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Key Concept: Real Numbers Example 1: Use Set-Builder Notation Five-Minute Check Then/Now New Vocabulary Key Concept: Real Numbers Example 1: Use Set-Builder Notation Example 2: Use Interval Notation Key Concept: Function Key Concept: Vertical Line Test Example 3: Identify Relations that are Functions Example 4: Find Function Values Example 5: Find Domains Algebraically Example 6: Real-World Example: Evaluate a Piecewise-Defined Function Lesson Menu

Find the value of x 2 + 4x + 4 if x = –2. B. 0 C. 4 D. 16 5–Minute Check 1

Solve 5n + 6 = –3n – 10. A. –8 B. –2 C. D. 2 5–Minute Check 2

Evaluate |x – 2y| – |2x – y| – xy if x = –2 and y = 7. B. 9 C. 19 D. 41 5–Minute Check 3

Factor 8xy 2 – 4xy. A. 2x(4xy 2 – y) B. 4xy(2y – 1) C. 4xy(y 2 – 1) D. 4y 2(2x – 1) 5–Minute Check 4

A. B. C. D. 5–Minute Check 5

Describe subsets of real numbers. You used set notation to denote elements, subsets, and complements. (Lesson 0-1) Describe subsets of real numbers. Identify and evaluate functions and state their domains. Then/Now

piecewise-defined function relevant domain set-builder notation interval notation function function notation independent variable dependent variable implied domain piecewise-defined function relevant domain Vocabulary

natural numbers: whole numbers: Integers: rational numbers: irrational numbers: real numbers:

Key Concept 1

A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation. Use Set-Builder Notation A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation. The set includes natural numbers greater than but equal to 2 and less than or equal to 7. This is read as the set of all x such that 2 is less than or equal to x and x is less than or equal to 7 and x is an element of the set of natural numbers. Answer: Example 1

B. Describe x > –17 using set-builder notation. Use Set-Builder Notation B. Describe x > –17 using set-builder notation. The set includes all real numbers greater than –17. Answer: Example 1

C. Describe all multiples of seven using set-builder notation. Use Set-Builder Notation C. Describe all multiples of seven using set-builder notation. The set includes all integers that are multiples of 7. Answer: Example 1

Describe {6, 7, 8, 9, 10, …} using set-builder notation. Example 1

This concept also describes the behavior of “x” in a function. Inequalities can be expressed geometrically or by using interval notation. This concept also describes the behavior of “x” in a function.

Interval Notation < > ≤ ≥ ( ) NOT included [ ] Included

Negative and Positive Infinity are never INCLUDED VALUES The symbols and Indicate that the corresponding line segments extend infinitely far to the left or right. Negative and Positive Infinity are never INCLUDED VALUES Real Numbers (7)

IMPORTANT WORD MEANINGS “AND” Intersection- what two things have in common.

Bounded intervals

Use Interval Notation   The set includes all real numbers greater than or equal to –2 AND less than but equal to 12. Answer: [–2, 12] Example 2

B. Write x > –4 using interval notation. Use Interval Notation B. Write x > –4 using interval notation. The set includes all real numbers greater than –4. Answer: (–4, ) Example 2

IMPORTANT WORD MEANINGS Union- two things combined to make one set

Unbounded intervals

A. Write –2 ≤ x ≤ 12 using interval notation. Use Interval Notation A. Write –2 ≤ x ≤ 12 using interval notation. The set includes all real numbers greater than or equal to –2 and less than or equal to 12. Answer: [–2, 12] Example 2

B. Write x > –4 using interval notation. Use Interval Notation B. Write x > –4 using interval notation. The set includes all real numbers greater than –4. Answer: (–4, ) Example 2

C. Write x < 3 OR x ≥ 54 using interval notation. Use Interval Notation C. Write x < 3 OR x ≥ 54 using interval notation. The set includes all real numbers less than 3 OR all real numbers greater than or equal to 54. Answer: Example 2

Write x > 5 OR x < –1 using interval notation. B. C. (–1, 5) D. Example 2

Concept of Function and Relation Relation: Any set of ordered pairs where Each value of the first variable is paired with one or more values of the second variable.

Concept of Function and Relation A FUNCTION is a relation in which each element of the domain is paired with exactly one element of the range.

Another way of saying it is that there is one and only one output (y) with each input (x). f(x) x y

Is the relation a function? {(2, 3), (3, 0), (5, 2), (4, 3)} YES, every domain is different! f(x) 2 3 f(x) 3 f(x) 5 2 f(x) 4 3

Is the relation a function. {(4, 1), (5, 2), (5, 3), (6, 6), (1, 9)} f(x) 4 1 f(x) 5 2 NO, 5 is paired with 2 numbers! f(x) 5 3 f(x) 6 f(x) 1 9

Identify Functions Verbally Determine whether the relation represents y as a function of x. The input value x is the height of a student in inches, and the output value y is the number of books that the student owns. Answer: No; there is more than one y-value for an x-value. Key Concept 3

Identify Functions Numerically Key Concept 3

Is this relation a function? {(1,3), (2,3), (3,3)} Yes No

B. Determine whether the table represents y as a function of x. Identify Relations that are Functions B. Determine whether the table represents y as a function of x. Answer: No; there is more than one y-value for an x-value. Example 3

Identify Functions Graphically Key Concept 3a

Vertical-Line Test If every vertical line intersects a graph only once, then the graph represents a function. function not a function

C. Determine whether the graph represents y as a function of x. Identify Relations that are Functions C. Determine whether the graph represents y as a function of x. Answer: Yes; there is exactly one y-value for each x-value. Any vertical line will intersect the graph at only one point. Therefore, the graph represents y as a function of x. Example 3

Identify Functions Algebraically D. Determine whether x = 3y 2 represents y as a function of x. To determine whether this equation represents y as a function of x, solve the equation for y. x = 3y 2 Original equation Divide each side by 3. Take the square root of each side.

Answer: No; there is more than one y-value for an x-value. Identify Relations that are Functions This equation does not represent y as a function of x because there will be two corresponding y-values, one positive and one negative, for any x-value greater than 0. Let x = 12. Answer: No; there is more than one y-value for an x-value. Example 3

Determine whether 12x 2 + 4y = 8 represents y as a function of x. A. Yes; there is exactly one y-value for each x-value. B. No; there is more than one y-value for an x-value. Example 3

Evaluate functions. f(x) = –2.5x + 11 where x = –1 Find f(-1): Key Skills

3x-2 3 7 3(3)-2 -2 -8 3x-2 3(-2)-2 f(3)= f(-2)= Given f(x) = 3x - 2, find: f(3)= f(-2)= 3 3x-2 3(3)-2 7 -2 3x-2 3(-2)-2 -8

Given h(z) = z2 - 4z + 9, find h(-3) (-3)2 – 4(-3) + 9 9+12+9 -3 30 h(-3) = 30

To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8. Find Function Values A. If f (x) = x 2 – 2x – 8, find f (3). To find f (3), replace x with 3 in f (x) = x 2 – 2x – 8. f (x) = x 2 – 2x – 8 Original function f (3) = 3 2 – 2(3) – 8 Substitute 3 for x. = 9 – 6 – 8 Simplify. = –5 Subtract. Answer: –5 Example 4

B. If f (x) = x 2 – 2x – 8, find f (–3d). Find Function Values B. If f (x) = x 2 – 2x – 8, find f (–3d). To find f (–3d), replace x with –3d in f (x) = x 2 – 2x – 8. f (x) = x 2 – 2x – 8 Original function f (–3d) = (–3d)2 – 2(–3d) – 8 Substitute –3d for x. = 9d 2 + 6d – 8 Simplify. Answer: 9d 2 + 6d – 8 Example 4

C. If f (x) = x2 – 2x – 8, find f (2a – 1). Find Function Values C. If f (x) = x2 – 2x – 8, find f (2a – 1). To find f (2a – 1), replace x with 2a – 1 in f (x) = x 2 – 2x – 8. f (x) = x 2 – 2x – 8 Original function f (2a – 1) = (2a – 1)2 – 2(2a – 1) – 8 Substitute 2a – 1 for x. = 4a 2 – 4a + 1 – 4a + 2 – 8 Expand (2a – 1)2 and 2(2a – 1). = 4a 2 – 8a – 5 Simplify. Answer: 4a 2 – 8a – 5 Example 4

If , find f (6). A. B. C. D. Example 4

Find the Domain of a Function Algebraically A. State the domain of the function . Because the square root of a negative number cannot be real, 4x – 1 ≥ 0. Therefore, the domain of g(x) is all real numbers x such that x ≥ , or . Answer: all real numbers x such that x ≥ , or

B. State the domain of the function . Find Domains Algebraically B. State the domain of the function . When the denominator of is zero, the expression is undefined. Solving t 2 – 1 = 0, the excluded values in the domain of this function are t = 1 and t = –1. The domain of this function is all real numbers except t = 1 and t = –1, or . Answer: Example 5

C. State the domain of the function . Find Domains Algebraically C. State the domain of the function . This function is defined only when 2x – 3 > 0. Therefore, the domain of f (x) is or . Answer: or Example 5

State the domain of g (x) = . A. or [4, ∞) B. or [–4, 4] C. or (− , −4] D. Example 5

Evaluate a Piecewise Function Piecewise Function –a function defined by two or more functions over a specified domain. What do they look like? f(x) = 2x + 1 , x  0 x – 1 , x  0

Evaluating Piecewise Functions: Evaluating piecewise functions is just like evaluating functions that you are already familiar with. f(x) = 2x + 1 , x  0 x – 1 , x  0 Let’s calculate f(2). You are being asked to find y when x = 2. Since 2 is  0, you will only substitute into the second part of the function. f(2) = 2 – 1 = 1

Evaluate the following: f(x) = 3x - 2, x  -2 -x , -2  x  1 2x – 7x, x  1 f(-2) = 2 f(3) = -12 f(-4) = -14 f(1) = -6

Evaluate a Piecewise-Defined Function A. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of 1400 square feet. Example 6

Because 1400 is between 1000 and 2600, use to find p(1400). Evaluate a Piecewise-Defined Function Because 1400 is between 1000 and 2600, use to find p(1400). Function for 1000 ≤ a < 2600 Substitute 1400 for a. Subtract. = 85 Simplify. Example 6

Answer: $85 per square foot Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 1400 square feet is $85. Answer: $85 per square foot Example 6

Evaluate a Piecewise-Defined Function A. FINANCE Realtors in a metropolitan area studied the average home price per square foot as a function of total square footage. Their evaluation yielded the following piecewise-defined function. Find the average price per square foot for a home with the square footage of 3200 square feet. Example 6

Because 3200 is between 2600 and 4000, use to find p(3200). Evaluate a Piecewise-Defined Function Because 3200 is between 2600 and 4000, use to find p(3200). Function for 2600 ≤ a < 4000. Substitute 3200 for a. Simplify. Example 6

Answer: $104 per square foot Evaluate a Piecewise-Defined Function According to this model, the average price per square foot for a home with a square footage of 3200 square feet is $104. Answer: $104 per square foot Example 6

ENERGY The cost of residential electricity use can be represented by the following piecewise function, where k is the number of kilowatts. Find the cost of electricity for 950 kilowatts. A. $47.50 B. $48.00 C. $57.50 D. $76.50 Example 6

HOMEWORK Pgs. 9-12 See Notes

End of the Lesson