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Section 2.1 - Functions of Intervals
Intervals - a set of points bound by two points on a line.
- open interval (a, b) - all real numbers greater than 'a' and less than 'b' (a < x < b)
- closed interval [c, d] - all real numbers greater than and equal to 'c' and less than and equal to 'd' (a ≤ x ≤ b)
- half-open interval [m, n) - all real numbers greater than and equal to 'm' and less than 'n' (m ≤ x < n)
- half-open interval
(m, n] - all real numbers greater than 'm' and less than and equal to 'n' (m < x ≤ n)
Cannot have an interval that includes infinity.
- [3, ∞] is not allowed because it includes infinity.
- [-∞, 1) is not allowed because it includes infinity.
- [-1, ∞) is allowed because it does not include infinity.
Functions
- M = 1000(1 + 1%), M1 = 1000(1 + 2%), M2 = 1000(1 + 3%)
- M = 1000 (1 + r) find interest on $1,000.
- C = 5/9(F - 32) function to convert Fahrenheit to Celcius.
- A = πr2 function to find area of a circle.
Definition - the relationship between x and y is a function if for any value of x, there exists one and only one value of y.
Examples of a Function
- y = x
- y = x2
- x2 + y2 = 1 is not a function because more than one value of y exists for any one value of x (creates a circle)
A formula can be found to be a function simply by looking at the graph.
y = f(x)
- y is a dependent variable
- x is an independent variable
- f is the function itself
Domains and Ranges
- Domain of a function - is the set of all values of the independent variable for which the dependent variable is defined
- y = 1/√x has a domain of x > 0
- y = √x has a domain of x ≥ 0
- Range of a function - is the set of all y values that can ocurr
- y = x2 has a range of [0, ∞) because no negatives can exist with a "squared"
- y = -√x has a range of (-∞, 0] because √x cannot be negative
Examples
- y = √(x2 - 1)
- Domain:
- x2 - 1 ≥ 0
- x2 ≥ 1
- |x| ≥ 1
- x ≥ 1 or x ≤ -1
- Range: y ≥ 0 because square root cannot be negative
- y = 1/(x-1)
- Domain: all numbers but 1
- Range: (-∞, ∞)
- y = x/(x2 - 1)
- f(x) = x2 + 3x3
- f(1) = 12 + 3(1)3
- f(1) = 4
- f(a2) = (a2)2 + 3(a2)3
- f(a2) = a4 + 3a6
- f(x - 2) = (x - 2)2 + 3(x - 2)3
- f(x - 2) = x2 - 4x + 4 + 3(x3 - 3x22 + 3x24 - 23)
- f(x - 2) = x2 - 4x + 4 + 3(x3 - 6x2 + 12x - 8)
- f(x - 2) = 3x3 + (1 - 18)x2 + (-4 + 36)x - 20
- f(x - 2) = 3x3 - 17x2 - 34x - 20
- f(x) = √(x + 1), g(x) = x2 - 1
- f(g(x)) = f(x2 - 1)
- f(x2 - 1) = √(x2 - 1 + 1)
- f(x2 - 1) = √(x2)
- f(x2 - 1) = x
Homework for Section 2.1 - questions 3, 8, 12, 18, 19, 20, 29, 33, 34, 35 |