Mathematical Analysis I Past Paper Notes PDF

MAT 201 Mathematical Analysis I TALLIE PEKA RONALD November 6, 2024 1 Polynomials A function P is called a polynomial1 if P (x) = a0 + a1 x + a2 x2 + · · · + an−1...

MAT 201 Mathematical Analysis I TALLIE PEKA RONALD November 6, 2024 1 Polynomials A function P is called a polynomial1 if P (x) = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 + an xn , where n is a nonnegative integer and the constants a0 , a1 , a2 , a · · · , an−1 , an are called the coefficients af the polynomial. The domain of any polynomial is the set of real numbers, R = (−∞, +∞). if the leading coefficient an 6= 0 then the degree of the polynomial is n. √ Example 1.1 The function P (x) = 2x7 − 3x5 + 41 x2 + 2 is a polynomial of degree 7. A polynomial of degree 1 has the form P (x) = mx + b so it is a linear function. A polynomial of degree 2 is of the form P (x) = ax2 + bx + c and is called a quadratic function. Exercise 1.2 Identify the following polynomials 1. F (x) = 12x6 + 2x3 − 6x 2. F (k) = 18k 2 + 2k + 1 3. P (y) = 81y 4 − 256. 1.1 Operations on polynomials Several operations can be done on polynomials such as addition, subtraction, multiplication and divi- sion of polynomials. 1.1.1 Addition and Subtraction of polynomials When addition or subtracting polynomials, we add coefficients of variables with the same powers. consider the following two polynomials P1 (x) = a0 + a1 x + a2 x2 + · · · + an xn , P2 (x) = b0 + b1 x + b2 x2 + · · · + bn xn. P1 (x) + P2 (x) = a0 + a1 x + a2 x2 + · · · + an xn + b0 + b1 x + b2 x2 + · · · + bn xn

= (a0 + b0 ) + (a1 + b1 )x + (a2 + b2 )x2 + · · · + (an + bn )xn. Example 1.3 Find the sum and difference of the following polynomials, 1. f (x) = 7x − 14x2 + 5, g(x) = 3x2 − 6x + 2. 2. f (x) = 4x2 + 12x, g(x) = 6x2 − 14x. 1 Tallie Peka Ronald, [email protected] 1  2 1.1.2 Multiplication of polynomials To find the product of two polynomials, you multiply eact term in one expression by the other expres- sion. Example 1.4 Consider the polynomials P1 (x) = x2 + 2x − 3 and P2 (x) = 2x2 − x + 4, P1 (x)P2 (x) = x2 + 2x − 3 2x2 − x + 4

= x2 2x2 − x + 4 + 2x 2x2 − x + 4 − 3 2x2 − x + 4


= 2x4 − x3 + 4x2 + 4x3 − 2x2 + 8x − 6x2 + 3x − 12 = 2x4 + 3x3 − 4x2 + 11x − 12. Exercise 1.5 Expand and simplify if possible: 1. (x + 3)(x − 5) 2. (2x − 7)(3x + 1) 3. (2x2 − x + 2)(3x2 + x − 1). 1.1.3 Division of polynomials Recall that a polynomialis √ a finite expression with positive whole number indice, so that 2x + 4 is a polynomial and x is not a polynomial. You can use long division to divide a polynomial by (x ± p), where p is a constant. Example 1.6. Example 1.7. Exercise 1.8 Write each of the followimg polynomials in the form (x ± p)(ax2 + bx + c) by dividing a. x3 + 6x2 + 8x + 3 by (x + 1) b. x3 + 10x2 + 25x + 4 by (x + 4)  3 3x3 − 8x − 8 Exercise 1.9 Simplify x−2 1.1.4 The factor theorem The factor theorem states that if f (x) is a polynomial then: 1. If f (p) = 0, then (x − p) is a factor of f (x). 2. If (x − p) is a factor of f (x), then f (p) = 0. Example 1.10. Example 1.11. Exercise 1.12. Use the factor theorem to show that: a. (x+3) is a factor of 5x4 − 45x2 − 6x − 18 b. (x − 1) is a factor of x3 + 6x2 + 5x − 12 and hence factorise the expression completely.  4 1.1.5 The Remainder theorem Let f (x) be a polynomial. Then the remainder when f (x) is divided by x − p is f (p). Exercise 1.13 1. Find the Remainder when f (x) = 10x3 + 43x2 − 2x − 10 is divided by (5x + 4). 2. Find the remainder when 3x4 − 8x3 + 10x2 − 3x − 25 is divided by (x + 1). Note 1 We note that when using algebraic division we have the following relationship F (x) = divisor × Q(x) + Remainder | {z } quotient to convert an improper fraction into a mixed fraction. Example 1.14. 1.2 Linear and quadratic polynomials 1.2.1 Linear polynomials A linear polynomial is any polynomial of the form P (x) = ax + b, where a and b are real numbers and a 6= 0 e.g f (x) = x − 1, P (x) = 4x + 9. Roots of linear polynomials Every linear polynomial has exactly one root. And finding the root is just a matter of basic algebra. Example 1.15. Find the roots of the following: a. f (x) = 2x − 6 b. f (x) = x + 3 c. f (x) = 3x − 7. 1.3 Quadratic polynomials A quadratic polynomial is any polynomial of the form P (x) = ax2 + bx + c, where a, b and c are real numbers with a 6= 0. e.g x2 , 2x2 − 3x + 6 and x2 + 1 are all quadratic polynomial.  5 1.3.1 Roots of quadratic polynomials Finding the roots of a quadratic polynomials is equivalent to solving the quadratic equation ax2 + bx + c = 0. We shall present here three methods for finding the roots of quadratic function,i.e. factorisation method, completing the square and the formula method. Factoring and solving quadratic polynomial To solve a quadratic equation by factorising: Write the equation in the form ax2 + bx + c = 0 Factorise the left-hand side Set each factor equal to zero and solve to find the values(s) of x. Example 1.16. Solve the following: a. x2 − 2x − 15 = 0 b. x2 = 9x c. 6x2 + 13x − 5 = 0 d. x2 − 5x + 18 = 2 + 3x. The quadratic formula The solution of the equation ax2 + bx + c = 0 are given by the formula: √ −b ± b2 − 4ac x=. 2a Example 1.17. Solve the following using the formula method: 1. 3x2 − 7x − 1 = 0 b. 4x2 − 4x − 1 = 0. Completing the square The method of completing the square finding the roots of a quadratic polynomial i.e. the values of x for which ax2 + bx + c = 0 is done in the following steps: 1. Divided althrough by the coefficient of x2 , we get x2 + ab x + c a = 0. 2. Subtract the constant term from both sides , we get x2 + ab x = − ac. b 2 b 2 b 2 on both sides, we have x2 + ab x + 2a = − ac + 2a


3. Add 2a. b 2 b 2 = − ac + 2a

4. factorise the left hand side, x + 2a. q b 2 5. Take the square root on both sides, x + 2a = ± − ac + 2a b
. q b b 2 − ac +
6. Make x the subject, x = − 2a ± 2a. Example 1.18. Solve the following using completing the square method 1. x2 + 4x + 1 2. 2x2 − 8x + 7 = 0 3. x2 + 12x + 3 = 0.  6 1.3.2 The discriminant For a quadratic polynomial f (x) = ax2 + bx + c, the expression b2 − 4ac is called the discriminant. The value of the discriminant shows how many roots f (x) has: If b2 − 4ac > 0 then f(x) has two distinct real roots. If b2 − 4ac = 0 then f (x) has one repeated root. If b2 − 4ac < 0 then f (x) has no real roots. 2 Functions Definition 2.1 A function is a rule that assigns to each object x in a set D exactly one object is a set E. The set D is called a domain and the set E (in which the assigned objects are found) is called the codomain. The set of assigned object is called the range. Example 2.2 Consider the rule ”square a number and add 4” (x2 + 4). We apply this rule on the set {0, 1, 2, 3}and we need to determine the values obtained in the set {3, · · · , 13}. The domain is D = {0, 1, 2, 3}, The codomain is E = {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}, The range is {4, 5, 8, 13}. Notation 2.3 We will denote by letters f, g, h, etc any function and we will write f : D → E. The domain of a function f is also denoted by D(f ) or Df and its range R(f ) of Rf Example 2.4 Find the domain and the range of the following functions: 1) f (x) = x + 1 2) g(a) = a2 √ 3) h(x) = x + 2 x−2 4) f (x) = x−1 √ 5) f (x) = x2 − 4  7 Remark 2.5 : 1 Unless otherwise specified, if the formula is used to define a function f , we use the set of real numbers as the codomain of the function f. 2 Determining the natural domain of a squareroot function is determining the set of all numbers √ that involve to divide it by zero or taking the of a < 0 number.  1  if x 0, a 6= 0, and x is a real number.  8 the base a = 1 is excluded since it yields f (x) = 1x = 1. This is a constant, not an exponential function. You have evaluated ax for integer and rational values of x. For example, you know that 1 43 = 16 and 4 2 = 2. Note 2. 1. The domain of the exponential functions is the set of real numbers R; the range is the set of positive real numbers. 2. The natural base denoted by e and has value e = 2.718281828 · · · 3. e0 = 1 3.1.2 Solving exponential equations Solve the following equations: 2 1. 3x+1 = 81 2. e−x = (ex )2 · 1 e3 3. 2(32t−5 − 4) = 11. 3.2 Logarithm functions 3.2.1 Definition defined for x > 0, a > 0 and x 6= 1 y = loga x if and only if x = ay. The function given by f (x) = loga x, is called the logarithhm function with base a. 3.2.2 Properties of logarithms 1. loga 1 = 0 because a0 = 1 2. loga a = 1 because a1 = a 3. loga ax = x and aloga x = x. 4. If loga x = loga y then x = y. From the above properties of the logarithm function, we present another set of properties that concerns the product, quotient and power property. Let a be a positive number such that a 6= 1, and let n be a real number. If u and v are positive real numbers, the following properties are true. Logarithm with base a Natural Logarithm 1.Product Property: loga (uv) = loga u + loga v In(uv) = Inu + Inv 2.Quotient Property: loga uv = loga u − loga v In uv = Inu − Inv 3.Power Property: loga un = nloga u Inun = nInu  9 3.2.3 Change of base formula Let a, b and x be positive real numbers such that a 6= 1 and b 6= 1. Then loga x can be converted to a different base as follows. Base b Base 10 Base e loga x = logbx logb a loga x = logx loga loga x = Inx Ina. Solving exponential equations Solve the following logarithmic equations: i) Inx = 2 ii) log3 (5x − 1) = log3 (x + 7) iii) log3 (3x + 14) − log6 5 = log6 2x iv)2log5 3x = 4. 1 Tallie Peka Ronald, [email protected]