Section 2.1: Functions and Their Graphspeople.ku.edu/~k088m880/Math115/Chapter0201.pdf · Section...
Transcript of Section 2.1: Functions and Their Graphspeople.ku.edu/~k088m880/Math115/Chapter0201.pdf · Section...
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Section 2.1: Functions and Their Graphs
Kevin Marshall
Math 115
19 January 2018
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Overview
1 Functions
2 Determining Domain
3 Graphs of Functions
4 Piecewise Function
5 Vertical Line Test
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Definitions
Definition (Function)
A function is a rule that assigns to each element in a set A toexactly one element in a set B. We call A the domain of thefunction and the values that the function takes the range(sometimes called image) of the function. Some places denotethe range by f (A).
Example
The function f : R→ R defined by f (x) = x2 + 1 has a domain ofR and a range of [1,∞)
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Find the value of f (a + h)− f (a) for f (x) = x2 − x − 1.f (a + h)− f (a) = (a + h)2 − (a + h)− 1− (a2 − a− 1)
= a2 + 2ah + h2 − a− h − 1− a2 − a− 1= 2ah + h2 − h = h(2a + h − 1)
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Find the value of f (a + h)− f (a) for f (x) = x2 − x − 1.f (a + h)− f (a) = (a + h)2 − (a + h)− 1− (a2 − a− 1)= a2 + 2ah + h2 − a− h − 1− a2 − a− 1
= 2ah + h2 − h = h(2a + h − 1)
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Find the value of f (a + h)− f (a) for f (x) = x2 − x − 1.f (a + h)− f (a) = (a + h)2 − (a + h)− 1− (a2 − a− 1)= a2 + 2ah + h2 − a− h − 1− a2 − a− 1= 2ah + h2 − h = h(2a + h − 1)
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Definitions
Definition (Independent and Dependent Variables)
Given a function y = f (x), then x is call the independentvariable and y is called the dependent variable since its valuedepends on x .
Example
Consider a profit function P(x) = −0.001x2 + 8x − 500 where x isthe number of widgets produced by a company and P is the profitin dollars. Then the number of widgets is the independent variableand the profit is the dependent variable.
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Some Tips
A lot of times a function will have a domain that is limited by thereal world or mathematical rules. Some possible limitations:
1 Division by zero is not defined.
2 Taking the square root of a negative number is not defined.
3 Volume and area are always positive values.
4 Number of students in a class room is always a nonnegativeinteger.
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Determine the domain of f (x) = 1√x2−1 .
We need x2 − 1 > 0 in order for the denominator to be nonzeroand the radical to be defined.We need to consider x2 − 1 = (x + 1)(x − 1) > 0.Testing points inthe intervals (−∞,−1), (−1, 1), and (1,∞) we see that thesolution set for the inequality is (−∞,−1) ∪ (1,∞).
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Determine the domain of f (x) = 1√x2−1 .
We need x2 − 1 > 0 in order for the denominator to be nonzeroand the radical to be defined.We need to consider x2 − 1 = (x + 1)(x − 1) > 0.
Testing points inthe intervals (−∞,−1), (−1, 1), and (1,∞) we see that thesolution set for the inequality is (−∞,−1) ∪ (1,∞).
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Determine the domain of f (x) = 1√x2−1 .
We need x2 − 1 > 0 in order for the denominator to be nonzeroand the radical to be defined.We need to consider x2 − 1 = (x + 1)(x − 1) > 0.Testing points inthe intervals (−∞,−1), (−1, 1), and (1,∞) we see that thesolution set for the inequality is (−∞,−1) ∪ (1,∞).
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Suppose we want to construct an open box from a rectangularpiece of cardboard that is 8 inches long and 5 inches wide bycutting out identical squares from the corners (x inches by xinches) then folding up the flaps. What is the function for thevolume of the box in terms of x? What is the function’s domain?
Solution: Take a moment to read through the problem and list outwhat you know. Making a picture sometimes helps. What variableswill we need and how will they relate to what we are given?
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
Suppose we want to construct an open box from a rectangularpiece of cardboard that is 8 inches long and 5 inches wide bycutting out identical squares from the corners (x inches by xinches) then folding up the flaps. What is the function for thevolume of the box in terms of x? What is the function’s domain?Solution: Take a moment to read through the problem and list outwhat you know. Making a picture sometimes helps. What variableswill we need and how will they relate to what we are given?
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).
We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0. Therefore the domain
of V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0. Therefore the domain
of V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0. Therefore the domain
of V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0. Therefore the domain
of V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0. Therefore the domain
of V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0.
Therefore the domainof V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example (cont.)
Example (cont.)
A diagram shows us that V (x) = x(8− 2x)(5− 2x).We expect volume to be nonnegative as well as our side lengths.Thus:
x ≥ 0
8− 2x ≥ 0
5− 2x ≥ 0
A little calculation shows us that 52 ≥ x ≥ 0. Therefore the domain
of V (x) is the interval [0, 52 ].
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Definitions
Definition
The graph of a function f is the set of all points (x , y) of theform (x , f (x)).
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Example
Example
−6 −4 −2 2 4 6
−50
50
x
y f (x) = x3 − 27x
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Definition
Definition
Sometimes a function is defined differently along its domain. In thiscase we say that our function is a piecewise-defined function.
Example
Consider the function f defined as follows:
f (x) =
{-x x ≤ 0
x x > 0
What does the graph of f look like? What familiar function does isit equivalent to?
Kevin Marshall University of Kansas
Section 2.1
Functions Determining Domain Graphs of Functions Piecewise Function Vertical Line Test
Vertical Line Test
Theorem (Vertical Line Test)
A curve in the xy -plane is the graph of a function f (x) if and onlyif each vertical line intersects the curve in at most one point.
Kevin Marshall University of Kansas
Section 2.1