Answer :
Sure, let's carefully analyze, plot, and examine the function [tex]\( f(x) \)[/tex].
The function [tex]\( f(x) \)[/tex] is defined in two different pieces:
[tex]\[ f(x) = \begin{cases} -x + 3 & \text{if } x < -1 \\ -3x + 1 & \text{if } x \geq -1 \end{cases} \][/tex]
### Step 1: Plotting the Function
1. For [tex]\( x < -1 \)[/tex]:
- The function is [tex]\( f(x) = -x + 3 \)[/tex].
- This is a linear function with a slope of -1 and a y-intercept of 3.
2. For [tex]\( x \geq -1 \)[/tex]:
- The function is [tex]\( f(x) = -3x + 1 \)[/tex].
- This is also a linear function, but with a slope of -3 and a y-intercept of 1.
### Step 2: Examine the Behavior at the Point [tex]\( x = -1 \)[/tex]
#### Left-Hand Side for [tex]\( x < -1 \)[/tex]:
[tex]\[ \lim_{{x \to -1^-}} f(x) = -(-1) + 3 = 1 + 3 = 4 \][/tex]
#### Right-Hand Side for [tex]\( x \geq -1 \)[/tex]:
[tex]\[ \lim_{{x \to -1^+}} f(x) = -3(-1) + 1 = 3 + 1 = 4 \][/tex]
Both the left-hand limit and the right-hand limit as [tex]\( x \)[/tex] approaches [tex]\(-1\)[/tex] are equal to 4. Thus,
[tex]\[ \lim_{{x \to -1^-}} f(x) = \lim_{{x \to -1^+}} f(x) = 4 \][/tex]
Also, we need to ensure the function value at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -3(-1) + 1 = 3 + 1 = 4 \][/tex]
Since [tex]\( \lim_{{x \to -1^-}} f(x) = \lim_{{x \to -1^+}} f(x) = f(-1) = 4 \)[/tex], the function is indeed continuous at [tex]\( x = -1 \)[/tex].
### Step 3: Graphing the Function
To graph the function properly, we consider the two intervals:
1. For [tex]\( x < -1 \)[/tex] (line segment for [tex]\( f(x) = -x + 3 \)[/tex]):
- It passes through the point [tex]\((-1, 4)\)[/tex] but does not include [tex]\((-1, 4)\)[/tex] (open circle).
2. For [tex]\( x \geq -1 \)[/tex] (line segment for [tex]\( f(x) = -3x + 1 \)[/tex]):
- It starts at [tex]\((-1, 4)\)[/tex] including the point (closed circle) and continues.
Below is a descriptive plot:
- An open circle at [tex]\((-1, 4)\)[/tex] for the first piece ([tex]\(x < -1\)[/tex]).
- A filled circle at [tex]\((-1, 4)\)[/tex] for the second piece ([tex]\(x \geq -1\)[/tex]).
### Final Result
- The function [tex]\( f(x) \)[/tex] is continuous for all [tex]\( x \in \mathbb{R} \)[/tex].
The correct graph of [tex]\( f(x) \)[/tex] includes:
- A line with slope -1 starting from [tex]\((-\infty, 3)\)[/tex], passing [tex]\((-1, 4)\)[/tex] but not including it (open circle).
- Another line with slope -3 starting at [tex]\((-1, 4)\)[/tex] including it (closed circle), and extending to [tex]\( (+\infty, -\infty) \)[/tex].
So, we have successfully graphed the function and confirmed its continuity at [tex]\( x = -1 \)[/tex].
The function [tex]\( f(x) \)[/tex] is defined in two different pieces:
[tex]\[ f(x) = \begin{cases} -x + 3 & \text{if } x < -1 \\ -3x + 1 & \text{if } x \geq -1 \end{cases} \][/tex]
### Step 1: Plotting the Function
1. For [tex]\( x < -1 \)[/tex]:
- The function is [tex]\( f(x) = -x + 3 \)[/tex].
- This is a linear function with a slope of -1 and a y-intercept of 3.
2. For [tex]\( x \geq -1 \)[/tex]:
- The function is [tex]\( f(x) = -3x + 1 \)[/tex].
- This is also a linear function, but with a slope of -3 and a y-intercept of 1.
### Step 2: Examine the Behavior at the Point [tex]\( x = -1 \)[/tex]
#### Left-Hand Side for [tex]\( x < -1 \)[/tex]:
[tex]\[ \lim_{{x \to -1^-}} f(x) = -(-1) + 3 = 1 + 3 = 4 \][/tex]
#### Right-Hand Side for [tex]\( x \geq -1 \)[/tex]:
[tex]\[ \lim_{{x \to -1^+}} f(x) = -3(-1) + 1 = 3 + 1 = 4 \][/tex]
Both the left-hand limit and the right-hand limit as [tex]\( x \)[/tex] approaches [tex]\(-1\)[/tex] are equal to 4. Thus,
[tex]\[ \lim_{{x \to -1^-}} f(x) = \lim_{{x \to -1^+}} f(x) = 4 \][/tex]
Also, we need to ensure the function value at [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = -3(-1) + 1 = 3 + 1 = 4 \][/tex]
Since [tex]\( \lim_{{x \to -1^-}} f(x) = \lim_{{x \to -1^+}} f(x) = f(-1) = 4 \)[/tex], the function is indeed continuous at [tex]\( x = -1 \)[/tex].
### Step 3: Graphing the Function
To graph the function properly, we consider the two intervals:
1. For [tex]\( x < -1 \)[/tex] (line segment for [tex]\( f(x) = -x + 3 \)[/tex]):
- It passes through the point [tex]\((-1, 4)\)[/tex] but does not include [tex]\((-1, 4)\)[/tex] (open circle).
2. For [tex]\( x \geq -1 \)[/tex] (line segment for [tex]\( f(x) = -3x + 1 \)[/tex]):
- It starts at [tex]\((-1, 4)\)[/tex] including the point (closed circle) and continues.
Below is a descriptive plot:
- An open circle at [tex]\((-1, 4)\)[/tex] for the first piece ([tex]\(x < -1\)[/tex]).
- A filled circle at [tex]\((-1, 4)\)[/tex] for the second piece ([tex]\(x \geq -1\)[/tex]).
### Final Result
- The function [tex]\( f(x) \)[/tex] is continuous for all [tex]\( x \in \mathbb{R} \)[/tex].
The correct graph of [tex]\( f(x) \)[/tex] includes:
- A line with slope -1 starting from [tex]\((-\infty, 3)\)[/tex], passing [tex]\((-1, 4)\)[/tex] but not including it (open circle).
- Another line with slope -3 starting at [tex]\((-1, 4)\)[/tex] including it (closed circle), and extending to [tex]\( (+\infty, -\infty) \)[/tex].
So, we have successfully graphed the function and confirmed its continuity at [tex]\( x = -1 \)[/tex].
