SMA 2101: Calculus I – University Lecture Notes with Worked Examples

This document provides complete lecture notes for Calculus I (SMA 2101), covering fundamental topics such as functions, limits, continuity, differentiation, and integration. Each section includes clear definitions, theorems, proofs, and numerous step-by-step solved examples to aid understanding. The notes also feature practice exercises for students to test comprehension, making it ideal for first-year university mathematics or engineering courses. SMA 2101: CALCULUS I Course content • Functions: definition, domain, range, codomain, composition (or composite), inverse. • Limits, continuity and differentiability of a function. • Differentiation by first principle and by rule for xn (integral and fractional n). • Other techniques of differentiation, i.e., sums, products, quotients, chain rule; their applications to algebraic, trigonometric, logarithmic, exponential, and inverse trigonometric functions all of a single variable. • Implicit and parametric differentiation. • Applications of differentiation to: rates of change, small changes, stationary points, equations of tangents and normal lines, kinematics, and economics and financial models (cost, revenue and profit). • Introduction to integration and its applications to area and volume. 1 1 Functions To understand the word function, we consider the following scenario and definitions. For example, the growth of a sidling is an instance of a functional relation, since the growth may be affected by variations in temperature, moisture, sunlight, etc. If all these factors remain constant, then the growth is a function of time. Definition 1.1 (Variables). A variable is an object, event, time period, or any other type of category you are trying to measure. Consider the formula used for calculating the volume of a sphere of radius r. V = πr3 (1) Then, i) V and r vary with different spheres. Hence, they are called variables. ii) π and are constants, irrespective of the size of the sphere. There are two types of variables, i.e., independent and dependent variables. Definition 1.2 (Independent and dependent variables). Independent variable refers to the input value while dependent variable refers to the output value. For example from formula (1), the volume, V , depends on the value of the radius, r, of the sphere. In this case, r is called the independent variable while V is called the dependent variable since it is affected by the variation of r. Similarly, for the function y = ax2 + bx + c, a,b and c are constants, x is the independent variable and y is the dependent variable. Definition 1.3 (Function). A function is a rule that assigns/associates each element in the independent set, say X, to a unique element in the dependent set, say Y . Examples of functions are i) Linear functions e.g., y = x + 5 ii) Quadratic functions e.g., y = x2 − 2x + 5 iii) Cubic functions e.g., y = x3 − 1 iv) Quartic functions e.g., y = 2x4 + x3 − 1 v) Trigonometric functions e.g., y = sin(2x + 5) vi) Logarithmic functions (log to base 10) e.g., y = log(3x + 1) vii) Natural logarithmic functions (log to base e ≈ 2.71828) e.g., y = ln(5x + 1) viii) Inverse of trigonometric functions e.g., y = tan−1(2x + 1) ix) Exponential functions e.g., y = e2x+1 x) Absolute value functions e.g., y = |x|. This function is defined as | | {−x, if x ≥ 0 y = x = x, if x 0 → Note: in the above examples the variable y depends on the variable x. Thus, we say that the dependent variable y is a function of the independent variable x. Using function notation, we write y = f(x), where f is a function. The function f(x) is read as f of x, meaning that f depends on x. 1.1 Domain, Range and Codomain 1.1 Domain, Range and Codomain Definition 1.4 (Domain). A domain consists of all the elements in the independent set (i.e., the set of inputs), X, for which the function is defined. Definition 1.5 (Range). A range refers to a set of all the images of the elements in the domain. Definition 1.6 (Codomain). A codomain consists of all the elements in the dependent set (i.e., the set of outputs), Y . For example, consider the diagram below Example(s): 1. Find the domain and range of the following functions. (a) f(x) = (x − 4)2 + 5 Solution Since f(x) is defined (or is a real number) for any real number x, the domain of f is the interval (−∞,∞). √ − Let y = (x − 4)2 + 5. Making x the subject, we have x = 4 ± y 5. This function is defined if y − 5 ≥ 0 or y ≥ 5. Therefore, the range is the interval [5,∞). (b) f(x) = 2x2 − 5x + 1 Solution Since f(x) is defined (or is a real number) for any real number x, the domain of f is the interval (−∞,∞). Let y = 2x2 5x + 1 or 2x2 5x + (1 y) = 0. Making x the subject (use quadratic 5 25 8(1 y) formula), we−have x ± √−− −− . This function is defined if 25−8(1−y) ≥ 0 or y . Therefore, the range is the interval . (c) f(x) = x2 − 5x + 6 Solution → Note: 4/0 = ∞ (infinity), vvvv large value, undefined, indeterminate. The function f(x) is defined when the denominator is nonzero, i.e., if x2 − 5x + 6 ≠ 0. Solving yields x ≠ 2 and x ≠ 3. Therefore, the domain of f includes all the real numbers of x except x = 2 and x = 3, i.e., the set (−∞,∞){2,3} or (−∞,2) ∪ (2,3) ∪ (3,∞). Let y x2 − 5x + 6 or x2 − 5x + (6 − y4) = 0. Making x the subject (use quadratic formula), we have x This function is defined if 25 0 or y ≥−16. Therefore, the range is the interval [−16,∞). 1.2 Evaluation of functions (d) f(x) = √x − 1 Solution Since f(x) is defined (or is a real number) if x−1 ≥ 0 or x ≥ 1, the domain of f is the interval [1√,∞−). Let y = x 1. Making x the subject, we have x = y2 + 1. This function is defined for any real number y. Therefore, the range is the interval (−∞,∞). (e) f(x) = 2|x − 3| + 4 Solution Since f(x) is defined for all real numbers, the domain of f is the interval (−∞,∞). Since for all |x − 3|≥ 0, the function f(x) = 2|x − 3| + 4 ≥ 4. Therefore, the range is all the values of y for which y ≥ 4 or the interval [4,∞). Exercise: 1. Find the domain and range of the following functions. (a) f(x) = 6 − x2.[ans: domain (−∞,∞), range (−∞,6]] 6 + 3x (b) f(x) =. [ans: domain (−∞,0.5) ∪ (0.5,∞), range (−∞,1.5) ∪ (1.5,∞)] x [ans: domain (−∞,2) ∪ (2,∞), range (−∞,1) ∪ (1,∞)] [ans: domain (−∞,2], range (−∞,∞ )] .[ans: domain (−∞,−4)∪[4,6)∪(6,∞), range 1.2 Evaluation of functions This involves replacing x in the function by the suggested value and retaining the rule of the function. Example(s): 1. Given f(x) = 2x+1. Find: (i) f(0), (ii) f(1), (iii) f(x+2), and (iv) for h = 0. h Solution f(x + h) − f(x) ̸ i) f(0) = 2(0) + 1 = 0 + 1 = 1 ii) f(1) = 2(1) + 1 = 2 + 1 = 3 iii) f(x + 2) = 2(x + 2) + 1 = 2x + 4 + 1 = 2x + 5 f(x + h) − f(x) [2(x + h) + 1] − [2x + 1] 2x + 2h + 1 − 2x − 1 2h iv) = = = = 2. h h h h f(x + h) f(x) 2. Given f(x) = 3x2 − 2x + 4. Find: (i) f(0), (ii) f(−1), (iii) f(x + 2), and (iv) − h for h ≠ 0. Solution i) f(0) = 3(0)2 − 2(0) + 4 = 0 + 0 + 4 = 4 ii) f(−1) = 3(−1)2 − 2(−1) + 4 = 3 + 2 + 4 = 9 iii) f(x + 2) = 3(x + 2)2 − 2(x + 2) + 4 = 3(x2 + 4x + 4) − 2x − 4 + 4 = 3x2 + 10x + 12 1.3 Composite functions iv) f(x + h) − f(x) [3(x + h)2 − 2(x + h) + 4] − [3x2 − 2x + 4] = hh ( 3x2 + 6hx + 3h2 − 2x − 2h + 4) − (3x2 − 2x + 4) 6hx + 3h2 − 2h == h h = 6x + 3h − 2 3. Given f(x) = x2 − 4x + 3. Find: (i) f(1), (ii) f(2), (iii) f(a), and (iv) f(a + h). Solution i) f(x) = x2 − 4x + 3 ⇒ f(1) = 12 − 4(1) + 3 = 0 ii) f(x) = x2 − 4x + 3 ⇒ f(2) = 22 − 4(2) + 3 = −1 iii) f(x) = x2 − 4x + 3 ⇒ f(a) = a2 − 4a + 3 iv) f(x) = x2 − 4x + 3 ⇒ f(a + h) = (a + h)2 − 4(a + h) + 3 4. Given ϕ(θ) = 2sinθ. Find: (i) ϕ( ), (ii) ϕ(0), and (iii) ϕ( ). Solution i) ϕ(θ) = 2sinθ ⇒ ii) ϕ(θ) = 2sinθ ⇒ iii) ϕ(θ) = 2sinθ ⇒ Exercise: (a) Given f(x) = x 3 + 2x + 1, find: (i) f(0), (ii) f(−a), (iii) f(x + 2), and (iv) − for h ≠ 0. 1 (b) Given g(x) = √ , find: (i) f(0), (ii) f(1), (iii) f(x+2), and (iv) x + 1 g(x + h) − g(x) ̸ for h = 0. h (c) Given p(x) = − , find: (i) f(0), (ii) f(−1), (iii) f(2−x), and (iv) − for h ≠ 0. 1 + 3x h (d) If f(x) = 2x2 − 4x + 1, find (i) f(1), (ii) f(0), (iii) f(2), (iv) f(a), and f(x + h). f(x + h) f(x) h (e) If f(x) = (x − 1)(x + 5), find (i) f(1), (ii) f(0), (iii) f(2), (iv) f(a + 1), and f( 1). (f) If f(θ) = cosθ, find ( i) ), and (v) f(π). (g) If f(x) = x 2, find (i) f(3), (ii) f(3.1), (iii) f(3.01), (iv) f(3.001), and − . f(3.001)f(3) 0.001 (h) If ϕ(x) = 2x, find (i) ϕ(0), (ii) ϕ(1), and (iii) ϕ(0.5). 1.3 Composite functions The composition of functions is a function of another function. Consider the function f with domain A and range B, and the function g with domain D and range E. If B is a subset of D, then the composite function (gof)(x) is the function with domain A and range E such that (gof)(x) = g(f(x)) For example, given f(x) = 2x + 1 and g(x) = 5x − 3. Then, (gof)(x) = g(f(x)) = g(2x + 1) = 5(2x + 1) − 3 = 10x + 2 6 2x p(x + h) p(x) Similarly, (fog)(x) = f(g(x)) = f(5x − 3) = 2(5x − 3) + 1 = 10x − 5 → Note: (fog)(x) ≠ (gof)(x). Exercise: 1. Given f(x) = x2 − 1, g(x) = x − 1 and h(x) = √x. Find: (a) (fog)(x) (b) (hog)(x) (c) (gog)(x) (d) (gohof)(x) 2. Consider the functions f(x) = x2 + 1 and g(x) = 1/x. Evaluate (a) (fog)(4) (b) (gof)(−1/2) 3. If f(x) = √x and g(x) = 4x + 2, find the domain of (fog)(x). [ans: x ≥−0.5 or (−∞,−0.5]] 2 Limits of functions Definition 2.1 (Basic limit definition). Let f(x) be a function and let a and L be real numbers. If f(x) approaches L as x approaches a from either RHS or LHS of a (but is not equal to a), then we say that f(x) has limit L as x approaches a, and is mathematically written as: Diagrammatically, we have → Note: lim f(x) is the value that f(x) approaches x→asa x approaches a, and a does not have to be in the domain of f. 2.1 Properties of limits Theorem 2.1. Suppose lim f(x) = L1 and lim g(x) = L2. Then, x→a x→a 1. [Addition/subtraction rule] lim [f(x) ± g(x)] = [lim f(x)] ± [lim g(x)] = L1 ± L2 x→a x→a x→a 2. [Scalar multiple] lim [λf(x)] = λ[lim f(x)] = λL1, where λ is a constant. x→a x→a 3. [Product rule] lim [f(x) · g(x)] = [lim f(x)] · [lim g(x)] = L1 · L2 x→a x→a x→a f(x) lim f( L1 4. [Quotient rule] lim [ ] = → = , provided g(a) ≠ 0. x→a g(x) xlim→a g(x) L2 1/n √ ( 5. lim n f(x) =lim f(x)) = (L1)1/n = √n L1 x→ax→a → Note: iff(x) = c(where c is a constant), then lim [f(x)] = lim [c] = c x→a x→a