Graph the function, insert solidcircles or open circles where necessary to indicate the true nature of the function
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Transcribed Image Text:### Piecewise Function Example
Consider the piecewise function \( K(x) \):
\[
K(x) =
\begin{cases}
1, & \text{if } x \leq -2 \\
x^2 - 4, & \text{if } -2 < x < 2 \\
\frac{1}{2}x, & \text{if } x \geq 2
\end{cases}
\]
This function is defined by three different expressions depending on the value of \( x \):
1. **For \( x \leq -2 \):** \( K(x) = 1 \)
- When \( x \) is less than or equal to -2, the value of the function \( K(x) \) is a constant 1.
2. **For \( -2 < x < 2 \):** \( K(x) = x^2 - 4 \)
- Between -2 and 2 (not inclusive), the function \( K(x) \) follows the quadratic expression \( x^2 - 4 \).
3. **For \( x \geq 2 \):** \( K(x) = \frac{1}{2}x \)
- When \( x \) is greater than or equal to 2, the function \( K(x) \) is a linear function with a slope of \(\frac{1}{2}\).
This piecewise function provides an example of how different equations can be used to define a function over different intervals of the domain. This is particularly useful in modeling scenarios where the behavior or rules change at certain points.
Transcribed Image Text:## Step 2
Consider the piecewise function \( K(x) \) defined as follows:
\[ K(x) = \begin{cases}
1 & \text{if } x \leq -2 \\
\frac{x^2 - 4}{x^2} & \text{if } -2 < x < 2 \\
1 & \text{if } x \geq 2
\end{cases} \]
We need to examine the continuity of the function at \( x = -2 \) and \( x = 2 \).
### Limits at \( x = -2 \) and \( x = 2 \):
1. \[ \lim_{{x \to -2^-}} K(x) = 1 \]
2. \[ \lim_{{x \to -2^+}} K(x) = \lim_{{x \to -2}} \frac{x^2 - 4}{x^2} = 0 \]
3. \[ \lim_{{x \to 2^-}} K(x) = \lim_{{x \to 2}} \frac{x^2 - 4}{x^2} = 0 \]
4. \[ \lim_{{x \to 2^+}} K(x) = 1 \]
### Continuity Analysis:
The function \( K(x) \) is not continuous at \( x = -2 \) and \( x = 2 \) because the limit from the left does not equal the limit from the right.
\[ K(-2) = 1, \quad K(2) = 1 \]
### Graph Explanation:
The graph provided shows a piecewise plot of \( K(x) \):
- There is a filled circle at \( (-2, 1) \) indicating the value of \( K(x) \) at \( x = -2 \).
- The curve corresponding to \( \frac{x^2 - 4}{x^2} \) represents \( K(x) \) for the interval \( -2 < x < 2 \). This curve approaches but does not include the points at \( x = -2 \) and \( x = 2 \).
- There is a filled circle at \( (2, 1) \) indicating the value of \( K(x) \) at \( x = 2 \
Two-dimensional figure measured in terms of radius. It is formed by a set of points that are at a constant or fixed distance from a fixed point in the center of the plane. The parts of the circle are circumference, radius, diameter, chord, tangent, secant, arc of a circle, and segment in a circle.
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![### Piecewise Function Example
Consider the piecewise function \( K(x) \):
\[
K(x) =
\begin{cases}
1, & \text{if } x \leq -2 \\
x^2 - 4, & \text{if } -2 < x < 2 \\
\frac{1}{2}x, & \text{if } x \geq 2
\end{cases}
\]
This function is defined by three different expressions depending on the value of \( x \):
1. **For \( x \leq -2 \):** \( K(x) = 1 \)
- When \( x \) is less than or equal to -2, the value of the function \( K(x) \) is a constant 1.
2. **For \( -2 < x < 2 \):** \( K(x) = x^2 - 4 \)
- Between -2 and 2 (not inclusive), the function \( K(x) \) follows the quadratic expression \( x^2 - 4 \).
3. **For \( x \geq 2 \):** \( K(x) = \frac{1}{2}x \)
- When \( x \) is greater than or equal to 2, the function \( K(x) \) is a linear function with a slope of \(\frac{1}{2}\).
This piecewise function provides an example of how different equations can be used to define a function over different intervals of the domain. This is particularly useful in modeling scenarios where the behavior or rules change at certain points.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe91e659-0274-4af0-b893-ea272a52ff02%2Fe8751c3c-c9b6-4315-8198-3a4430374708%2Fkawf88x_processed.jpeg&w=3840&q=75)
![## Step 2
Consider the piecewise function \( K(x) \) defined as follows:
\[ K(x) = \begin{cases}
1 & \text{if } x \leq -2 \\
\frac{x^2 - 4}{x^2} & \text{if } -2 < x < 2 \\
1 & \text{if } x \geq 2
\end{cases} \]
We need to examine the continuity of the function at \( x = -2 \) and \( x = 2 \).
### Limits at \( x = -2 \) and \( x = 2 \):
1. \[ \lim_{{x \to -2^-}} K(x) = 1 \]
2. \[ \lim_{{x \to -2^+}} K(x) = \lim_{{x \to -2}} \frac{x^2 - 4}{x^2} = 0 \]
3. \[ \lim_{{x \to 2^-}} K(x) = \lim_{{x \to 2}} \frac{x^2 - 4}{x^2} = 0 \]
4. \[ \lim_{{x \to 2^+}} K(x) = 1 \]
### Continuity Analysis:
The function \( K(x) \) is not continuous at \( x = -2 \) and \( x = 2 \) because the limit from the left does not equal the limit from the right.
\[ K(-2) = 1, \quad K(2) = 1 \]
### Graph Explanation:
The graph provided shows a piecewise plot of \( K(x) \):
- There is a filled circle at \( (-2, 1) \) indicating the value of \( K(x) \) at \( x = -2 \).
- The curve corresponding to \( \frac{x^2 - 4}{x^2} \) represents \( K(x) \) for the interval \( -2 < x < 2 \). This curve approaches but does not include the points at \( x = -2 \) and \( x = 2 \).
- There is a filled circle at \( (2, 1) \) indicating the value of \( K(x) \) at \( x = 2 \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffe91e659-0274-4af0-b893-ea272a52ff02%2Fe8751c3c-c9b6-4315-8198-3a4430374708%2Fx6yr8mh_processed.jpeg&w=3840&q=75)
