What is the greatest number of x’s all the terms have in common? PLEASE HELP!
2 answers:
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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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Learn more about piecewise-defined functions here:
<h3>Answer: HF = 5</h3>
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Explanation:
Refer to the diagram below. I'm adding in points A and B, which are marked in blue. I'm also adding in variables x, y, and z.
The goal is to find HF, so let's make that equal to x. This is also equal to HB because they are radii of the same circle.
Let y be the length of segments JA and JB.
Let z be the length of segments CF and CA.
So in short we have
- x = HF = HB
- y = JA = JB
- z = CF = CA
We know that
- CH = 13
- HJ = 19
- CJ = 22
This must mean
- CH = CF+HF = z+x = 13
- HJ = HB+JB = x+y = 19
- CJ = CA+JA = z+y = 22
Or in short,
- z+x = 13
- x+y = 19
- z+y = 22
Let's solve the first equation for z to get z = -x+13. I subtracted x from both sides.
Now plug this into the third equation to get
z+y = 22
(z) + y = 22
(-x+13) + y = 22 ... replace z with -x+13
-x+13+y = 22
-x+y+13 = 22
-x+y = 22-13 .... subtract 13 from both sides
-x+y = 9
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So we have this reduced system of equations of two variables and two equations
- x+y = 19 ... found earlier
- -x+y = 9 .... what we just found above
Add the equations straight down. This is the elimination property.
Doing so leads to the x terms canceling out since x + (-x) = 0x = 0
The y terms add to y+y = 2y
The terms on the right hand side add to 19+9 = 28
We are left with 2y = 28 which solves to y = 28/2 = 14.
If y = 14, then,
x+y = 19
x+14 = 19
x = 19-14
x = 5
In which we can then say
z = -x+13
z = -5+13
z = 8
Though we don't need to find z (unless you're curious about it).
Since x = 5, and we set HF equal to x, this means that HF = 5.