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Module 1 Lecture Notes Module 1 Lecture Notes Contents 1.1 Identifying Functions.............................1 1.2 Algebraically Determining the Domain of a Function..........4 1.3 Evaluating Functions.............................6 1.4 Function Operations..............................7 1.5 The Difference Quotient...........................9 1.6 Applications of Function Operations.................... 10 1.7 Determining the Domain and Range of a Function Graphically.... 12 1.8 Reading the Graph of a Function...................... 14 1.1 Identifying Functions In previous classes, you should have studied a variety basic functions. For example, 1 p f(x) = 3x − 5; g(x) = 2x2 − 1; h(x) = ; j(x) = 5x + 2 x − 5 We will begin this course by studying functions and their properties. As the course progresses, we will study inverse, composite, exponential, logarithmic, polynomial and rational functions. Math 111 Module 1 Lecture Notes Definition 1: A relation is a correspondence between two variables. A relation can be ex- pressed through a set of ordered pairs, a graph, a table, or an equation. A set containing ordered pairs (x; y) defines y as a function of x if and only if no two ordered pairs in the set have the same x-coordinate. In other words, every input maps to exactly one output. We write y = f(x) and say \y is a function of x." For the function defined by y = f(x), • x is the independent variable (also known as the input) • y is the dependent variable (also known as the output) • f is the function name Example 1: Determine whether or not each of the following represents a function. Table 1.1 Chicken Name Egg Color Emma Turquoise Hazel Light Brown George(ia) Chocolate Brown Isabella White Yvonne Light Brown (a) The set of ordered pairs of the form (chicken name, egg color) shown in Table 1.1. This defines a function. For each input, there is exactly one output. More specifically, if you are told the name of the chicken, you know for certain what the egg color will be. (b) The set of ordered pairs of the form (egg color, chicken name) shown in Table 1.1. This DOES NOT define a function. For every input, there is not exactly one output. More specifically, if you are told the color is Light Brown, you have no way of knowing if the egg belongs to Hazel or Yvonne. Example 2: In Table 1.2, determine for which relations y is a function of x. Notice in Table 1.2 that there are a lot of similarities between the equations x2 + y = 4 and x + y2 = 4. However, we are determining if y is a function of x and thus are determining if every x-value is associated with exactly one y-value|not the reverse. Instructor: A.E.Cary Page 2 of 15 Math 111 Module 1 Lecture Notes Table 1.2: Identifying Functions Relation Function? Justification x y 1 1 No The input 2 has more than one output. 2 1 2 3 x y 8 49 Yes Every input has exactly one output. 12 60 16 49 x + y = 4 Yes There is exactly one y-value associated with every x-value. Note that the equation can be written explicitly in terms of y: y = 4 − x. x2 + y = 4 Yes There is exactly one y-value associated with every x-value. Note that the equation can be written explicitly in terms of y: y = 4 − x2. 2 x + y = 4 No The outputs 2 and −2 bothp correspond to the in- put 0. Furthermore, y = ± 4 − x and y cannot be written explicitly in terms of x. 2 2 x + y = 25 No The outputs 3 and −3 bothp correspond to the in- put 4. Furthermore, y = ± 25 − x2 and y cannot be written explicitly in terms of x. Instructor: A.E.Cary Page 3 of 15 Math 111 Module 1 Lecture Notes 1.2 Algebraically Determining the Domain of a Function Definition 2: Domain and Range • The domain of a function is the set of all possible inputs. • The range of a function is the set of all possible outputs. The largest possible set for each the domain and the range is the set of all real numbers. So far, there are two possible circumstances that restrict the domain of a function. When a number causes either of the following to occur, it must be excluded from the domain of a function: • Division by zero • The square root (or any even root) of a negative number Example 3: State the domain of each of the following functions in both set-builder and interval notation. p (a) f(x) = 7 − 2x Since this function contains a radical, we know that 7 − 2x ≥ 0. Solving for x, we find: 7 − 2x ≥ 0 7 ≥ 2x 7 ≥ x 2 7 x ≤ 2 7 7 The domain of f is x j x ≤ 2 . In interval notation, we write −∞; 2 (b) s(t) = 5t2 + 4t + 3 Noting that s is a quadratic function (and does not have division by zero or contain a radical), we know that the domain of s is all real numbers. In set-builder notation, we write ft j t is a real numberg. In interval notation, we write (−∞; 1). Instructor: A.E.Cary Page 4 of 15 Math 111 Module 1 Lecture Notes x2 − 7x + 10 (c) g(x) = x2 + x − 6 Since this function contains x2 + x − 6 in the denominator, we know that we must exclude values of x such that x2 + x − 6 = 0 from the domain: x2 + x − 6 = 0 (x + 3)(x − 2) = 0 x + 3 = 0 or x − 2 = 0 x = −3 or x = 2 The domain of f is fx j x 6= −3; 2g. In interval notation, we write (−∞; −3) S (−3; 2) S (2; 1). x + 2 (d) m(x) = x2 − 1 Since this function contains x2 − 1 in the denominator, we know that we must exclude values of x such that x2 − 1 = 0 from the domain: x2 − 1 = 0 x2 = 1 p x = ± 1 x = ±1 The domain of m is fx j x 6= −1; 1g. In interval notation, we write (−∞; −1) S (−1; 1) S (1; 1). p 3 t − 2 (e) k(t) = 5t + 40 Since this function contains 5t + 40 in the denominator, we know that we must exclude values of t such that 5t + 40 = 0 from the domain: 5t + 40 = 0 5t = −40 t = −8 The domain of k is ft j t 6= −8g. In interval notation, we write (−∞; −8) S (−8; 1). Note that the radical with an odd power does NOT limit the domain in any way. Instructor: A.E.Cary Page 5 of 15 Math 111 Module 1 Lecture Notes 4t + 1 (f) g(t) = p 5 − t p Since this functionp contains 5 − t in the denominator, we know that we must exclude values of t such that 5 − t = 0. Furthermore, since 5 − t is inside the radical, we must have 5 − t ≥ 0. Combined, this means that 5 − t > 0. Solving this for t: 5 − t > 0 5 > t t < 5 The domain of g is ft j t < 5g. In interval notation, we write (−∞; 5). p x − 5 (g) q(x) = x − 9 Since the function q contains x − 9 in the denominator, we know that values of x such that x − 9 = 0 must be excluded from the domain. Thus x 6= 9. Furthermore, since the function also has x − 5 inside a radical, we know that x − 5 ≥ 0. Thus x ≥ 5. Combined, we have x ≥ 5 and x 6= 9. The domain of q is then fx j x ≥ 5 and x 6= 9g. In interval notation, we write [5; 9) S (9; 1). 1.3 Evaluating Functions Recall that to evaluate a function is to find the function's value at a given input. One of the primary motivations of using function notation is that it is a means of communicating both the input and the output. For example, if we define a function by f(x) = x2, when we evaluate the function at x = −3, we see f(−3) = (−3)2 = 9 The statement f(−3) = 9 communicates both the input and the output. Conversely, if we had let y = x2 and found the value of y when x = −3, we would have seen y = (−3)2 = 9 Our result is the same: y = 9. But this does not tell us what the original input was! x2 Example 4: Let f(x) = . Evaluate the following, simplifying each expression as much as x + 1 possible: (a) f(−3) (c) f(0) (e) f(−x) (g) f(2x) (i) f(x + h) (b) f(1) (d) f(t) (f) −f(x) (h) f(x − 2) Instructor: A.E.Cary Page 6 of 15 Math 111 Module 1 Lecture Notes (a) (f) (−3)2 f(−3) = x2 −3 + 1 −f(x) = − 9 x + 1 = − 2 (g) (b) (1)2 (2x)2 f(1) = f(2x) = 1 + 1 2x + 1 1 4x2 = = 2 2x + 1 (c) (0)2 (h) f(0) = 0 + 1 (x − 2)2 = 0 f(x − 2) = (x − 2) + 1 (d) x2 − 4x + 4 = x − 1 (t)2 f(t) = t + 1 (i) (e) (−x)2 (x + h)2 f(−x) = f(x + h) = −x + 1 (x + h) + 1 x2 x2 + 2xh + h2 = = −x + 1 x + h + 1 1.4 Function Operations Function Operations • The sum of f and g, f + g, is defined by (f + g)(x) = f(x) + g(x). • The difference of f and g, f − g, is defined by (f − g)(x) = f(x) − g(x).
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