Answer:
The definition for the given piecewise-defined function is: [tex]\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}[/tex].
Step-by-step explanation:
General Concepts:
- Piecewise-defined functions.
- Interval notations.
What is a piecewise-defined function?
A piecewise-defined function represents specific rules over different intervals of the domain.
Symbols used in expressing interval notations:
Open interval: This means that the endpoint is not included in the interval.
We can use the following symbols to indicate the exclusion of endpoints in the interval:
- Left or right parenthesis, "( )" (or both).
- Greater than (>) or less than (<) symbols.
- Open dot "[tex]\circ[/tex]" is another way of expressing the exclusion of an endpoint in the graph of a piecewise-defined function.
Closed interval: This implies the inclusion of endpoints in the interval.
We can use the following symbols to indicate the inclusion of endpoints in the interval:
- Open- or closed brackets (or both), "[ ]."
- Greater than or equal to (≥) or less than or equal to (≤) symbols.
- Closed circle or dot, "•" is another way of expressing the inclusion of the endpoint in the graph of a piecewise-defined function.
Determine the appropriate function rule that defines different parts of the domain.
The best way to determine which piecewise-defined function represents the graph is by observing the endpoints and orientation of both partial lines.
- Open circle on (-1, 2): The graph shows that one of the partial lines has an excluded endpoint of (-1, 2) extending towards the right. This implies that its domain values are defined when x > -1.
- Closed circle on (-1, 1): The graph shows that one of the partial lines has an included endpoint of (-1, 1) extended towards the left. Hence, its domain values are defined when x ≤ -1.
Based on our observations from the previous step, we can infer that x > -1 or x ≤ -1 apply to piecewise-defined functions A or D. However, only one of those two options represent the graph.
Solution:
a) Test option A:
[tex]\boxed{\displaystyle\sf Option\:A)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ 2x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ x + 4 & \sf\:{if\:\:x > -1}\end{cases}}[/tex]
Piece 1: If x ≤ -1, then it is defined by f(x) = 2x + 2.
We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a closed dot.
Substitute x = -2 into f(x) = 2x + 2:
- f(-2) = 2(-2) + 2
- f(-2) = -4 + 2
- f(-2) = -2 ⇒ False statement.
⇒ The output value of f(-2) = -2 is not included in the graph of the partial line whose endpoint is at (-1, 1).
Piece 2: If x > -1, then it is defined by f(x) = x + 4.
We must choose a domain value that falls within the interval of x > -1 whose output is included in the graph of the partial line with an open dot.
Substitute x = 0 into f(x) = x + 4:
- f(x) = x + 4
- f(0) = (0) + 4
- f(0) = 4 ⇒ True statement.
⇒ The output value of f(0) = 4 is included in the graph of the partial line whose endpoint is at (-1, 2).
Conclusion for Option A:
Option A is not the correct piecewise-defined function because one of the pieces, f(x) = 2x + 2, does not specify the interval (-∞, -1].
b) Test option D:
[tex]\boxed{\displaystyle\sf Option\:D)\:\:\:f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}[/tex]
Piece 1: If x ≤ -1, then it is defined by f(x) = x + 2.
We must choose a domain value that falls within the interval of x ≤ -1 whose output is included is included in the graph of the partial line with a closed dot.
Substitute x = -2 into f(x) = x + 2:
- f(x) = x + 2
- f(-2) = (-2) + 2
- f(-2) = 0 ⇒ True statement.
⇒ The output value of f(-2) = 0 is included the graph of the partial line whose endpoint is at (-1, 1).
Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.
We must choose a domain value that falls within the interval of x > -1 whose output is included is included in the graph of the partial line with an open dot.
Substitute x = 0 into f(x) = 2x + 4:
- f(x) = 2x + 4
- f(0) = 2(0) + 4
- f(0) = 0 + 4 = 0 ⇒ True statement.
⇒ The output value of f(0) = 4 is included in the graph of the partial line whose endpoint is at (-1, 2).
Final Answer:
We can infer that the piecewise-defined function that represents the graph is:
[tex]\boxed{\displaystyle\sf\ Option\:D:\:\: f(x) = \begin{cases}\displaystyle\sf\ x + 2 & \sf\:{if\:\:x \leq -1} \\\displaystyle\sf\ 2x + 4 & \sf\:{if\:\:x > -1}\end{cases}}[/tex].
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