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PreCalculus Section 1-2 Functions Objectives • Determine whether a relation between two variables is a function. • Use function notation and evaluate functions. • Find the domain of functions. • Use functions to model and solve real-life problems. Definition of Relations A relation is a rule of correspondence that relates two sets. For instance, the formula I = 500r describes a relation between the amount of interest I earned in one year and the interest rate r. In mathematics, relations are represented by sets of ordered pairs (x, y) . Definition of Function A function from a set A to a set B is a relation that assigns to each element x in the set A exactly one element y in the set B (for each x value there is one and only one y value). The set A (x values) is called the domain of the function. The set B (y values) is called the range of the function. Ways to Write a Domain/Range Let’s use the function y = 2x, for example, the domain and range of this function is any real number. The domain and range of a function can be expressed using set, interval, and inequality notation. • Set Notation - Domain: {x| x∈ℝ} and Range: {y| y∈ℝ} , where ∈ means “element of” and ℝ means “real numbers”. • Interval Notation - Domain: (-∞, ∞) and Range: (-∞, ∞), where -∞ represents “negative infinity” and ∞ represents “positive infinity”. Open parentheses are used to indicate the interval is infinite, does not include the endpoints. A square bracket ‘[ ]’ indicates that the interval includes the endpoints and must never be used with -∞ or ∞. • Inequality Notation - Domain: -∞< x <∞ and Range: -∞< y <∞. Set Builder Notation • A method of representing points on the real number line. “such that” : or | “set of” • {x: 2x + 3 > 5, xєR} Variable “all x” Rule(s) that defines the variable Read “the set of all x such that 2x + 3 > 5 and x is an element of real numbers” π Set Builder Notation • Other useful set notation – U Union (or statement) – ∩ Intersection (and statement) – \ excludes whatever follows – Ø empty set or null set π Graphing a set on the real number line • If the end point of a set is included, then a “closed circle” is used. – {x: x ≥ 3} • If the end point of a set is excluded, then an “open circle” is used. – {x: x > 3} π Examples • Write the following intervals on the number line in set builder notation. 1) | | | | | | | | –4 –3 –2 –1 0 1 2 3 { x : x >-2, x𝛜ℝ } If the variable is on the left side of the inequality, then the inequality sign is the same direction as the arrow. | 4 Continued 2) | | | | | | | | | –4 –3 –2 –1 0 1 2 3 4 { y | y ≤ 1, y𝛜ℤ } Another 3) | | | | | | | | | –4 –3 –2 –1 0 1 2 3 4 Conjunction “and statement” x ≥-3 and x < 2 As compound inequality -3 ≤ x < 2 Set builder { x | -3 ≤ x < 2, x𝛜ℝ } One more 4) | | | | | | | | | –4 –3 –2 –1 0 1 2 3 4 Disjunction “or statement” x ≤-3 or x > 0 Disjunction must be written with the word or (or symbol ∪). Set builder { x : x ≤-3 ∪ x > o, x𝛜ℝ } Last one 5) | | | | | | | | | –4 –3 –2 –1 0 1 2 3 4 x equals all real numbers except 0 { x : x𝛜ℝ\0 } Interval Notation • Uses an interval, lower endpoint coma upper endpoint, and “square brackets or square brackets & round brackets” to indicate if the endpoint is included or excluded. – Endpoint included: lower [ or upper ] – Endpoint excluded: lower (, ] or upper ),[ – Note, ∞ is never included. π Notation Dilemma ],( & ),[ mean the same as Both notations mean that the number IS NOT a solution! [ & ] mean the same as Both notations mean that the number IS a solution. Intervals on the real number line. Interval type Notation Inequality Closed Open [ a, b] ( a, b) [ a, b) ( a, b] a xb a xb Graph [ ] a b ( ) a b [ ) a b a xb a xb ( ] a b Intervals on the real number line. Interval Type Notation Inequality [ a, ) xa Graph [ a Open ( a, ) (, b] xa ( xb a ] b Open Entire real line (, b) (, ) xb ) b Interval Notation Summary: Example 1) Write an inequality to represent the interval (2, 10]. This notation represents all real numbers x between 2 and 10, including 10. Inequality notation 2 x 10 2) Write an inequality to represent the interval [-5, ∞). This notation represents all real numbers x greater than or equal to –5. Inequality notation x 5 Function Diagram Set B Function Set A y y x y x y x x Domain y Range Example: Determine whether the relation represents y as a function of x. a) {(-2, 3), (0, 0), (2, 3), (4, -1)} Function b) {(-1, 1), (-1, -1), (0, 3), (2, 4)} Not a Function Your Turn: Determine whether each relation is a function? {(2, -2), (3, 0), (4, 1), (5, -1)} Function {(4, -3), (2, 0), (5, -2), (3, 1), (2, -1)} Not Function {(2, -1), (3, -1), (4, -1), (5, -1)} Function {(0, 0), (3, -2), (5, 0), (2, -3), (0, 5)} Not Function Four Ways to Represent a Function • Verbally (description in words) • Numerically (by a table of values) • Visually (by a graph) • Algebraically (by an equation or formula) A Function Verbally The amount of sales tax depends on the amount of the purchase. OR Divide x seconds by 5 to obtain y miles. A Function Numerically Use a table of values: x (seconds) 1 2 3 4 y (miles) 0.2 0.4 0.6 0.8 5 1 6 7 1.2 1.4 A Function Visually Any function can be visually represented by a graph. Dependent variable Independent variable A Function Algebraically Any function can be algebraically represented by a equation in two variables. y = 3x + 10 OR f(x) = 3x + 10 Testing for Functions 1. Set of ordered pairs or points on a graph. • Use definition of a function and determine whether each input value (x) is matched with exactly one output value (y). 2. Function represented algebraically (equation). • • Solve for y in terms of x. Determine whether each value of x corresponds to exactly one value of y. Testing for Functions Example: Determine whether the relation represents y as a function of x. y 4 3 2 1 x 4 3 2 1 2 3 4 1 2 3 4 The ordered pairs in the graph are: (– 3, 0), (– 2, 2), (– 1, 4), (0, 1), (1 – 2), and (2, 1). y is a function of x. Your Turn Determine whether the relation represents y as a function of x. y 5 4 3 2 1 0 -4 -3 -2 -1 0 1 2 3 4 -1 -2 -3 y is not a function of x. x Example Determine whether y is a function of x. 1. x2+y=3 y=-x2+3 solve for y each value of x corresponds to exactly one value of y, so y is a function of x. 2. x2+y2=8 y2=8-x2 subtract x2 from both sides y 8 x2 solve for y each value of x corresponds to 2 values of y, so y is not a function of x. Your Turn 1. y2-x=0 Not Function 2. y=√x Function 3. y3+x=1 Function Function Notation - f(x) : “f of x” Functions represented by equations are often named using a letter such as f or g. The symbol f (x), read as “the value of f at x” or simply as “f of x”, is the element in the range of f that corresponds with the domain element x. That is, y = f (x) Inputs & Outputs The domain elements, x, can be thought of as the inputs and the range elements, f (x), can be thought of as the outputs. Function Input Output f x f (x) Independent variable Dependent variable Evaluating Functions To evaluate a function f (x) at x = a, substitute the specified value a for x into the given function. 2 Example: Let f (x) = x – 3x – 1. Find f (–2). 2 f (x) = x – 3x – 1 2 f (–2) = (–2) – 3(–2) – 1 Substitute –2 for x. f (–2) = 4 + 6 – 1 Simplify. f (–2) = 9 The value of f at –2 is 9. Your Turn 2 Let f (x) = 4x – x . Find f (x + 2). 2 f (x) = 4x – x f (x + 2) = 4(x + 2) – (x + 2)2 Substitute x + 2 for x. f (x + 2) = 4x + 8 – (x2 + 4x + 4) Expand (x + 2)2. 2 f (x + 2) = 4x + 8 – x – 4x – 4 f (x + 2) = 4 – x2 Distribute –1. The value of f at x + 2 2 is 4 – x . Piecewise Function • A piecewise function is an equation with multiple parts in which each part matches up with a different part of the domain. Piecewise Function A piecewise function is an equation with multiple parts or pieces. It is made up of a series of different equations. Each different equation covers a different set of input values (or numbers on the x-axis) over the domain. A different piece of the domain is paired with each equation. The domain determines if the end points on the line are included or not. If the value is stated as equal to, then the value is represented with a closed circle on the line. If the value in the equation is stated as less than or greater than, but not equal too, then it is represented by an open circle. Piecewise Example: x-1 if x<-1 f(x)= -1 if x=-1 x+1 if x>-1 A closed dot indicates that the graph does not extend beyond this point, and the point belongs to the graph. An open dot indicates that the graph does not extend beyond this point and the point does not belong to the graph. An arrow indicates that the graph extends indefinitely in the direction in which the arrow points. Piecewise Functions 2 x 1,if x 1 f x 3x 1,if x 1 •One equation gives the value of f(x) when x ≤ 1 •And the other when x>1 Example: Evaluate f(x) when x=0, x=2, x=4 x 2,if x 2 f ( x) 2 x 1,if x 2 •First you have to figure out which equation to use •You NEVER use both X=0 So: This one fits 0+2=2 into the top f(0)=2 equation X=2 This So: one fits here 2(2) + 1 = 5 f(2) = 5 X=4 So: one fits here This 2(4) + 1 = 9 f(4) = 9 Your Turn Evaluate the function when x=-2, 0, and 2. 3x 4, x 0 f ( x) 3x 1, x 0 f(-2)=-10 f(0)=1 f(2)=7 The Domain • When defining a function you MUST specify the set of input values to be used (the domain). – Example: y=3x2-5, x≥1 • A function is only defined over it’s domain, outside of it’s domain the function does not exist. • If the domain is omitted we assume an implied domain. Implied Domain • When no domain or restrictions on the xvalues have been provided, we assume the largest possible set of x-values for which the relation is defined as the implied domain. Determining Implied Domain • Look for restrictions on the input or values of x where the relation is undefined. • These restrictions on x are caused by; – Dividing by zero – Taking the square root of a negative number Example Find the domain of the function 3x 10 x 8 g ( x) 2 x 3x 28 2 Solution: We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0. Solve x2 3x 28 = 0. (x 7)(x + 4) = 0 x 7 = 0 or x + 4 = 0 x = 7 or x = 4 The domain consists of the set of all real numbers except 4 and 7 or {x|x 4 and x 7}. Example Find the domain of the function f ( x) x 3 The function is defined only for x-values for which x – 3 0. Solving the inequality yields x–30 x3 Domain: {x| x 3} Your Turn x2 Find the domain of the function g ( x) 2 x 1 The x values for which the function is undefined are excluded from the domain. The function is 2 undefined when x – 1 = 0. 2 x –1=0 (x + 1)(x – 1) = 0 x=1 Domain: {x| x 1} Your Turn Find the domain of the function f ( x) x 1 The function is defined only for x-values for which x – 1 0. Solving the inequality yields x–10 x1 Domain: {x| x 1} Applications Example: The total volume V (in billions of dollars) of farm real estate in the United States from 1980 through 1997 can be modeled by 13.826t 2 59.96t 775.4, 0 t 6 V (t ) 7 t 17 382.4 28.7t , where t=0 represents 1980. Use this model to approximate the value of farm real estate 1n 1982, 1987, and 1996. 1982 : t 2 use 13.826t 2 59.96t 775.4 V (2) 13.826(2) 2 59.96(2) 775.4 V (2) 840 1982 : $840 billion 1987 : t 7 use 382.4+28.7t V (7) 382.4 28.7(7) V (7) 583 1987 : $583 billion 1996 : t 16 use 382.4+28.7t V (16) 382.4 28.7(16) V (16) 842 1996 : $842 billion Your Turn A baseball is hit at a point 3 feet above ground at a velocity of 100 feet per second and an angle of 45˚. The path of the baseball is modeled by y = -0.0032x2 + x + 3, where x and y are measured in feet. Will the baseball clear a 20-foot fence located 280 feet from home plate? Yes, the ball will be 32 feet off the ground when it passes over the fence. The Difference Quotient The difference quotient of a function f is an expression of the form where h ≠ 0. Where does it come from? f (x h) f (x) h Example: Calculating a Difference Quotient f(x+h)-f(x) 2 Find for f(x)=x 2 x 5 h First find f(x+h) f(x+h)=(x+h) 2(x+h)-5 2 x 2hx h 2 x 2h 5 2 2 f(x+h)-f(x) Find for f(x)=x 2 2 x 5 h Use f(x+h) from the previous slide f(x+h)-f(x) Second find h 2 2 2 x 2 hx h 2 x 2 h 5 x 2 x 5 f(x+h)-f(x) h h x 2 2hx h 2 2 x 2h 5 x 2 2 x 5 h 2hx h 2 2h h h 2x h 2 h 2x+h-2 Your Turn: Find and simplify the expressions if f ( x) 2 x 1 Find f(x+h) f(x+h)-f(x) Find , h0 h f ( x h) 2 x 2h 1 f ( x h) f ( x ) 2 h Example: Calculating a Difference Quotient Let the distance d in feet that a racehorse runs in t seconds be d(t) = 2t2 for 0 ≤ t ≤ 10. (a) Find d(t +h). (b) Find the difference quotient of d and simplify the result. (c) Evaluate the difference quotient for t =7 and h=0.1. Interpret your results. (d) Evaluate the difference quotient for t =4 and h=1. Then sketch a graph that illustrates the result. (a) Find d(t +h). (b) Find the difference quotient of d and simplify the result. Solution (a) d t h 2 t h 2 t 2 2th h 2 2 2t 4th 2h 2 (b) d t h d t h 2 4th h h h 4t h h 4t h 2 2t 4ht 2h 2t h 2 2 2 (c) Evaluate the difference quotient for t =7 and h=0.1. Interpret your results. Solution (c) If t = 7 and h = 0.1, then the difference quotient becomes 4t + 2h = 4(7) +2(0.1) = 28.2. The average rate of change, or average velocity, of the horse from 7 seconds to 7 + 0.1 = 7.1 seconds is 28.2 feet per second. (d) Evaluate the difference quotient for t =4 and h=1. Solution (d) If t = 4 and h = 1, then 4t + 2h = 4(4) + 2(1) = 18 If t = 4 and d = 2(4)2 = 32, the first point is (4, 32). If t = 4 and h = 1, it follows that t + h = 4 + 1 = 5 so d = 2(5)2 = 50, the second point is (5, 50). Thus the slope of the line passing through (4, 32) and (5, 50) is m = 18. REVIEW PROBLEMS Problems 1. Let A = {2, 3, 4, 5} and B = {-3, -2, -1, 0, 1}. Which of the following sets of ordered pairs represent functions from set A to set B? a) – b) – c) – d) – {(2, -2), (3, 0), (4, 1), (5, -1)} Each element of A is matched with exactly one element of B, Function. {(4, -3), (2, 0), (5, -2), (3, 1), (2, -1)} The element 2 in A is matched with two elements, 0 and -1, in B, Not a Function. {(2, -1), (3,-1), (4, -1), (5, -1)} Each element of A is matched with exactly one element of B, Function. {(3, -2), (5, 0), (2, -3)} The element 4 in A is not matched with an element of B, Not a Function. Problems 2. Determine whether the equation x2 + y2 = 8 represents y as a function of x. • y is not a function of x. 3. Let f(x) = 10 – 3x2 and find the following: a) f(2) – f(2) = -2 b) f(-4) – f(-4) = -38 c) f(x – 1) – f(x – 1) = -3x2 + 6x +7 Problems 4. Evaluate the function below at x = -2, 0, and 2. 3x 4, x 0 f ( x) 3x 1, x 0 • f(-2) = -10, f(0) = 1, f(2) = 7 5. Find the domain of g ( x) x 16 • x ≥ 16 or [16, ∞) . Summary of Function Terminology • Function: A function is a relationship between two variables such that to each value of the independent variable there corresponds exactly one value of the dependent variable. • Function Notation: y = f(x) – – – – f is the name of the function. y is the dependent variable, or the output value. x is the independent variable, or the input value. f(x) is the value of the function at x, or the output value. • Domain: The domain of a function is the set of all values (inputs) of the independent variable for which the function is defined. • Range: The range of a function is the set of all values (outputs) assumed by the dependent variable. • Implied Domain: If f is defined by an algebraic expression and the domain is not specified, the implied domain consists of all real numbers for which the expression is defined. Homework • Section 1.2, pg. 24 – 29: Vocabulary Check #1 – 5 all Exercises: #1-23 odd, 27-39 odd, 47-59odd, 83, 87, 89 • Read Section 1.3, pg. 30-37