is given by the difference of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A. Example 2.5.1. Find the determinant of the matrix 2 4 −3 1 Solution. Applying the above theorem we get

163. 2 4 −3 1

167. = (2 × 1) − (4 × (−3)) = 2 + 12 = 14 J The determinant of a 3 × 3 matrix is defined in terms of determinants of 2 × 2 matrices. The determinant of a 4 × 4 matrix is defined in terms of determinants of 3 × 3 matrices, and so on. For these definitions we need the following concepts of minor and cofactor. Ambo University DEPARTMENT OF MATHEMATICS 24

168. Definition 2.5.2. Let A be a square matrix. The minor of the element aij is denoted Mij and is the determinant of the matrix that remains after deleting row i and column j of A. The cofactor of aij is denoted Cij and is given by Cij = (−1)i+j Mij Note that the minor and cofactor differ in at most sign. Example 2.5.2. Determine the minors and cofactors of the elements a11 and a31 of the following matrix A. A =   1 0 3 4 −1 2 0 −2 1   Solution. Applying the above definitions we get the following. Minor of a11: M11 =

174. 1 0 3 4 −1 2 0 −2 1

180. =

184. −1 2 −2 1

188. = (−1 × 1) − (2 × (−2)) = 3 (2.5.0) Cofactor of a11 : C11 = (−1)1+1 M11 = (−1)2 3 = 3 J Definition 2.5.3. The determinant of a square matrix is the sum of the products of the elements of the first row and their cofactors. If A is 3 × 3, |A| = a11C11 + a12C12 + a13C13 If A is 4 × 4, |A| = a11C11 + a12C12 + a13C13 + a14C14 . . . If A is n × n, |A| = a11C11 + a12C12 + · · · + a1nC1n These equations are called cofactor expansions of |A| Example 2.5.3. Evaluate the determinant of the following matrix A. A =   1 2 −1 3 0 1 4 2 1   Solution. Using the elements of the first row and their corresponding cofactors we get |A| = a11C11 + a12C12 + a13C13 = 1(−1)2

192. 0 1 2 1

196. + 2(−1)3

200. 3 1 4 1

204. + (−1)(−1)4

208. 3 0 4 2

212. = [(0 × 1) − (1 × 2)] − 2[(3 × 1) − (1 × 4)] − [(3 × 2) − (0 × 4)] = −2 + 2 − 6 = −6 J Theorem 2.5.1. The determinant of a square matrix is the sum of the products of the elements of any row or column and their cofactors. ith row expansion: |A| = ai1Ci1 + ai2Ci2 + · · · + ainCin jth column expansion: |A| = a1jC1j + a2jC2j + · · · + anjCnj There is a useful rule that can be used to give the sign part, (−1)i+j , of the cofactors in these expansions. The rule is summarized in the following array      + − + − . . . − + − + . . . + − + − . . . . . .      Example 2.5.4. Find the determinant of the following matrix using the second row. A =   1 2 −1 3 0 1 4 2 1   Solution. Expanding the determinant in terms of the second row we get |A| = a21C21 + a22C22 + · · · + a23C23 = −3

216. 2 −1 2 1

220. + 0

224. 1 −1 4 1

228. − 1

232. 1 2 4 2

236. = −3[(2 × 1) − (−1 × 2)] + 0[(1 × 1) − (−1 × 4)] − 1[(1 × 2) − (2 × 4)] = −12 + 0 + 6 = −6 J Ambo University DEPARTMENT OF MATHEMATICS 25

237. Exercise 2.5.2. Evaluate the determinant of the following 4 × 4 matrix. A =     2 1 0 4 0 −1 0 2 7 −2 3 5 0 1 0 −3     Computing Determinants of 2 × 2 and 3 × 3 Matrices The determinants 2 × 2 and 3 × 3 matrices can be found quickly using diagonals. For a 2 × 2 matrix the actual diagonals are used while in the case of a 3 × 3 matrix the diagonals of an array consisting of the matrix with the two first columns added to the right are used. A determinant is equal to the sum of the diagonal products that go from left to right minus the sum of the diagonal products that go from right to left, as follows. 2×2 matrix A a1 b1 a2 b2 + − 3×3 matrix A   a1 b1 c1 a2 b2 c2 a3 b3 c3   a1 b1 a2 b2 a3 b3 + + + − − − 2 × 2 matrix: |A| = a1b2 − a2b1 3 × 3 matrix: |A| = a1b2c3 + b1c2a3 + c1a2b3 (diagonal products from left to right) − c1b2a3 − a1c2b3 − b1a2c3 (diagonal products from right to left) For example: A 2 3 4 1 + − B   1 2 3 4 0 1 5 2 6   1 2 4 0 5 2 + + + − − − |A| = 2 − 12 = −10 |B| = 0 + 10 + 24 − 0 − 2 − 48 There are no such short cuts for computing determinants of larger matrices. Theorem 2.5.3 (Properties of determinants). Let A be an n × n matrix and c be a non-zero scalar a. If a matrix B is obtained from A by multiplying the elements of a row (column) by c then |B| = c|A| b. If a matrix B is obtained from A by interchanging two rows (columns) then |B| = −|A| c. If a matrix B is obtained from A by adding a multiple of one row (column) to another row (column), then |B| = |A| Example 2.5.5. Evaluate the determinant

243. 3 4 −2 −1 −6 3 2 9 −3

249. Solution. We examine the rows and columns of the determinant to see if we can create zeros in a row or column using the above operations. Note that we can create zeros in the second column by adding twice the third column to it:

255. 3 4 −2 −1 −6 3 2 9 −3

261. = C2+2C3

267. 3 0 −2 −1 0 3 2 3 −3

273. Expand this determinant in terms of the second column to take advantage of the zeros. = (−3)

277. 3 −2 −1 3

281. = (−3)(9 − 2) = −21 J Definition 2.5.4. A square matrix A is said to be singular if |A| = 0. A is nonsingular if |A| 6= 0. Theorem 2.5.4. Let A be a square matrix. A is singular if a. all the elements of a row (column) are zero. b. two rows (columns) are equal. c. two rows (columns) are proportional. [Note that (b) is a special case of (c), but we list it to give it special emphasis.] Example 2.5.6. Show that the following matrices are singular. (a) A =   2 0 −7 3 0 1 −4 0 9   , (b) B =   2 −1 3 1 2 4 2 4 8   Ambo University DEPARTMENT OF MATHEMATICS 26

282. Solution. (a) All the elements in column 2 of A are zero. Thus |A| = 0. (b) Observe that every element in row 3 of B is twice the corresponding element in row 2. We write (row 3) = 2(row 2) row 2 and row 3 are proportional. Thus |B| = 0. J Theorem 2.5.5. Let A and B be n × n matrices and c be a nonzero scalar. a. Determinant of a scalar multiple: |cA| = cn |A| b. Determinant of a product: |AB| = |A||B| c. Determinant of a transpose: |At | = |A| d. Determinant of an inverse: |A−1 | = 1 |A| (Assuming A−1 exists.) Exercise 2.5.6. Prove that |A−1 At A| = |A| Remark 2.5.1. The determinant of a triangular matrix is the product of its diagonal elements. Example 2.5.7. Evaluate the determinant

288. 2 4 1 −2 −5 4 4 9 10

294. Solution. We create zeros below the main diagonal, column by column.

300. 2 4 1 −2 −5 4 4 9 10

306. = R2 + R1 R3 − 2R1

312. 2 4 1 0 −1 5 0 1 8

318. = R3+R2

324. 2 4 1 0 −1 5 0 0 13

330. = 2×(−1)×13 = −26 J Exercise 2.5.7. Evaluate the following 4 × 4 determinant using elimination method

338. 2 1 3 1 −2 3 −1 2 2 1 2 3 −4 −2 0 −1

346. Definition 2.5.5. Let A be an n × n matrix and Cij be the cofactor of aij. The matrix whose (i, j)th element is Cij is called the matrix of cofactors of A. The transpose of this matrix is called the adjoint of A and is denoted adj(A).      C11 C12 . . . C1n C21 C22 . . . C2n . . . . . . . . . Cn1 Cn2 . . . Cnn      matrix of cofactors      C11 C21 . . . Cn1 C12 C22 . . . Cn2 . . . . . . . . . C1n C2n . . . Cnn      adjoint matrix Determinants and Matrix Inverses Theorem 2.5.8. Let A be a square matrix with |A| 6= o. A is invertible with A−1 = 1 |A| adj(A) Theorem 2.5.9. A square matrix A is invertible if and only if |A| 6= 0. Example 2.5.8. Use the formula for the inverse of a matrix to compute the inverse of the matrix A =   2 0 3 −1 4 −2 1 −3 5   Solution. |A| = 25. Thus the inverse of A exists. adj(A) =   14 −9 −12 3 7 1 −1 6 8   A−1 = 1 25 adj(A) =   14 25 −9 25 −12 25 3 25 7 25 1 25 −1 25 6 25 8 25   J 2.6 Solving system of linear equations One of the application of determinant is to find the solution to the linear systems Ax = b when A is an invertible square matrix. Ambo University DEPARTMENT OF MATHEMATICS 27

347. 2.6.1 Cramer’s rule Before stating the theorem, we need to introduce some notation. If A = [a1 a2 · · · an] is an n×n matrix and b is in Rn , then let Ai denote the matrix A after replacing ai with b. That is, Theorem 2.6.1 (CRAMER’S RULE ). Let A be an invertible n × n matrix. Then the components of the unique solution x to Ax = b are given by xi = det(Ai) det(A) for i = 1, 2, . . . , n [Ai = a1 · · · ai−1 b ai+1 · · · an] Example 2.6.1. Use Cramer’s Rule to find the solution to the system 3x1 + x2 = 5 −x1 + 2x2 + x3 = −2 −x2 + 2x3 = −1 Solution. The system is equivalent to Ax = b, where A =   3 1 0 −1 2 1 0 −1 2   and b =   5 −2 −1   We have A1 =   5 1 0 −2 2 1 −1 −1 2   , A2 =   3 5 0 −1 −2 1 0 −1 2   , A3 = A =   3 1 5 −1 2 −2 0 −1 −1   Computing determinants gives us det(A) = 17, det(A1) = 28, det(A2) = 1 and det(A3) = −8. Therefore, by CramerâĂŹs Rule, the solution to Ax = b is x1 = det(A1) det(A) = 28 17 , x2 = det(A2) det(A) = 1 17 , x3 = det(A3) det(A) = −8 17 J 2.6.2 Gaussian method When row reduction is applied to the augmented matrix of a system of linear equations, we create an equivalent system that can be solved by back substitution. The entire process is known as Gaussian elimination. Ambo University DEPARTMENT OF MATHEMATICS 28

348. Gaussian Elimination 1. Write the augmented matrix of the system of linear equations. 2. Use elementary row operations to reduce the augmented matrix to row echelon form. 3. Using back substitution, solve the equivalent system that corresponds to the row-reduced matrix. Example 2.6.2. Solve the system 2x2 + 3x3 = 8 2x1 + 3x2 + x3 = 5 x1 − x2 − 2x3 = −5 Solution. The augmented matrix is   0 2 3 8 2 3 1 5 1 −1 −2 −5   reduce the matrix to row echelon form   0 2 3 8 2 3 1 5 1 −1 −2 −5   R1↔R3 −→   1 −1 −2 −5 2 3 1 5 0 2 3 8   We now create a second zero in the first column, using the leading 1 : R2−2R1 −→   1 −1 −2 −5 0 5 5 15 0 2 3 8   1 5 R2 −→   1 −1 −2 −5 0 1 1 3 0 2 3 8   We now need another zero at the bottom of column 2: R3−2R2 −→   1 −1 −2 −5 0 1 1 3 0 0 1 2   The augmented matrix is now in row echelon form, and we move to step 3. The corresponding system is x1 − x2 − 2x3 = −5 x2 + x3 = 3 x3 = 2 and back substitution gives x3 = 2, then x2 = 3 − x3 = 3 − 2 = 1, and finally x1 = −5 + x2 + 2x3 = −5 + 1 + 4 = 0. We write the solution in vector form as   0 1 2   J Gauss-Jordan Elimination 1. Write the augmented matrix of the system of linear equations. 2. Use elementary row operations to reduce the augmented matrix to reduced row echelon form. 3. If the resulting system is consistent, solve for the leading variables in terms of any remaining free variables Example 2.6.3. Solve the system in Example 2.6.2 by Gauss-Jordan elimination. Solution. The reduction proceeds as it did in Example 2.6.2 until we reach the echelon form:   1 −1 −2 −5 0 1 1 3 0 0 1 2   J 2.6.3 Inverse matrix method We now see that matrix inverse enables us to conveniently express the solutions to certain systems of linear equations. Ambo University DEPARTMENT OF MATHEMATICS 29

349. Theorem 2.6.2. Let AX = Y be a system of n linear equations in n variables. If A−1 exists, the solution is unique and is given by X = A−1 Y Example 2.6.4. Solve the following system of equations using the inverse of the matrix of coefficients. x1 − x2 − 2x3 = 1 2x1 − 3x2 − 5x3 = 3 −x1 + 3x2 + 5x3 = −2 Solution. This system can be written in the following matrix form,   1 −1 −2 2 −3 −5 −1 3 5     x1 x2 x3   =   1 3 −2   If the matrix of coefficients is invertible, the unique solution is   x1 x2 x3   =   1 −1 −2 2 −3 −5 −1 3 5   −1   1 3 −2   This inverse has already been found in Example 2.4.3 Using that result we get   x1 x2 x3   =   0 1 1 5 −3 −1 −3 2 1   −1   1 −2 1   The unique solution is x1 = 1, x2 = −1, x3 = 1 J 2.7 Eigenvalues and Eigenvectors The focus of this section is eigenvalues and eigenvectors, which are charac- teristics of matrices and linear transformations. Eigenvalues and eigenvectors arise in a wide range of fields, including finance, quantum mechanics, image processing, and mechanical engineering. Definition 2.7.1. Let A be an n × n matrix. Then a nonzero vector u is an eigenvector of A if there exists a scalar λ such that Au = λu (2.7.0) The scalar λ is called an eigenvalue of A. When λ and u are related as in equation 2.7.1, we say that λ is the eigenvalue associated with u and that u is an eigenvector associated with λ. The next theorem shows how to use determinants to find eigenvalues. Theorem 2.7.1. Let A be an n × n matrix. Then λ is an eigenvalue of A if and only if det(A − λIn) = 0. Theorem 2.7.2. Let A be a square matrix, and suppose that u is an eigenvector of A associated with eigenvalue λ. Then for any scalar c 6= 0, cu is also an eigenvector of A associated with λ. Example 2.7.1. Find the eigenvalues for A = 3 3 6 −4 . Solution. Our aim is to determine the values of λ that satisfy det(A−λIn) = 0. We have A − λIn = 3 3 6 −4 − λ 0 0 λ = 3 − λ 3 6 −4 − λ . Next, we compute the determinant, det(A − λIn) = (3 − λ)(−4 − λ) − 18 = λ2 + λ − 30 Setting det(A − λI) = 0, we have λ2 + λ − 30 = 0 ⇒ (λ − 5)(λ + 6) = 0 ⇒ λ = 5 or λ = −6 Thus the eigenvalues for A are λ = 5 and λ = −6. J Exercise 2.7.3. Find the eigenvalues and eigenvectors of the following matrices. (A) L =   4 4 −2 1 4 −1 3 6 −1   (B) G = 3 −4 1 3 (C) L =   −1 3 −4 −2 3 −4 1 1 3   . Ambo University DEPARTMENT OF MATHEMATICS 30

350. Chapter 3 Limit and continuity 3.1 Definition of limit Definition 3.1.1 (Informal definition of limit). Suppose f(x) is defined when x is near the number a . (these means that f is defined on some open interval that contains a, except possibly at a itself.) Then we write lim x→a f(x) = L and say “the limit of f(x), as x approaches a, equals L If we can make the values of f(x) arbitrarily close to L (as close to L as we like) by taking x to be sufficiently close to a (on either side of a) but not equal to a. Example 3.1.1. Guess the value of lim x→1 x−1 x2−1 . Solution. Note that the function f(x) = x−1 x2−1 is not defined at x = 1. Consider the following table: x 1 f(x) x 1 f(x) 0.5 0.66667 1.5000 0.400000 0.9 0.526316 1.1000 0.476190 0.99 0.502513 1.0010 0.499700 0.999 0.500250 1.0001 0.499975 Thus, lim x→1 x−1 x2−1 = 0.5. Example 3.1.1 is illustrated by the graph of f in Figure below. Now let’s change f slightly by giving it the value 2 when x = 1 and calling the resulting function g: g(x) = ( x−1 x2−1 if x 6= 1 2 if x = 1 This new function g still has the same limit as x approaches 1. lim x→1 g(x) = 0.5 J Example 3.1.2. Investigate lim x→0 sin(π x ) Exercise 3.1.1. Describe the behavior of the the function f(x) = x2 −1 x−1 near x = 1 and find the limit of f(x) at x = 1. 31

351. −4. −3. −2. −1. 1. 2. 3. −2. −1. 1. 2. 3. 4. 5. 6. 0 → 1 ← B Definition 3.1.2 (The formal definition of a limit). Let f be a function defined on some open interval that contains the number a, except possibly at a itself .Then we say that the limit of f(x) as x approaches a is L and we write lim x→a f(x) = L if for every number 0 there is a number δ 0 such that ,if 0 |x − a| δ then |f(x) − L| . Example 3.1.3. Show that lim x→3 (4x − 5) = 7. Solution. Let be a given positive number. We want to find a number δ such that ,if 0 |x − 3| δ then |4x − 5 − 7| . But |(4x − 5) − 7| = |4x − 12| = 4|x − 3|.Therefore we want δ such that if 0 |x − 3| δ then 4|x − 3| . that is ; if 0 |x − 3| δ then |x − 3| 4 . This suggests that we should choose δ = 4 Now showing that this δ works , given 0 choose δ = 4 if 0 |x − 3| δ then |(4x − 5) − 7| = |4x − 12| = 4|x − 3| 4δ = 4( 4 ) = . Ambo University DEPARTMENT OF MATHEMATICS 32

352. Thus if 0 |x − 3| δ then |(4x − 5) − 7| . Therefore by a definition of a limit lim x→3 (4x − 5) = 7. J Example 3.1.4. Use Formal Definition of limit to prove that lim x→4 x2 −2x−8 x−4 = 6. 3.2 Basic limit theorems Theorem 3.2.1 (limit laws). Suppose that c is a constant and the limits lim x→a f(x) and lim x→a g(x) exists. Then 1. Limit of a sum: lim x→a [f(x) + g(x)] = lim x→a f(x) + lim x→a g(x) 2. Limit of a difference: lim x→a [f(x) − g(x)] = lim x→a f(x) − lim x→a g(x) 3. Limit of a multiple: lim x→a cf(x) = c lim x→a f(x) 4. Limit of a product: lim x→a f(x)g(x) = lim x→a f(x) lim x→a g(x) 5. Limit of a quotient: lim x→a f(x) g(x) = lim x→a f(x) lim x→a g(x) if lim x→a g(x) 6= 0 Example 3.2.1. Evaluate the following limits a. lim x→0 (x2 + cos x) b. lim x→9 1 2 √ x c. lim x→0 x cos x d. lim x→0 x cos x x2+cos x Theorem 3.2.2. 6. Power law: lim x→a [f(x)]n = [lim x→a f(x)]n where n is a positive integer. 7. lim x→a c = c 8. lim x→a x = a 9. lim x→a xn = an where n is a positive integer 10. lim x→a n √ x = n √ a where n is a positive integer (If n is even, we assume that a 0). 11. Root Law: lim x→a n p f(x) = n q lim x→a f(x) where n is a positive integer (If n is even, we assume that lim x→a f(x) 0). Example 3.2.2. Evaluate lim x→−2 x3 +3x+1 x2−3 √ 5x Theorem 3.2.3 (Limits of Polynomials and Rational Functions). 1. If P(x) is a polynomial and a is any number, then lim x→a P(x) = P(a). 2. If P(x) and Q(x) are polynomials and Q(a) 6= 0, then lim x→a P(x) Q(x) = P(a) Q(a) . Example 3.2.3. Evaluate lim x→−1 (4x3 − 6x2 − 9x) Remark 3.2.1. If f(x) = g(x) when x 6= a, then lim x→a f(x) = lim x→a g(x) provided the limits exist. Example 3.2.4. Find lim x→1 x−1 x2−1 Example 3.2.5. Find lim x→−2 x3 +2x2 −x−2 x2−4 Ambo University DEPARTMENT OF MATHEMATICS 33

353. Theorem 3.2.4. If f(x) ≤ g(x) when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then lim x→a f(x) ≤ lim x→a g(x) Example 3.2.6. Find the lim x→0 √ 1 − x2 Theorem 3.2.5 (The Squeeze theorem). Assume f(x) ≤ g(x) ≤ h(x) for all x in some open interval about a, except possibly a itself. If lim x→a f(x) = lim x→a h(x) = L, then lim x→a g(x) exists and lim x→a g(x) = L. Example 3.2.7. Show that a. lim x→0 sin x x = 1. b. lim x→0 cos x−1 x = 0. Exercise 3.2.6. Evaluate the following a lim x→−2 x2 +x−2 x2+5x+6 b lim x→a 1 x − 1 a x−a c lim x→4 √ x−2 x2−16 3.3 One sided limits Definition 3.3.1 ( Left-hand-limit). We write lim x→a− f(x) = L and say the left-hand limit of f(x) as x approaches a [or the limit of f(x) as x approaches a from the left] is equal to L if we can make the values of f(x) arbitrarily close to L by taking x to be sufficiently close to a and x less than a. In this case we consider only for x a. Definition 3.3.2 (Right-hand-limit). If the values of f(x) can be made as close as we like to L by making x sufficiently close to a (but greater than a).Then we write lim x→a+ f(x) = L which is read as “the limit of f(x) as x approaches a from the right is L . lim x→a f(x) = L iff lim x→a− f(x) = L = lim x→a+ f(x). Example 3.3.1. Show that lim x→0 |x| = 0 Ambo University DEPARTMENT OF MATHEMATICS 34

354. Example 3.3.2. Let f(x) = |x| x . Find a. lim x→0− f(x) b. lim x→0+ f(x) c. lim x→0 f(x) if it exists. 3.4 Infinite limits,limit at infinity and asymptotes 3.4.1 Limits at infinity (negative infinity) Definition 3.4.1. If the function f is defined on an interval (a, ∞) and if we can ensure that f(x) is as close as we want to the number L by taking x large enough, then we say that f(x) approaches the limit L as x approaches infinity ,and we write lim x→∞ f(x) = L If f is defined on an interval (−∞, b) and if we can ensure that f(x) is close as we want to the number M by taking x negative and large enough in absolute value ,then we say that f(x) approaches the limit M as x approaches negative infinity ,and we write lim x→−∞ f(x) = M. Example 3.4.1. Evaluate lim x→∞ f(x) and lim x→−∞ f(x) for f(x) = x √ x2+1 3.4.2 Infinite limits Example 3.4.2 (A two sided infinite limit). Describe the behaviour of the function f(x) = 1 x2 near x = 0. Solution. As x approaches 0 from either side, the values of f(x) are positive and grow larger and larger.Thus lim x→0 f(x) = lim x→0 1 x2 = ∞ J Example 3.4.3 (One sided infinite limits). Describe the behaviour of the function f(x) = 1 x near x = 0. −3. −2. −1. 1. 2. 3. −1. 1. 2. 3. 4. 0 y = 1 x2 Figure 3.1: Graph of y = 1 x2 Solution. As x approaches 0 from the right ,the values of f(x) become larger and larger positive number, and we say that fhas right -hand limit infinity at x = 0. ⇒ lim x→0+ f(x) = ∞ Similarly , the values of f(x) become larger and larger negative numbers as x approaches 0 from the left, so f has left hand limit ∞ at x = 0. lim x→0− f(x) = −∞ These statements do not say that the one-sided limits exists ;they do not exists because ∞ and −∞ are not numbers. J Example 3.4.4. polynomial behaviour at infinity a. lim x→0 (3x3 − x2 + 2) = ∞ b. lim x→−∞ (3x3 − x2 + 2) = −∞ Ambo University DEPARTMENT OF MATHEMATICS 35

355. −3. −2. −1. 1. 2. 3. −1. 1. 2. 3. 4. 0 y = 1 x Figure 3.2: Graph of y = 1 x The highest degree term of a polynomial dominates the other terms as |x| grows large , so the limit of this term at ∞ and −∞ determine the limits of the whole polynomial. Thus 3x3 − x2 + 2 = 3x3 (1 − 1 3x + 2 3x3 ) ⇒ lim x→∞ (3x3 − x2 + 2) = ( lim x→∞ 3x3 )( lim x→∞ (1 − 1 3x + 2 3x3 )) = ∞ 3.4.3 Asymptotes Definition 3.4.2. The line x = a is called a vertical asymptote of the curve y = f(x) if at least one of the following statements is true. a. lim x→a f(x) = ∞ b. lim x→a− f(x) = ∞ c. lim x→a+ f(x) = ∞ d. lim x→a f(x) = −∞ e. lim x→a− f(x) = −∞ f. lim x→a+ f(x) = −∞ For instance in examples 3.4.2 and 3.4.3 above the y-axis is a vertical asymptote and the x-axis is a horizontal asymptote. Exercise 3.4.1. a. Determine the infinite limit lim x→−3+ x+2 x+3 and lim x→−3− x+2 x+3 . b. Find the vertical asymptotes of the function y = x2 +1 3x−2x2 . c. Evaluate lim x→0− |x| x and lim x→0+ |x| x . 3.4.4 Continuity One sided continuity Continuity from the left and right Definition 3.4.3. A function f is continuous from the left at a point c if lim x→c− f(x) = f(c) and is continuous from the right at a point c if lim x→c+ f(x) = f(c) Definition 3.4.4. A function f is said to be continuous at a point c if the following conditions are satisfied 1. f(c) is defined 2. lim x→c f(x) exists 3. lim x→c f(x) = f(c) Example 3.4.5. Determine whether the following functions are continuous at the point x = 2. a. f(x) = x2 −4 x−2 Ambo University DEPARTMENT OF MATHEMATICS 36

356. b. g(x) = ( x2 −4 x−2 if x 6= 2 3 if x = 2 c. h(x) = ( x2 −4 x−2 if x 6= 2 4 if x = 2 Theorem 3.4.2. Polynomials are continuous every where. Example 3.4.6. Show that |x| is continuous every where , |x| =      x if x 0 0 if x = 0 −x if x 0 So |x| is continuous every where because lim x→c |x| = |c|. Theorem 3.4.3. If the functions f and g are continuous at c, then • f + g is continuous at c • f − g is continuous at c • f × g is continuous at c • f ÷g is continuous at c if g(c) 6= 0 and has a discontinuity at c if g(c) = 0 lim x→c f(x) g(x) = f(c) g(c) since f and g are continuous at c, lim x→c f(x) = f(c) and lim x→c g(x) = g(c). Thus lim x→c f(x) g(x) = lim x→c f(x) lim x→c g(x) = f(c) g(c) Theorem 3.4.4. A rational function is continuous every where except at the points where the denominator is zero. Example 3.4.7. For what values of x is there a hole or a gap in the graph of y = x2 −9 x2−5x+6 . Theorem 3.4.5. If f is continuous at b and lim x→a g(x) = b, then lim x→a f(g(x)) = f(b). In other words lim x→a f(g(x)) = f(lim x→a g(x)) Example 3.4.8. If g(x) = 5 − x2 and f(x) = |x| then show that f(g(x)) is continuous at x = 3. Theorem 3.4.6. If g is continuous at a and f is continuous at g(a), then the composite function f ◦ g given by (f ◦ g)(x) = f(g(x)) is continuous at a. Continuity on a closed interval Definition 3.4.5. A function f is said to be continuous on a closed interval [a, b] if the following conditions are satisfied. 1. f is continuous on (a, b). 2. f is continuous from the right at a. 3. f is continuous from the left at b. But f is continuous on (a, b) if f is continuous at every element in (a, b). Example 3.4.9. What can you say about the continuity of the function f(x) = √ 9 − x2? 1. If c ∈ (−3, 3), then lim x→c f(x) = lim x→c √ 9 − x2 = √ 9 − c2 = f(c) which shows f is continuous at each point in (−3, 3). 2. lim x→−3+ f(x) = lim x→−3+ √ 9 − x2 = q lim x→−3+ (9 − x2) = 0 = f(−3) 3. lim x→3− f(x) = lim x→3− √ 9 − x2 = q lim x→3− (9 − x2) = 0 = f(3). Thus f is continuous on the closed interval [−3, 3]. 3.5 Intermediate value theorem Theorem 3.5.1 (Intermediate value theorem). If f is continuous on a closed interval [a, b] and k is any number between f(a) and f(b) inclusive then there is at least one number x in the interval [a, b] such that f(x) = N. Ambo University DEPARTMENT OF MATHEMATICS 37

357. Theorem 3.5.2. If f is continuous on [a, b] and if f(a) and f(b) are non- zero and have positive signs, then there is at least one solution of the equation f(x) = 0 in the interval (a, b). Ambo University DEPARTMENT OF MATHEMATICS 38

358. Chapter 4 Derivatives and application of derivatives 4.1 Definition of derivatives; basic rules Many real-world phenomena involve changing quantities - the speed of a rocket, the inflation of currency, the number of bacteria in a culture, the shock intensity of an earthquake, the voltage of an electrical signal, and so forth. In this chapter we will develop the concept of a “derivative, which is the mathematical tool for studying the rate at which one quantity changes relative to another. The study of rates of change is closely related to the geometric concept of a tangent line to a curve, so we will also be discussing the general definition of a tangent line and methods for finding its slope and equation. Definition 4.1.1. The derivative of a function f at a number a , denoted by f0 (a), is f0 (a) = lim h→0 f(a + h) − f(a) h if this limit exists. If we write x = a + h , then we have h = x − a and h approaches 0 if and only if x approaches a. Therefore an equivalent way of stating the definition of the derivative, as we saw in finding tangent lines, is f0 (a) = lim x→a f(x) − f(a) x − a . Two interpretations of the derivative are as follows. 1. Geometric Interpretation of the Derivative: The derivative f0 of a function f is a measure of the slope of the tangent line to the graph of f at any point P(x, f(x)), provided that the derivative exists. Example 4.1.1. Find an equation of the tangent line to the parabola y = x2 at the P(1, 1). 2. Physical Interpretation of the Derivative: The derivative f0 of a function f measures the instantaneous rate of change of f at x . Example 4.1.2. Find the derivative of the function f(x) = x at the number a. 39

359. Exercise 4.1.1. Find the derivative of the the following functions with respect to x using Definition 4.1.1. A. f(x) = √ x B. g(x) = 1 x + 1 C. h(x) = x3 − x D. h(x) = 1 − x 2 + x Theorem 4.1.2 (Derivative of a Constant Function). If c is a constant, then d dx (c) = 0. Example 4.1.3. a) If f(x) = 6, then f0 (x) = d dx (6) = 0. b) If f(x) = π2 , then f0 (x) = d dx (π2 ) = 0. Theorem 4.1.3 (The Power Rule). If r is any real number and f(x) = xr , then f0 (x) = d dx (xr ) = rxr−1 . Exercise 4.1.4. Find the derivative of the following functions with respect to x using power rule. A. f(x) = x100 B. g(x) = 3 √ x C. h(x) = 1 x2 D. l(x) = xπ Theorem 4.1.5 (The Constant Multiple Rule). If f is a differentiable function and c is a constant, then d dx [cf(x)] = cf0 (x). Exercise 4.1.6. Find the derivative of the following functions with respect to x using power rule. A. f(x) = 5 3x5 B. g(x) = π x C. 1 x2 Theorem 4.1.7 ( The Sum and Difference Rule). If f and g are differentiable functions, then d dx [f(x) ± g(x)] = f0 (x) ± g0 (x) Theorem 4.1.8 (The Product Rule). If f and g are differentiable functions, then d dx [f(x)g(x)] = f0 (x)g(x) + g0 (x)f(x). Example 4.1.4. a) Find dy/dx if y = (4x2 − 1)(7x3 + x). b) Find ds/dt if s = (1 + t) √ t . c) Find dy/dx if y = (2 √ x + 3 x )(2 √ x − 2 x ) d) Let y = uv be the product of the functions u and v. Find y0 (2) if u(2) = 2, u0 (2) = −5, v(2) = 1 and v0 (2) = 3. Theorem 4.1.9 (The Quotient Rule). If f and g are differentiable functions and g(x) 6= 0 , then d dx f(x) g(x) = g(x)f0 (x) − f(x)g0 (x) [g(x)]2 . Example 4.1.5. a) Find dy/dx if y = 1 x2+1 . b) Find ds/dt if s = √ t 1−5t . c) Find df/dθ if f(θ) = a+bθ m+nθ d) Find equations of any lines that pass through the point (−1, 0) and any tangent to the curve y = x−1 x+1 . Theorem 4.1.10 (Chain Rule). If g is differentiable at x and f is differentiable at g(x), then the composite function F(x) = f ◦ g(x) defined by F(x) = f(g(x)) is differentiable at x and F0 is given by the product F0 (x) = f0 (g(x)) · g0 (x) In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions,then dy dx = dy du du dx . Example 4.1.6. a) Differentiate y = cos 4x b) Find F0 (x) if F(x) = √ x2 + 3 Ambo University DEPARTMENT OF MATHEMATICS 40

360. The Power Rule Combined with the Chain Rule: If n is any real number and u = g(x) is differentiable, then d dx un = nun−1 du dx Alternatively, d dx [g(x)]n = n[g(x)]n−1 · g0 (x) Example 4.1.7. Differentiate A. y = (x3 −1)100 B. g(x) = sin(cos(tan x)) C. h(x) = 1 3 √ x2 + x + 1 D. l(x) = esin x 4.2 Derivatives of inverse functions Definition 4.2.1. A function f is called a one-to-one function if it never takes on the same value twice; that is, f(x1) 6= f(x2) whenever x1 6= x2. HORIZONTAL LINE TEST: A function f is one-to-one if and only if no horizontal line intersects its graph more than once. Definition 4.2.2. Let f be a one-to-one function with domain A and range B. Then its inverse function f−1 (x) has domain B and range A and is defined by f−1 (y) = x ⇔ f(x) = y for any y in B. domain of f−1 = range f image of f = domain of f−1 Example 4.2.1. Find the inverse of f(x) = √ 3x − 2. 4.2.1 Inverse trigonometric functions Definition 4.2.3 (Inverse Trigonometric Functions). Domain y = sin−1 x if and only if x = sin y [−1, 1] y = cos−1 x if and only if x = cos y [−1, 1] y = tan−1 x if and only if x = tan y (−∞, ∞) y = csc−1 x if and only if x = csc y (−∞, −1] ∪ [1, ∞) y = sec−1 x if and only if x = sec y (−∞, −1] ∪ [1, ∞) y = cot−1 x if and only if x = cot y (−∞, ∞) Theorem 4.2.1 (Derivatives of inverse trigonometric functions ). Let u be a differentiable function of x. • d dx (sin−1 u) = u0 √ 1−u2 • d dx (cos−1 u) = −u0 √ 1−u2 • d dx (tan−1 u) = u0 1+u2 • d dx (cot−1 u) = u0 1+u2 • d dx (sec−1 u) = u0 |u| √ u2−1 • d dx (csc−1 u) = −u0 |u| √ u2−1 4.2.2 Hyperbolic and inverse hyperbolic functions Definition 4.2.4 (DEFINITION OF THE HYPERBOLIC FUNCTIONS). sinh x = ex − e−x 2 csc x = 1 sinh x cosh x = ex + e−x 2 sec x = 1 cosh x tanh x = sinh x cosh x coth x = cosh x sinh x Ambo University DEPARTMENT OF MATHEMATICS 41

361. Hyperbolic Identities sinh(−x) = − sinh(x), cosh x = cosh x cosh2 x − sinh2 x = 1, 1 − tanh2 x = sech2 x sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y Derivatives of Hyperbolic Functions • d dx (sinh x) = cosh x • d dx (cosh x) = sinh x • d dx (tanh x) = sech2 x • d dx (cschx) = cschx coth x • d dx (sechx) = −sechx tanh x • d dx (cothx) = −csch2 x Inverse Hyperbolic Functions y = sinh−1 x ⇔ sinh y = x y = cosh−1 x ⇔ cosh y = x and y ≥ 0 y = tanh−1 x ⇔ tanh y = x Since the hyperbolic functions are defined in terms of exponential functions, itâĂŹs not surprising to learn that the inverse hyperbolic functions can be expressed in terms of logarithms. • sinh−1 x = ln(x + √ x2 + 1) x ∈ R • cosh−1 x = ln(x + √ x2 − 1) x ∈ [1, ∞) • tanh−1 x = 1 2 ln 1+x 1−x x ∈ (−1, 1) • coth−1 x = 1 2 ln x+1 x−1 x ∈ (−∞, −1) ∪ (1, ∞) • sech−1 x = ln 1+ √ 1−x2 x x ∈ (0, 1] • csch−1 x = ln 1 x + √ 1+x2 |x| x ∈ (−∞, 0) ∪ (0, ∞) Example 4.2.2. Show that sinh−1 x = ln(x + √ x2 + 1). Derivatives of Inverse Hyperbolic Functions • d dx (sinh−1 x) = 1 √ 1+x2 • d dx (cosh−1 x) = 1 √ x2−1 • d dx (tanh−1 x) = 1 1−x2 • d dx (coth−1 x) = 1 1−x2 • d dx (sech−1 x) = − 1 x √ 1−x2 • d dx (csch−1 x) = − 1 |x| √ x2+1 Example 4.2.3. Show that d dx (sinh−1 x) = 1 √ 1+x2 . 4.3 Higher order derivatives The derivative f0 of a function f is itself a function. As such, we may consider differentiating the function f0 . The derivative of f0 , if it exists, is denoted by f00 and is called the second derivative of f. Continuing in this fashion, we are led to the third, fourth, fifth, and higher-order derivatives of f, whenever they exist. Notations for the first, second, third, and in general, the n−th derivative of f are f0 , f00 , f000 , . . . , f(n) or d dx [f(x)], d2 f dx2 , d3 f dx3 , . . . , dn f dxn or Dxf, D2 xf, D3 xf, . . . , Dn x f Example 4.3.1. Find the derivatives of all orders of f(x) = x4 − 3x3 + x2 − 2x + 8. Example 4.3.2. Find the third derivative of y = 1 x . 4.4 Implicit differentiation Suppose y = f(x) is differentiable function of x, we have seen that dy dx = f0 (x). It was a case when y is written explicitly in terms of x. On the other hand, Ambo University DEPARTMENT OF MATHEMATICS 42

362. assume y is a diffentiable function of x expressed by equation F(x, y) = 0, where y is not expressed explicitly. For instance, x3 + y3 = 2xy. The process of differentiating such functions without the need of first writing y in terms of x, is called Implicit differentiation. Guidelines for implicit Differentiation 1. Differentiate both sides of the equation with respect to x. 2. Collect all terms involving dy dx on the left side of the equation and move all other terms to the right side of the equation. 3. Factor dy dx out of the left side of the equation. 4. Solve for dy dx . Example 4.4.1. Use implicit differentiation to find dy dx if 5y2 + sin y = x2 . Exercise 4.4.1. (a) Use implicit differcntiation to find dy dx for the Folium of Descartes , x3 + y3 = 3xy. (b) Find an equation for the tangent line to the Folium of Descartes at the point (3 2 , 3 2 ). (c) At what points is the tangent line to the Folium of Descartes horizontal? 4.5 Application of derivatives 4.5.1 Extrema of a function Definition 4.5.1 (Extrema of a Function). A function f has an absolute maximum at c if for all x in the domain D of f. The number f(c) is called the maximum value of f on D. Similarly, f has an absolute minimum at c if for all x in D. The number f(c) is called the minimum value of f on D. The absolute maximum and absolute minimum values of f on D are called the extreme values, or extrema, of f on D. Definition 4.5.2 (Relative Extrema of a Function). A function f has a relative (or local) maximum at c if f(c) ≥ f(x) for all values of x in some open interval containing c . Similarly, f has a relative (or local) minimum at c if f(c) ≤ f(x) for all values of x in some open interval containing c. Theorem 4.5.1 (Fermat’s Theorem). If f has a relative extremum at c, then either f0 (c) = 0 or f0 (c) does not exist. Definition 4.5.3 (Critical Number of f). A critical number of a function f is any number c in the domain of f at which f0 (c) = 0 or f0 (c) does not exist. Finding the Extreme Values of a Continuous Function on a Closed Interval Theorem 4.5.2 (The Extreme Value Theorem). If f is continuous on a closed interval [a, b], then attains an absolute maximum value f(c) for some number c in [a, b] and an absolute minimum value f(d) for some number d in [a, b]. Guidelines for Finding the Extrema of a Continuous Function f on [a, b] 1. Find the critical numbers of f that lie in (a, b). 2. Compute the value of f at each of these critical numbers, and also compute f(a) and f(b). 3. The absolute maximum value of f and the absolute minimum value of f are precisely the largest and the smallest numbers found in Step 2. Example 4.5.1. Find the extreme values of the function f(x) = 3x4 −4x3 −8 on [−1, 2]. Example 4.5.2. Find the extreme values of the function f(x) = 2 cos x − x on [0, 2π] 4.5.2 Mean value theorem Theorem 4.5.3 (ROLLE’S THEOREM). Let f be a function that satisfies the following three hypotheses: 1. f is continuous on the closed interval [a, b]. 2. f is differentiable on the open interval (a, b). 3. f(a) = f(b). 4. Then there is a number c in (a, b) such that f0 (c) = 0. Example 4.5.3. Let f(x) = x3 − x for x in [−1, 1]. Ambo University DEPARTMENT OF MATHEMATICS 43

363. • Show that f satisfies the hypotheses of Rolle’s Theorem on [−1, 1]. • Find the number(s) c in (−1, 1) such that f0 (c) = 0 as guaranteed by Rolle’s Theorem. Theorem 4.5.4 ( THE MEAN VALUE THEOREM). Let f be a function that satisfies the following hypotheses: 1. f is continuous on the closed interval [a, b]. 2. f is differentiable on the open interval (a, b) Then there is a number c in (a, b) such that f0 (c) = f(b) − f(a) b − a or, equivalently, f(b) − f(a) = f0 (c)[b − a] Example 4.5.4. Let f(x) = x3 . a. Show that f satisfies the hypotheses of the Mean Value Theorem on [1, −1]. b. Find the number(s) c in (−1, 1) that satisfy Equation (1) as guaranteed by the Mean Value Theorem. 4.6 First and second derivative tests Definition 4.6.1 (Increasing and Decreasing Functions). A function f is increasing on an interval I, if for every pair of numbers x1 and in x2, x1 x2 implies that f(x1) f(x2), f is decreasing on I if, for every pair of numbers x1 and x2 in I, x1 x2 implies that f(x1) f(x2), f is monotonic on I if it is either increasing or decreasing on I. Theorem 4.6.1. Suppose f is differentiable on an open interval (a, b). a. If f0 (x) 0 for all x in (a, b) , then f is increasing on (a, b). b. If f0 (x) 0 for all x in (a, b) , then f is decreasing on (a, b). c. If f0 (x) = 0 for all x in (a, b), then f is constant on (a, b). Determining the Intervals Where a Function Is Increasing or Decreasing 1. Find all the values of x for which f0 (x) = 0 or f0 (x) does not exist. Use these values of x to partition the domain of f into open intervals. 2. Select a test number c in each interval I found in Step 1, and determine the sign of f0 (c) in that interval. a. If f0 (c) 0 then f is increasing on that interval. b. If f0 (c) 0 then f is decreasing on that interval. c. If f0 (c) = 0, then f is constant on that interval. Example 4.6.1. Determine the intervals where the function f(x) = x3 − 3x2 + 2 is increasing and where it is decreasing. We will now see how the derivative of a function can be used to help us find the relative extrema of f. Theorem 4.6.2 (The First Derivative Test). Let c be a critical number of a continuous function f in the interval (a, b) and suppose that f is differentiable at every number c in (a, b) with the possible exception of c itself. a. If f0 (x) 0 on (a, c) and f0 (x) 0 on (c, b), then f has a relative maximum at c. b. If f0 (x) 0 on (a, c) and f0 (x) 0 on (c, b), then has a relative minimum at c. c. If f0 (x) has the same sign on (a, c) and (c, b), then f does not have a relative extremum at c. Theorem 4.6.3 (The Second Derivative Test). Suppose that f has a continuous second derivative on an interval (a, b) containing a critical number c of f. a. If f00 (c) 0 , then f has a relative maximum at c. Ambo University DEPARTMENT OF MATHEMATICS 44

364. b. If f00 (c) 0, then has a relative minimum at c. c. If f00 (c) = 0, then the test is inconclusive. Example 4.6.2. Find the relative extrema of f(x) = x3 − 3x2 − 24x + 32 using the Second Derivative Test. 4.6.1 Concavity and inflection point Definition 4.6.2. Concavity of the Graph of a Function Suppose f is differentiable on an open interval I. Then a. the graph of f is concave upward on I if f0 is increasing on I. b. the graph of f is concave downward on I if f0 is decreasing on I. Theorem 4.6.4. Suppose f has a second derivative on an open interval I. a. If f00 (x) 0 for all x in I , then the graph of f is concave upward on I. b. If f00 (x) 0 for all x in I, then the graph of f is concave downward on I. Determining the Intervals of Concavity of a Function 1. Find all values of x for which f00 (x) = 0 or f00 (x) does not exist. Use these values of x to partition the domain of f into open intervals. 2. Select a test number c in each interval found in Step 1 and determine the sign of f00 (c) in that interval. a. If f00 (c) 0 , the graph of f is concave upward on that interval. b. If f00 (c) 0, the graph of f is concave downward on that interval. Definition 4.6.3. A point P on a curve y = f(x) is called an inflection point if f is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P. Finding Inflection Points 1. Find all numbers c in the domain of f for which f00 (c) = 0 or f00 (c) does not exist. These numbers give rise to candidates for inflection points. 2. Determine the sign of f00 (x) to the left and to the right of each number c found in Step 1. If the sign of f00 (x) changes, then the point P(c, f(c)) is an inflection point of f, provided that the graph of f has a tangent line at P. Example 4.6.3. Find the points of inflection of f(x) = x4 − 4x3 + 12. 4.6.2 Curve sketching We have seen on many occasions how the graph of a function can help us to visualize the properties of the function. From a practical point of view, the graph of a function also gives, at one glance, a complete summary of all the information captured by the function. Guidelines for Curve Sketching 1. Find the domain of f. 2. Find the x− and y− intercepts of f. The y-intercept is f(0) and this tells us where the curve intersects the y-axis. To find the x-intercepts, we set y = 0 and solve for x. (You can omit this step if the equation is difficult to solve.) 3. Determine whether the graph of f is symmetric with respect to the y-axis or the origin. i ) If f(x) = f(−x) for all x in D, that is, the equation of the curve is unchanged when x is replaced by −x, then f is an even function and the curve is symmetric about the y-axis. This means that our work is cut in half. If we know what the curve looks like for x ≥ 0, then we need only reflect about the y-axis to obtain the complete curve. ii) If f(x) = −f(−x) for all x in D, then f is an odd function and the curve is symmetric about the origin. Again we can obtain the complete curve if we know what it looks like for x ≥ 0. Rotate 180◦ about the origin. iii) If f(x + p) = f(x) for all x in D, where p is a positive constant, then f is called a periodic function and the smallest such number p is called the period. 4. Determine the behavior of f for large absolute values of x. Ambo University DEPARTMENT OF MATHEMATICS 45

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