Question
The graph of a piece-wise function is shown.The graph of a piece-wise function is shown. A linear segment with negative slope terminates at a filled circle at left-parenthesis negative 1 comma 6 right-parenthesis. A parabolic segment begins at an open circle at left-parenthesis negative 1 comma 4 right-parenthesis, descends to the y-intercept at left-parentheses 0 comma 1 right-parentheses, and continues up and to the right.
Question
Use the graph to answer the question.
For which value of a is f left-parenthesis x right-parenthesis discontinuous?
(1 point)
Responses
negative 1
Image with alt text: negative 1
0
0
4
4
6
The following function is continuous at all points in the domain of real numbers. What is the value of n?
f left-parenthesis x right-parenthesis equals enlarged left brace start layout 1st row 2 x plus 2 comma if x greater than n 2nd row 4 x comma if x less than or equals n end layout
(1 point)
Responses
negative 1
Image with alt text: negative 1
0
0
one half
Image with alt text: one half
1
Question
Use the graph to answer the question.
For which value of a is f left-parenthesis x right-parenthesis discontinuous?
(1 point)
Responses
negative 1
Image with alt text: negative 1
0
0
4
4
6
The following function is continuous at all points in the domain of real numbers. What is the value of n?
f left-parenthesis x right-parenthesis equals enlarged left brace start layout 1st row 2 x plus 2 comma if x greater than n 2nd row 4 x comma if x less than or equals n end layout
(1 point)
Responses
negative 1
Image with alt text: negative 1
0
0
one half
Image with alt text: one half
1
Answers
GPT-4o mini
To determine the point of discontinuity for the piecewise function based on the graph described:
1. The function has a linear segment that ends at a filled circle at \((-1, 6)\), indicating that at \(x = -1\), \(f(-1) = 6\).
2. There is also a parabolic segment that begins at an open circle at \((-1, 4)\) which means that at \(x = -1\), the value approaches \(4\) but does not include it. Thus, \(f(-1)\) does not equal \(4\).
Since the value from the left exceeds the value from the right at \(x = -1\), the function is discontinuous at this point.
Therefore, the answer is **negative 1**.
For the second part regarding the function \(f(x) = \begin{cases} 2x + 2, & \text{if } x > n \\ 4x, & \text{if } x \leq n \end{cases}\):
1. To find continuity at \(x = n\), we need:
\[
\lim_{x \to n^+} f(x) = \lim_{x \to n^-} f(x)
\]
2. As \(x\) approaches \(n\) from the right (\(f(n^+) = 2n + 2\)) and from the left (\(f(n^-) = 4n\)).
3. For these to be equal:
\[
2n + 2 = 4n
\]
Rearranging gives:
\[
2 = 2n \implies n = 1
\]
Thus, the value of \(n\) for which the function is continuous is **1**.
Summarizing the answers:
1. Discontinuity at: **negative 1**
2. Value of \(n\): **1**
1. The function has a linear segment that ends at a filled circle at \((-1, 6)\), indicating that at \(x = -1\), \(f(-1) = 6\).
2. There is also a parabolic segment that begins at an open circle at \((-1, 4)\) which means that at \(x = -1\), the value approaches \(4\) but does not include it. Thus, \(f(-1)\) does not equal \(4\).
Since the value from the left exceeds the value from the right at \(x = -1\), the function is discontinuous at this point.
Therefore, the answer is **negative 1**.
For the second part regarding the function \(f(x) = \begin{cases} 2x + 2, & \text{if } x > n \\ 4x, & \text{if } x \leq n \end{cases}\):
1. To find continuity at \(x = n\), we need:
\[
\lim_{x \to n^+} f(x) = \lim_{x \to n^-} f(x)
\]
2. As \(x\) approaches \(n\) from the right (\(f(n^+) = 2n + 2\)) and from the left (\(f(n^-) = 4n\)).
3. For these to be equal:
\[
2n + 2 = 4n
\]
Rearranging gives:
\[
2 = 2n \implies n = 1
\]
Thus, the value of \(n\) for which the function is continuous is **1**.
Summarizing the answers:
1. Discontinuity at: **negative 1**
2. Value of \(n\): **1**