Single variable linear equation that has one solution
1 answer:
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Answer:
The definition for the given piecewise-defined function is: .
Step-by-step explanation:
<h3>General Concepts:</h3>- Piecewise-defined functions.
- Interval notations.
<h3>What is a piecewise-defined function?</h3>
A piecewise-defined function represents specific rules over different intervals of the domain.
<h3>Symbols used in expressing interval notations:</h3>Open interval: This means that the endpoint is <em>not</em> included in the interval.
We can use the following symbols to indicate the <u>exclusion</u> of endpoints in the interval:
- Left or right parenthesis, "( )" (or both).
- Greater than (>) or less than (<) symbols.
- Open dot "
" 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 <u>inclusion</u> 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 <em>inclusion</em> of the endpoint in the graph of a piecewise-defined function.
The best way to determine which piecewise-defined function represents the graph is by observing the <u>endpoints</u> and <u>orientation</u> of both partial lines.
- Open circle on (-1, 2): The graph shows that one of the partial lines has an <em>excluded</em> endpoint of (-1, 2) extending towards the <u>right</u>. 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 <em>included</em> endpoint of (-1, 1) extended towards the <u>left</u>. 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.
<h2>Solution:</h2><h3>a) Test option A:</h3>
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 <u>closed dot</u>.
Substitute x = -2 into f(x) = 2x + 2:
- f(x) = 2x + 2
- f(-2) = 2(-2) + 2
- f(-2) = -4 + 2
- f(-2) = -2 ⇒ <em>False statement</em>.
⇒ The output value of f(-2) = -2 is <u>not</u> included in the graph of the partial line whose endpoint is at (-1, 1).
<h3>Piece 2: If x > -1, then it is defined by f(x) = x + 4. </h3>
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 <u>open dot</u>.
Substitute x = 0 into f(x) = x + 4:
- f(x) = x + 4
- f(0) = (0) + 4
- f(0) = 4 ⇒ <em>True statement</em>.
⇒ The output value of f(0) = 4 <u>is</u> 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].
<h3>b) Test option D:</h3>
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 <u>closed dot</u>.
Substitute x = -2 into f(x) = x + 2:
- f(x) = x + 2
- f(-2) = (-2) + 2
- f(-2) = 0 ⇒ <em>True statement</em>.
⇒ The output value of f(-2) = 0 <u>is</u> included the graph of the partial line whose endpoint is at (-1, 1).
<h3>Piece 2: If x > -1, then it is defined by f(x) = 2x + 4.</h3>
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 <u>open dot</u>.
Substitute x = 0 into f(x) = 2x + 4:
- f(x) = 2x + 4
- f(0) = 2(0) + 4
- f(0) = 0 + 4 = 0 ⇒ <em>True statement</em>.
⇒ The output value of f(0) = 4 <u>is</u> included in the graph of the partial line whose endpoint is at (-1, 2).
<h2>Final Answer: </h2>
We can infer that the piecewise-defined function that represents the graph is:
.
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