Answer :
Let's analyze the given piecewise function:
[tex]\[ f(x) = \begin{cases} 6 & \text{if } x < -4 \\ -10 + x^2 & \text{if } -4 \leq x < 4 \\ 2x - 2 & \text{if } x \geq 4 \end{cases} \][/tex]
### Graphing the Function
Let's graph the function by considering each piece separately.
#### 1. For [tex]\( x < -4 \)[/tex]:
The function [tex]\( f(x) = 6 \)[/tex] is a horizontal line at [tex]\( y = 6 \)[/tex] for all [tex]\( x < -4 \)[/tex].
#### 2. For [tex]\( -4 \leq x < 4 \)[/tex]:
The function [tex]\( f(x) = -10 + x^2 \)[/tex] is a parabola that opens upwards (since the coefficient of [tex]\( x^2 \)[/tex] is positive). It starts from [tex]\( x = -4 \)[/tex] and ends at [tex]\( x = 4 \)[/tex] without including 4.
#### 3. For [tex]\( x \geq 4 \)[/tex]:
The function [tex]\( f(x) = 2x - 2 \)[/tex] is a linear equation with a slope of 2 and a y-intercept of -2. It starts from [tex]\( x = 4 \)[/tex] and goes to infinity.
### Determining Continuity
To determine the continuity of the function at [tex]\( x = -4 \)[/tex] and [tex]\( x = 4 \)[/tex], we need to check if the left-hand limit, right-hand limit, and the value of the function at these points are all equal.
#### 1. At [tex]\( x = -4 \)[/tex]:
- Left-hand limit as [tex]\( x \to -4^- \)[/tex]:
[tex]\[ \lim_{{x \to -4^-}} f(x) = 6 \][/tex]
Since [tex]\( f(x) = 6 \)[/tex] for [tex]\( x < -4 \)[/tex].
- Right-hand limit as [tex]\( x \to -4^+ \)[/tex]:
[tex]\[ \lim_{{x \to -4^+}} f(x) = -10 + (-4)^2 = -10 + 16 = 6 \][/tex]
Since [tex]\( f(x) = -10 + x^2 \)[/tex] for [tex]\( -4 \leq x < 4 \)[/tex].
- Value at [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = -10 + (-4)^2 = -10 + 16 = 6 \][/tex]
Since the left-hand limit, right-hand limit, and the actual value of the function at [tex]\( x = -4 \)[/tex] are all equal to 6, the function is continuous at [tex]\( x = -4 \)[/tex].
#### 2. At [tex]\( x = 4 \)[/tex]:
- Left-hand limit as [tex]\( x \to 4^- \)[/tex]:
[tex]\[ \lim_{{x \to 4^-}} f(x) = -10 + 4^2 = -10 + 16 = 6 \][/tex]
Since [tex]\( f(x) = -10 + x^2 \)[/tex] for [tex]\( -4 \leq x < 4 \)[/tex].
- Right-hand limit as [tex]\( x \to 4^+ \)[/tex]:
[tex]\[ \lim_{{x \to 4^+}} f(x) = 2 \cdot 4 - 2 = 8 - 2 = 6 \][/tex]
Since [tex]\( f(x) = 2x - 2 \)[/tex] for [tex]\( x \geq 4 \)[/tex].
- Value at [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 2 \cdot 4 - 2 = 8 - 2 = 6 \][/tex]
Since the left-hand limit, right-hand limit, and the actual value of the function at [tex]\( x = 4 \)[/tex] are all equal to 6, the function is continuous at [tex]\( x = 4 \)[/tex].
### Conclusion
The function [tex]\( f \)[/tex] is continuous for all [tex]\( x \)[/tex] because it is continuous at both critical points [tex]\( x = -4 \)[/tex] and [tex]\( x = 4 \)[/tex], and it is composed of continuous functions on their respective intervals.
### Graph of [tex]\( f(x) \)[/tex]:
Here is a rough sketch of the graph of [tex]\( f(x) \)[/tex]:
- For [tex]\( x < -4 \)[/tex], the graph is a horizontal line at [tex]\( y = 6 \)[/tex].
- For [tex]\( -4 \leq x < 4 \)[/tex], the graph is a parabola opening upwards, starting at [tex]\( ( -4,6) \)[/tex] and ending at [tex]\( (4, 6) \)[/tex].
- For [tex]\( x \geq 4 \)[/tex], the graph is a line with slope 2, starting from [tex]\( (4, 6) \)[/tex].
This combined graph represents a continuous function across the entire real line.
[tex]\[ f(x) = \begin{cases} 6 & \text{if } x < -4 \\ -10 + x^2 & \text{if } -4 \leq x < 4 \\ 2x - 2 & \text{if } x \geq 4 \end{cases} \][/tex]
### Graphing the Function
Let's graph the function by considering each piece separately.
#### 1. For [tex]\( x < -4 \)[/tex]:
The function [tex]\( f(x) = 6 \)[/tex] is a horizontal line at [tex]\( y = 6 \)[/tex] for all [tex]\( x < -4 \)[/tex].
#### 2. For [tex]\( -4 \leq x < 4 \)[/tex]:
The function [tex]\( f(x) = -10 + x^2 \)[/tex] is a parabola that opens upwards (since the coefficient of [tex]\( x^2 \)[/tex] is positive). It starts from [tex]\( x = -4 \)[/tex] and ends at [tex]\( x = 4 \)[/tex] without including 4.
#### 3. For [tex]\( x \geq 4 \)[/tex]:
The function [tex]\( f(x) = 2x - 2 \)[/tex] is a linear equation with a slope of 2 and a y-intercept of -2. It starts from [tex]\( x = 4 \)[/tex] and goes to infinity.
### Determining Continuity
To determine the continuity of the function at [tex]\( x = -4 \)[/tex] and [tex]\( x = 4 \)[/tex], we need to check if the left-hand limit, right-hand limit, and the value of the function at these points are all equal.
#### 1. At [tex]\( x = -4 \)[/tex]:
- Left-hand limit as [tex]\( x \to -4^- \)[/tex]:
[tex]\[ \lim_{{x \to -4^-}} f(x) = 6 \][/tex]
Since [tex]\( f(x) = 6 \)[/tex] for [tex]\( x < -4 \)[/tex].
- Right-hand limit as [tex]\( x \to -4^+ \)[/tex]:
[tex]\[ \lim_{{x \to -4^+}} f(x) = -10 + (-4)^2 = -10 + 16 = 6 \][/tex]
Since [tex]\( f(x) = -10 + x^2 \)[/tex] for [tex]\( -4 \leq x < 4 \)[/tex].
- Value at [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = -10 + (-4)^2 = -10 + 16 = 6 \][/tex]
Since the left-hand limit, right-hand limit, and the actual value of the function at [tex]\( x = -4 \)[/tex] are all equal to 6, the function is continuous at [tex]\( x = -4 \)[/tex].
#### 2. At [tex]\( x = 4 \)[/tex]:
- Left-hand limit as [tex]\( x \to 4^- \)[/tex]:
[tex]\[ \lim_{{x \to 4^-}} f(x) = -10 + 4^2 = -10 + 16 = 6 \][/tex]
Since [tex]\( f(x) = -10 + x^2 \)[/tex] for [tex]\( -4 \leq x < 4 \)[/tex].
- Right-hand limit as [tex]\( x \to 4^+ \)[/tex]:
[tex]\[ \lim_{{x \to 4^+}} f(x) = 2 \cdot 4 - 2 = 8 - 2 = 6 \][/tex]
Since [tex]\( f(x) = 2x - 2 \)[/tex] for [tex]\( x \geq 4 \)[/tex].
- Value at [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 2 \cdot 4 - 2 = 8 - 2 = 6 \][/tex]
Since the left-hand limit, right-hand limit, and the actual value of the function at [tex]\( x = 4 \)[/tex] are all equal to 6, the function is continuous at [tex]\( x = 4 \)[/tex].
### Conclusion
The function [tex]\( f \)[/tex] is continuous for all [tex]\( x \)[/tex] because it is continuous at both critical points [tex]\( x = -4 \)[/tex] and [tex]\( x = 4 \)[/tex], and it is composed of continuous functions on their respective intervals.
### Graph of [tex]\( f(x) \)[/tex]:
Here is a rough sketch of the graph of [tex]\( f(x) \)[/tex]:
- For [tex]\( x < -4 \)[/tex], the graph is a horizontal line at [tex]\( y = 6 \)[/tex].
- For [tex]\( -4 \leq x < 4 \)[/tex], the graph is a parabola opening upwards, starting at [tex]\( ( -4,6) \)[/tex] and ending at [tex]\( (4, 6) \)[/tex].
- For [tex]\( x \geq 4 \)[/tex], the graph is a line with slope 2, starting from [tex]\( (4, 6) \)[/tex].
This combined graph represents a continuous function across the entire real line.
