Key Concept 1
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Transcript of Key Concept 1
Use Set-Builder Notation
A. Describe {2, 3, 4, 5, 6, 7} using set-builder notation. The set includes natural numbers greater than or 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:
Use Set-Builder Notation
B. Describe x > –17 using set-builder notation.
The set includes all real numbers greater than –17.
Answer:
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:
Describe {6, 7, 8, 9, 10, …} using set-builder notation.
A.
B.
C.
D.
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]
Use Interval Notation
B. Write x > –4 using interval notation.
The set includes all real numbers greater than –4.
Answer: (–4, )
Use Interval Notation
C. Write x < 3 or x ≥ 54 using interval notation.
The set includes all real numbers less than 3 and all real numbers greater than or equal to 54.
Answer:
Write x > 5 or x < –1 using interval notation.
A.
B.
C. (–1, 5)
D.
Identify Relations that are Functions
A. 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.
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.
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.
Identify Relations that are Functions
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.
Identify Relations that are Functions
Answer: No; there is more than one y-value for an x-value.
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.
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.
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
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
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
If , find f (6).
A.
B.
C.
D.
Find Domains 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
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:
Find Domains Algebraically
C. State the domain of the function .
Answer: or
This function is defined only when 2x – 3 > 0.
Therefore, the domain of f (x) is
or .
State the domain of g (x) = .
A. or [4, ∞)
B. or [–4, 4]
C. or (− , −4]
D.
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.
Evaluate a Piecewise-Defined Function
Evaluate a Piecewise-Defined Function
Because 1400 is between 1000 and 2600,
use to find p(1400).
Subtract.
= 85 Simplify.
Function for 1000 ≤ a < 2600
Substitute 1400 for a.
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.
B. 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.
Evaluate a Piecewise-Defined Function
Evaluate a Piecewise-Defined Function
Because 3200 is between 2600 and 4000, use
to find p(3200).
Function for
2600 ≤ a < 4000.
Simplify.
Substitute 3200 for a.
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.
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