Chapter 0 Functions.

Chapter 0 Functions

Chapter Outline Functions and Their Graphs Some Important Functions
The Algebra of Functions Zeros of Functions – The Quadratic Formula and Factoring Exponents and Power Functions Functions and Graphs in Applications

§ 0.1 Functions and Their Graphs

Section Outline Rational and Irrational Numbers The Number Line
Open and Closed Intervals Applications of Functions Domain of a Function Graphs of Functions The Vertical Line Test Graphing Calculators Graphs of Equations

Rational & Irrational Numbers
Definition Example Rational Number: A number that may be written as a finite or infinite repeating decimal, in other words, a number that can be written in the form m/n such that m, n are integers Irrational Number: A number that has an infinite decimal representation whose digits form no repeating pattern

The Number Line The Number Line
A geometric representation of the real numbers is shown below.

Open & Closed Intervals
Definition Example Open Interval: The set of numbers that lie between two given endpoints, not including the endpoints themselves Closed Interval: The set of numbers that lie between two given endpoints, including the endpoints themselves [-1, 4]

Functions in Application
EXAMPLE (Response to a Muscle) When a solution of acetylcholine is introduced into the heart muscle of a frog, it diminishes the force with which the muscle contracts. The data from experiments of the biologist A. J. Clark are closely approximated by a function of the form where x is the concentration of acetylcholine (in appropriate units), b is a positive constant that depends on the particular frog, and R(x) is the response of the muscle to the acetylcholine, expressed as a percentage of the maximum possible effect of the drug. (a) Suppose that b = 20. Find the response of the muscle when x = 60. (b) Determine the value of b if R(50) = 60 – that is, if a concentration of x = 50 units produces a 60% response. SOLUTION (a) This is the given function.

CONTINUED Replace b with 20 and x with 60. Simplify the numerator and denominator. Divide. Therefore, when b = 20 and x = 60, R (x) = 75%. (b) This is the given function. Replace x with 50. Replace R(50) with 60.

CONTINUED Simplify the numerator. Multiply both sides by b + 50 and cancel. Distribute on the left side. Subtract 3000 from both sides. Divide both sides by 60. Therefore, when R (50) = 60, b = 33.3.

Functions EXAMPLE SOLUTION If , find f (a - 2).
This is the given function. Replace each occurrence of x with a – 2. Evaluate (a – 2)2 = a2 – 4a + 4. Remove parentheses and distribute. Combine like terms.

Domain Definition Example
Domain of a Function: The set of acceptable values for the variable x. The domain of the function is

Graphs of Functions Definition Example
Graph of a Function: The set of all points (x, f (x)) where x is the domain of f (x). Generally, this forms a curve in the xy-plane.

The Vertical Line Test Definition Example
Vertical Line Test: A curve in the xy-plane is the graph of a function if and only if each vertical line cuts or touches the curve at no more than one point. Although the red line intersects the graph no more than once (not at all in this case), there does exist a line (the yellow line) that intersects the graph more than once. Therefore, this is not the graph of a function.

Graphing Using a Graphing Calculator
Graphing Calculators Graphing Using a Graphing Calculator Step Display 1) Enter the expression for the function. 2) Enter the specifications for the viewing window. 3) Display the graph.

Graphs of Equations EXAMPLE SOLUTION false
Is the point (3, 12) on the graph of the function ? SOLUTION This is the given function. Replace x with 3. Replace f (3) with 12. Simplify. false Multiply. Since replacing x with 3 and f (x) with 12 did not yield a true statement in the original function, we conclude that the point (3, 12) is not on the graph of the function.

§ 0.2 Some Important Functions

Section Outline Linear Equations Applications of Linear Functions
Piece-Wise Functions Quadratic Functions Polynomial Functions Rational Functions Power Functions Absolute Value Function

Linear Equations y = mx + b x = a Equation Example
(This is a linear function) x = a (This is not the graph of a function)

since it assigns the same number b to every value of x
Linear Equations CONTINUED Equation Example y = b In this case, f(x) is called a constant function, since it assigns the same number b to every value of x

Applications of Linear Functions
EXAMPLE (Enzyme Kinetics) In biochemistry, such as in the study of enzyme kinetics, one encounters a linear function of the form , where K and V are constants. If f (x) = 0.2x + 50, find K and V so that f (x) may be written in the form, Find the x-intercept and y-intercept of the line in terms of K and V. SOLUTION (a) Since the number 50 in the equation f (x) = 0.2x + 50 is in place of the term 1/V (from the original function), we know the following. 50 = 1/V Explained above. 50V = 1 Multiply both sides by V. V = 0.02 Divide both sides by 50. Now that we know what V is, we can determine K. Since the number 0.2 in the equation f (x) = 0.2x + 50 is in place of K/V (from the original function), we know the following.

CONTINUED 0.2 = K/V Explained above. 0.2V = K Multiply both sides by V. 0.2(0.02) = K Replace V with 0.02. 0.004 = K Multiply. Therefore, in the equation f (x) = 0.2x + 50, K = and V = 0.02. (b) To find the x-intercept of the original function, replace f (x) with 0. This is the original function. Replace f (x) with 0. Solve for x by first subtracting 1/V from both sides.

CONTINUED Multiply both sides by V/K. Simplify. Therefore, the x-intercept is -1/K. To find the y-intercept of the original function, we recognize that this equation is in the form y = mx + b. Therefore we know that 1/V is the y-intercept.

Piece-Wise Functions EXAMPLE SOLUTION
Sketch the graph of the following function SOLUTION We graph the function f (x) = 1 + x only for those values of x that are less than or equal to 3. Notice that for all values of x greater than 3, there is no line.

Piece-Wise Functions CONTINUED
Now we graph the function f (x) = 4 only for those values of x that are greater than 3. Notice that for all values of x less than or equal to 3, there is no line.

Piece-Wise Functions CONTINUED
Now we graph both functions on the same set of axes.

Quadratic Functions Quadratic Function: A function of the form
Definition Example Quadratic Function: A function of the form where a, b, and c are constants and a 0.

Polynomial Functions Polynomial Function: A function of the form
Definition Example Polynomial Function: A function of the form where n is a nonnegative integer and a0, a1, ...an are given numbers.

Rational Functions Definition Example Rational Function: A function expressed as the quotient of two polynomials.

Power Functions Power Function: A function of the form Definition
Example Power Function: A function of the form

Absolute Value Function
Definition Example Absolute Value Function: The function defined for all numbers x by such that |x| is understood to be x if x is positive and –x if x is negative

§ 0.3 The Algebra of Functions

Section Outline Adding Functions Subtracting Functions
Multiplying Functions Dividing Functions Composition of Functions

Adding Functions EXAMPLE SOLUTION
Given and , express f (x) + g (x) as a rational function. SOLUTION f (x) + g (x) = Replace f (x) and g (x) with the given functions. Multiply to get common denominators. Evaluate. Add. Simplify the numerator.

Adding Functions CONTINUED Evaluate the denominator.
Simplify the denominator.

Subtracting Functions
EXAMPLE Given and , express f (x) - g (x) as a rational function. SOLUTION f (x) - g (x) = Replace f (x) and g (x) with the given functions. Multiply to get common denominators. Evaluate. Subtract. Simplify the numerator.

Subtracting Functions
CONTINUED Evaluate the denominator. Simplify the denominator.

Multiplying Functions
EXAMPLE Given and , express f (x)g (x) as a rational function. SOLUTION f (x)g (x) = Replace f (x) and g (x) with the given functions. Multiply the numerators and denominators. Evaluate.

Dividing Functions EXAMPLE SOLUTION
Given and , express [f (x)]/[g (x)] as a rational function. SOLUTION f (x)/g (x) = Replace f (x) and g (x) with the given functions. Rewrite as a product (multiply by reciprocal of denominator). Multiply the numerators and denominators. Evaluate.

Composition of Functions
EXAMPLE (Conversion Scales) Table 1 shows a conversion table for men’s hat sizes for three countries. The function converts from British sizes to French sizes, and the function converts from French sizes to U.S. sizes. Determine the function h (x) = f (g (x)) and give its interpretation. SOLUTION h (x) = f (g (x)) This is what we will determine. In the function f, replace each occurrence of x with g (x). Replace g (x) with 8x + 1.

Composition of Functions
CONTINUED Distribute. Multiply. Therefore, h (x) = f (g (x)) = x + 1/8. Now to determine what this function h (x) means, we must recognize that if we plug a number into the function, we may first evaluate that number plugged into the function g (x). Upon evaluating this, we move on and evaluate that result in the function f (x). This is illustrated as follows. g (x) f (x) British French French U.S. h (x) Therefore, the function h (x) converts a men’s British hat size to a men’s U.S. hat size.

§ 0.4 Zeros of Functions – The Quadratic Formula and Factoring

Section Outline Zeros of Functions Quadratic Formula
Graphs of Intersecting Lines Factoring

Zeros of Functions Definition Example Zero of a Function: For a function f (x), all values of x such that f (x) = 0.

Quadratic Formula Definition Example Quadratic Formula: A formula for solving any quadratic equation of the form The solution is: There is no solution if These are the solutions/zeros of the quadratic function

Graphs of Intersecting Functions
EXAMPLE Find the points of intersection of the pair of curves. SOLUTION The graphs of the two equations can be seen to intersect in the following graph. We can use this graph to help us to know whether our final answer is correct.

CONTINUED To determine the intersection points, set the equations equal to each other, since they both equal the same thing: y. Now we solve the equation for x using the quadratic formula. This is the equation to solve. Subtract x from both sides. Add 9 to both sides. We now recognize that, for the quadratic formula, a = 1, b = -11, and c =18. Use the quadratic formula. Simplify.

CONTINUED Simplify. Simplify. Rewrite. Simplify. We now find the corresponding y-coordinates for x = 9 and x = 2. We can use either of the original equations. Let’s use y = x – 9.

CONTINUED Therefore the solutions are (9, 0) and (2, -7). This seems consistent with the two intersection points on the graph. A zoomed in version of the graph follows.

Factoring EXAMPLE SOLUTION Factor the following quadratic polynomial.
This is the given polynomial. Factor 2x out of each term. Rewrite 3 as Now I can use the factorization pattern: a2 – b2 = (a – b)(a + b). Rewrite

Factoring EXAMPLE SOLUTION Solve the equation for x.
This is the given equation. Multiply everything by the LCD: x2. Distribute. Multiply. Subtract 5x + 6 from both sides. Factor.

Factoring CONTINUED Set each factor equal to zero. Solve.
Since no denominator from the original equation is zero when x = -1 or when x = 6, these are our solutions.

§ 0.5 Exponents and Power Functions

Section Outline Exponent Rules Applications of Exponents
Compound Interest

Exponents Definition Example

Applications of Exponents
EXAMPLE Use the laws of exponents to simplify the algebraic expression. SOLUTION This is the given expression.

Applications of Exponents
CONTINUED Subtract. Divide.

Compound Interest - Annual
Definition Example Compound Interest Formula: A = the compound amount (how much money you end up with) P = the principal amount invested i = the compound interest rate per interest period n = the number of compounding periods If $700 is invested, compounded annually at 8% for 8 years, this will grow to: Therefore, the compound amount would be $1,

Compound Interest - General

Compound Interest - General
EXAMPLE (Quarterly Compound) Assume that a $500 investment earns interest compounded quarterly. Express the value of the investment after one year as a polynomial in the annual rate of interest r. SOLUTION Since interest is not being compounded annually, we must use this formula. Replace P with 500, m with 4 (interest is compounded 4 times each year), and t with 1 (interest is being compounded for 1 year). Simplify.

§ 0.6 Functions and Graphs in Application

Section Outline Geometric Problems Cost, Revenue, and Profit
Surface Area Functions and Graphs

Geometric Problems EXAMPLE SOLUTION
(Fencing a Rectangular Corral) Consider a rectangular corral with two partitions, as shown below. Assign letters to the outside dimensions of the corral. Write an equation expressing the fact that the corral has a total area of 2500 square feet. Write an expression for the amount of fencing needed to construct the corral (including both partitions). SOLUTION First we will assign letters to represent the dimensions of the corral. y x x x x y

Geometric Problems CONTINUED
Now we write an equation expressing the fact that the corral has a total area of 2500 square feet. Since the corral is a rectangle with outside dimensions x and y, the area of the corral is represented by: Now we write an expression for the amount of fencing needed to construct the corral (including both partitions). To determine how much fencing will be needed, we add together the lengths of all the sides of the corral (including the partitions). This is represented by:

Cost Problems EXAMPLE SOLUTION
(Cost of Fencing) Consider the corral of the last example. Suppose the fencing for the boundary of the corral costs $10 per foot and the fencing for the inner partitions costs $8 per foot. Write an expression for the total cost of the fencing. SOLUTION This is the diagram we drew to represent the corral. y x x x x y Since the boundary of the fence is represented by the red part of the diagram, the length of fencing for this portion of the corral is x + x + y + y = 2x + 2y. Therefore the cost of fencing the boundary of the fence is (2x + 2y)(cost of boundary fencing per foot) = (2x + 2y)(10) = 20x + 20y.

Cost Problems CONTINUED
Since the inner partitions of the fence are represented by the blue part of the diagram, the length of fencing for this portion of the corral is x + x = 2x. Therefore the cost of fencing the inner partitions of the fence is (2x)(cost of inner partition fencing per foot) = (2x)(8) = 16x. Therefore, an expression for the total cost of the fencing is: (cost of boundary fencing) + (cost of inner partition fencing) (20x + 20y) + (16x) 36x + 20y

Surface Area EXAMPLE SOLUTION
Assign letters to the dimensions of the geometric box and then determine an expression representing the surface area of the box. SOLUTION First we assign letters to represent the dimensions of the box. z y x

Surface Area CONTINUED z y x
Now we determine an expression for the surface area of the box. Note, the box has 5 sides which we will call Left (L), Right (R), Front (F), Back (B), and Bottom (Bo). We will find the area of each side, one at a time, and then add them all up. L: yz R: yz F: xz B: xz Bo: xy Therefore, an expression that represents the surface area of the box is: yz + yz + xz + xz + xy = 2yz +2xz +xy.

Cost, Revenue, & Profit EXAMPLE SOLUTION
(Cost, Revenue, and Profit) An average sale at a small florist shop is $21, so the shop’s weekly revenue function is R(x) = 21x where x is the number of sales in 1 week. The corresponding weekly cost is C(x) = 9x dollars. What is the florist shop’s weekly profit function? How much profit is made when sales are at 120 per week? If the profit is $1000 for a week, what is the revenue for the week? SOLUTION (a) Since Profit = Revenue – Cost, the profit function, P(x), would be: P(x) = R(x) – C(x) P(x) = 21x – (9x + 800) P(x) = 21x – 9x - 800 P(x) = 12x - 800

Cost, Revenue, & Profit CONTINUED
(b) Since x represents the number of sales in one week, to determine how much profit is made when sales are at 120 per week, we will replace x with 120 in the profit function and then evaluate. P(120) = 12(120) - 800 P(120) = 1, P(120) = 640 Therefore, when sales are at 120 per week, profit is $640 for that week. (c) To determine the revenue for the week when the profit is $1000 for that week, we use an equation that contains profit, namely the profit function: P(x) = 12x - 800 Now we replace P(x) with 1000 and solve for x. 1000 = 12x - 800 1800 = 12x 150 = x Therefore x, the number of units sold in a week, is 150 when profit is $1000.

Cost, Revenue, & Profit CONTINUED
Now, to determine the corresponding revenue, we replace x with 150 in the revenue function. R(x) = 21x R(150) = 21(150) R(150) = 3,150 Therefore, when profit is $1000 in a week, the corresponding revenue is $3,150. NOTE: In order to determine the desired revenue value in part (c), we needed to solve for R(x). But in order to do that, we needed to have a value for x to plug into the R(x) function. In order to acquire that value for x, we needed to use the given information – profit is $1000.

Functions & Graphs EXAMPLE
The function f (r) gives the cost (in cents) of constructing a 100-cubic-inch cylinder of radius r inches. The graph of f (r) is given. What is the cost of constructing a cylinder of radius 6 inches? Interpret the fact that the point (3, 162) is on the graph of the function. Interpret the fact that the point (3, 162) is the lowest point on the graph of the function. What does this say in terms of cost versus radius?

Functions & Graphs CONTINUED SOLUTION
To determine the cost of constructing a cylinder of radius 6 inches, we look on the graph where r = 6. The corresponding y value will be the cost we are seeking. The red arrow is emphasizing the point in which we are interested. The y value of that point is Therefore, the cost of constructing a cylinder of radius 6 inches is 270 cents or $2.70.

Functions & Graphs CONTINUED
(b) The fact that the point (3, 162) is on the graph tells us that the cost to make 100-cubic-inch cylinders with a radius as small as 3 inches is 162 cents or $1.62. (c) The fact that the point (3, 162) is the lowest point on the graph tells us that the least expensive 100-cubic-inch cylinder that can be made is a 3 inch cylinder at a cost of $ Therefore, the 3 inch cylinder is the most cost-effective one that is offered.

Chapter 0 Functions.

Similar presentations

Presentation on theme: "Chapter 0 Functions."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chapter 0 Functions.

Similar presentations

Presentation on theme: "Chapter 0 Functions."— Presentation transcript:

Similar presentations

About project

Feedback