Answer :
To sketch the graph of the given piecewise function
[tex]\[ f(x) = \left\{ \begin{array}{lll} 2 & \text{if} & x \leq -2 \\ -2x - 1 & \text{if} & -2 < x \leq 2 \\ -2 & \text{if} & x > 2 \end{array} \right. \][/tex]
we analyze each piece of the function separately and then combine them on the same coordinate plane.
### 1. For [tex]\(x \leq -2\)[/tex]
In this interval, the function [tex]\(f(x) = 2\)[/tex]. This means the value of the function is a constant 2 for all [tex]\(x\)[/tex] less than or equal to -2.
Sketch:
- This is a horizontal line at [tex]\(y = 2\)[/tex] from negative infinity up to [tex]\(x = -2\)[/tex].
### 2. For [tex]\(-2 < x \leq 2\)[/tex]
In this interval, the function is given by [tex]\(f(x) = -2x - 1\)[/tex]. This represents a linear function with a slope of -2 and a y-intercept of -1.
#### Steps to sketch this line:
- Find the value of the function at the boundaries:
[tex]\[ f(-2) = -2(-2) - 1 = 4 - 1 = 3 \][/tex]
[tex]\[ f(2) = -2(2) - 1 = -4 - 1 = -5 \][/tex]
- Draw the line through the points [tex]\((-2, 3)\)[/tex] and [tex]\((2, -5)\)[/tex].
### 3. For [tex]\(x > 2\)[/tex]
In this interval, the function [tex]\(f(x) = -2\)[/tex]. This is another horizontal line, but this time at [tex]\(y = -2\)[/tex] for all [tex]\(x\)[/tex] greater than 2.
Sketch:
- This is a horizontal line at [tex]\(y = -2\)[/tex] from [tex]\(x = 2\)[/tex] to positive infinity.
### Combining all parts:
1. For [tex]\(x \leq -2\)[/tex]: Draw a horizontal line at [tex]\(y = 2\)[/tex] from [tex]\(x = -\infty\)[/tex] to [tex]\(x = -2\)[/tex].
2. For [tex]\(-2 < x \leq 2\)[/tex]: Draw the line segment connecting [tex]\((-2, 3)\)[/tex] and [tex]\((2, -5)\)[/tex], ensuring it passes through the linear function’s path.
3. For [tex]\(x > 2\)[/tex]: Draw a horizontal line at [tex]\(y = -2\)[/tex] from [tex]\(x = 2\)[/tex] to [tex]\(x = +\infty\)[/tex].
In summary:
- The function jumps from 2 at [tex]\(x = -2\)[/tex] to approximately 3.
- It then follows the line [tex]\(y = -2x - 1\)[/tex] from just above [tex]\(x = -2\)[/tex] down through the points [tex]\((0, -1)\)[/tex] to [tex]\((2, -5)\)[/tex].
- Finally, it drops to -2 after [tex]\(x = 2\)[/tex].
Here is a visual representation based on the described steps:
[tex]\[ \begin{array}{c|c|c} \text{x-values} & \text{Function} & \text{y-values} \\ \hline (-\infty, -2] & f(x) = 2 & 2 \\ (-2, 2] & f(x) = -2x - 1 & 3 \text{ to } -5 \\ (2, \infty) & f(x) = -2 & -2 \\ \end{array} \][/tex]
Each part transitions smoothly, except at [tex]\(x = -2\)[/tex] and [tex]\(x = 2\)[/tex], where the functions meet but do not continue as smooth curves. Place hollow (open) dots at these transition points to indicate jumps in function values.
[tex]\[ f(x) = \left\{ \begin{array}{lll} 2 & \text{if} & x \leq -2 \\ -2x - 1 & \text{if} & -2 < x \leq 2 \\ -2 & \text{if} & x > 2 \end{array} \right. \][/tex]
we analyze each piece of the function separately and then combine them on the same coordinate plane.
### 1. For [tex]\(x \leq -2\)[/tex]
In this interval, the function [tex]\(f(x) = 2\)[/tex]. This means the value of the function is a constant 2 for all [tex]\(x\)[/tex] less than or equal to -2.
Sketch:
- This is a horizontal line at [tex]\(y = 2\)[/tex] from negative infinity up to [tex]\(x = -2\)[/tex].
### 2. For [tex]\(-2 < x \leq 2\)[/tex]
In this interval, the function is given by [tex]\(f(x) = -2x - 1\)[/tex]. This represents a linear function with a slope of -2 and a y-intercept of -1.
#### Steps to sketch this line:
- Find the value of the function at the boundaries:
[tex]\[ f(-2) = -2(-2) - 1 = 4 - 1 = 3 \][/tex]
[tex]\[ f(2) = -2(2) - 1 = -4 - 1 = -5 \][/tex]
- Draw the line through the points [tex]\((-2, 3)\)[/tex] and [tex]\((2, -5)\)[/tex].
### 3. For [tex]\(x > 2\)[/tex]
In this interval, the function [tex]\(f(x) = -2\)[/tex]. This is another horizontal line, but this time at [tex]\(y = -2\)[/tex] for all [tex]\(x\)[/tex] greater than 2.
Sketch:
- This is a horizontal line at [tex]\(y = -2\)[/tex] from [tex]\(x = 2\)[/tex] to positive infinity.
### Combining all parts:
1. For [tex]\(x \leq -2\)[/tex]: Draw a horizontal line at [tex]\(y = 2\)[/tex] from [tex]\(x = -\infty\)[/tex] to [tex]\(x = -2\)[/tex].
2. For [tex]\(-2 < x \leq 2\)[/tex]: Draw the line segment connecting [tex]\((-2, 3)\)[/tex] and [tex]\((2, -5)\)[/tex], ensuring it passes through the linear function’s path.
3. For [tex]\(x > 2\)[/tex]: Draw a horizontal line at [tex]\(y = -2\)[/tex] from [tex]\(x = 2\)[/tex] to [tex]\(x = +\infty\)[/tex].
In summary:
- The function jumps from 2 at [tex]\(x = -2\)[/tex] to approximately 3.
- It then follows the line [tex]\(y = -2x - 1\)[/tex] from just above [tex]\(x = -2\)[/tex] down through the points [tex]\((0, -1)\)[/tex] to [tex]\((2, -5)\)[/tex].
- Finally, it drops to -2 after [tex]\(x = 2\)[/tex].
Here is a visual representation based on the described steps:
[tex]\[ \begin{array}{c|c|c} \text{x-values} & \text{Function} & \text{y-values} \\ \hline (-\infty, -2] & f(x) = 2 & 2 \\ (-2, 2] & f(x) = -2x - 1 & 3 \text{ to } -5 \\ (2, \infty) & f(x) = -2 & -2 \\ \end{array} \][/tex]
Each part transitions smoothly, except at [tex]\(x = -2\)[/tex] and [tex]\(x = 2\)[/tex], where the functions meet but do not continue as smooth curves. Place hollow (open) dots at these transition points to indicate jumps in function values.
