Chapter 2

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## Chapter 2

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**Chapter 2**Functions and Graphs**Functions**• Definition: A function is a rule that assigns to each input number exactly one output number. The set of all input numbers to which the rule applies is called the domain of the function. The set of all (corresponding) output numbers is called the range. • Notation: if y is a function of x, we write • y = f(x) • Other common symbols for functions include but are not limited to g, h, F, G**Example #1**• y = x + 2 • Is a one-to–one mapping because we get a unique value of y for any chosen value of x, so satisfies the definition of a function • Domain of the function is any real number; x can take on any value • Range is 2 plus the value of x and can be any real number as well.**Example #2**• y = 2x • Is a function because values of x are mapped uniquely to single values of y • Domain of the function is the set of real numbers; x can be any real number • Range of the function is the set of real numbers; y is always twice the value of x • Note that x can be written as a function of y, x = y/2.**Example #3**• y2 = x • Is not a function. Suppose x = 9. Then, y2 = 9, but y either could equal 3 or –3 • y = 1/(x-6) • Is a function because x is mapped to a unique value of y • Domain of function is the set of real numbers except x = 6, because division by zero is undefined.**Equality of Functions**• Two functions, f and g are equal (f = g) if • The domain of f is equal to the domain of g • For every x in the domain of f and g, the values of the two functions are the same; that is f(x) = g(x)**Example #1 (p. 79-80)**• Which of the following functions are equal • f(x) = (x + 2)(x + 1)/(x – 1) • g(x) = x + 2 • h(x) = x + 2 if x 1; = 0 if x = 1 • k(x) = x + 2 if if x 1; = 3 if x = 1 • Domains of g, h, and k the set of all real numbers and are equal, but the domain of f is the set of all real numbers except x = 1 • Function values are equal for g, h, k when x 1. When x = 1, only function values for g and k are equal**Example #2a (p.80)**• Find the domain of f(x) = x/(x2 –x – 2) • The domain would be the set of all real numbers except those values of x which set the denominator equal to zero • These values are found by factoring • (x2 –x – 2) = (x + 1)(x - 2) • X = -1, 2 • So the domain is the set of all real numbers , except x =-1, 2**Domain and Function Values**• Let g(x) = 3x2 – x +5, and note that the domain of g is the set of all real numbers • Then • g(z) = 3z2 – z + 5 • G(r2) = 3(r2)2 – r2 + 5 • So, g(r2) = 3r4 – r2 + 5 • G(x + h) = 3(x + h)2 - (x + h) +5 • So, g(x + h) = 3x2 + 3h2 + 6xh – x – h + 5**Difference Quotient**• Let f(x) = x2. Find [f(x + h) – f(x)]/h • [f(x + h) – f(x)]/h = [(x + h)2 - x2]/h • = [x2 + h2 + 2xh – x2]/h • = [h2 + 2xh]/h • = h[h + 2x]/h • = h + 2x**Demand Function**• Suppose the demand function for pizzas at a local pizza parlor is • p = f(q) = 26 – (q/40) • Mathematically, the domain of the function is all real values of q, but economically makes sense to restrict the domain to q 0**Demand (cont.)**• If 200 pizzas are sold each week, what is the current price? $21 • If the owner wants to double the number of pizzas sold, what should the price be? $16 • If the price is $18.50, how many pizzas are sold? 300**Homework**• P. 83: 3,7,31,33 and maybe a few more of the odd numbered exercises • P.84: 47, 49, 53**Special Functions**• Constant functions: h(x) = c • Polynomial functions: • y = f(x) = cnxn + cn-1xn-1 + + c1x +c0 where n is a non-negative integer and the ci are constants with cn 0. The number n is called the degree of the polynomial and cn is called the leading coefficient.**Special functions (cont.)**• A rational function is the ratio of two polynomial functions. • Case defined functions • 1 if –1 s 1 • F(s) = 0 if 1 s 2 • s-3 if 2 < s <8**Special Functions (cont.)**• Absolute value function: Already have encountered this one; actually an example of a case defined function • x = x if x 0; x = -x if x < 0 • Factorials: The symbol r!, with positive integer r is read “r factorial.” It represents the product of the first r positive integers; e.g., 3! = 6 • Note: 0! = 1, by definition**Examples**• Suppose monthly health insurance premiums for an individual are $125? • Write monthly premiums as a function of the number of visits the individual makes to the doctor: P = 125 + av, where a=0. • How do the health insurance premiums change as number of doctor visits increase? • What kind of function is this?**More examples**• The manager of a baseball team has 9 players available for a particular game. How many different lineups are possible? 9! = 362,880 • The function d(t) = 3t2 represents the distance in meters a car will travel in t seconds when it has constant acceleration of 6 m/sec. • What kind of function is this? What is its degree? What is the leading coefficient?**One More Example**• Let f(x) = 3/x3. Is f(x) a polynomial function? • f(x) = 3/x3 = 3x-3 • Not a polynomial because the exponent of x is not a non-negative integer. • Similarly, g(x) = x1/2 is not a polynomial either because the exponent is not an integer**Homework**• pp. 88-89: 3,5,11,29,31,33**Combinations of Functions**• Addition, multiplication, subtraction, division, multiplication by a constant • Example: f(x) = x2 and g(x) = 3x • f(x) + g(x) = x2 + 3x • f(x)g(x) = 3x3 • f(x) – g(x) = x2 – 3x • f(x)/g(x) = x2/3x = x/3 • cf(x) = cx2**Composition**• Define two functions of x: f(x) and g(x). A composite function then can be defined as: f(g(x)), where the domain is the set of x in the domain of g(x) such that g(x) is in the domain of f (p.91). • Example: Let f(x) = x2 and g(x) = x + 1 • f(g(x) = f(x + 1) = (x + 1)2 • g(f(x)) = g(x2) = x2 + 1 • Note that composition is not the same as multiplication: f(x)g(x) = x2(x + 1)**Examples**• Suppose for a particular business, fixed cost is $50/yr and variable cost is depends on the number of units produced, q, such that V(q) = 2q2. Find the total cost function and the total cost of producing 10 units. • TC(q) = 2q2 + 50 • TC(10) = 200 +50 = 250**Example**• In the previous problem, find average fixed cost, average variable cost, and average total cost. • AFC(q) = 50/q • AVC(q) = 2q2/q = 2q • ATC(q) = 2q + 50/q**Example**• The demand curve for a particular product is: p = f(q) = 12 – q. Find total revenue and average revenue • TR(q) = q(f(q)) = q(12 – q) = 12q – q2 • AR(q) = TR(q)/q = f(q) = 12 – q • So, the average revenue curve is just the demand curve for the product**Example**• A realtor needs to determine the square footage yard space for a piece of property that is for sale. The entire lot is shaped as rectangle and is 200 feet x 300 feet. The home on the lot is 2000 sq. ft. And there is a 10ft. x 20ft. Driveway. Also, here is a circular lake with diameter 30 ft., half of which is on the property and half of which is on the adjacent property.**Solution**• Let lot size = length x width of property: LS = l(q)w(q), where q = feet. • Then LS = 300q x 200q = 60,000q2 • YS = LS – 2000 –200 – ½(r2) • YS = 60,000 – 2000 – 200 – ½ (3.1416)152 So, YS = 57,446.57 sq. ft.**Example (#19, p. 94)**• A manufacturer finds that the total number of units of output per day, q, is a function of the number of employees, m. • q = q(m) = (m – m2)/4 • Total revenue for the manufacturer is: r(q)= $40q. • Write total revenue as a function of the number of employees • r(q) = r(q(m)) = 40(m – m2)/4 = 10(m – m2) • This relationship describes the total revenue from selling the output of m employees.**Homework**• P. 94: 3,7,9,13,19 (worked out on this set of slides)**Inverse Functions**• Turns out that this is just an application of composition, the topic just treated. • Define the function I(x) = x such that I(f(x)) = f(I(x)) = f(x), where we can refer to I(x) as the identity function • Then, the inverse of the function f(x) is f-1(x). f-1(x) has the property that • f(f-1(x)) = f-1(f(x)) = I(x) • Note that f-1(x) is not the reciprocal of f(x)**Inverse Functions (cont.)**• Under what conditions will the function f(x) have an inverse? • First need a definition • A function that satisfies the following condition is called a one-to-one function • For all a and b, if f(a) = f(b), then a = b • This means that if we insert two distinct input values, we get distinct output values from the function • A function has an inverse if it is one-to-one**How to Find the Inverse**• x = 1 +**How to Find the Inverse**• To find the inverse of a one-to-one function, y = f(x), solve for x in terms of y obtaining g(y). Then f-1(x) = g(x) • Example • y = (x - 1)2; restrict domain to x 1 • Solve for x in terms of y; x = y1/2 +1 • So, f-1(x) = x1/2 + 1 • f(f-1(x) = [(x1/2 + 1) -1]2 = I(x) = x**Graphical Interpretation**f(x) = (x – 1)2 f(x) I(x) = x 45° x 1**Examples**• Suppose that the demand curve for a product is: p = 50 – 2q. Find the inverse of this demand function, if it exists (restrict the domain to q 0) • So, p = f(q) = 50 – 2q • Solve for q in terms of p; q = 25 – ½ p • Then f-1(q) = 25 – ½ q • f-1(f(q)) = [25 – ½ (50 – 2q)] = q = I(q)**Another Example**p TR MR D q**Example (cont.)**• Given the demand curve, find the values of p and q that maximize total revenue • p = f(q) = 50 - 2q “demand” • P = g(q) = 50 – 4q “marginal revenue” • TR maximized when g(q) = 0 • g-1(g(q)) = g-1(0) • So, find g-1(q) and set q = 0 • Solve for q in terms of p; q = 12.5 – 0.25p • g-1(q) = 12.5 – 0.25q**Example (Cont.)**• We find that the solution of g-1(q) = 0 is q = 12.5 • If q = 12.5, what is p? • Look at the demand curve: p = 50 – 2q • P = 25 • Total revenue is maximized by setting p = $25