Answer :
To determine if the function
[tex]\[ f(x) = \begin{cases} -4x + 3, & \text{if } x \leq 1 \\ x - 3, & \text{if } x > 1 \end{cases} \][/tex]
is continuous, we need to examine the behavior of the function around the point where the definition of the function changes, which is at [tex]\( x = 1 \)[/tex].
The criteria for continuity at [tex]\( x = 1 \)[/tex] require three conditions to be satisfied:
1. The function [tex]\( f(x) \)[/tex] must be defined at [tex]\( x = 1 \)[/tex].
2. The limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 1 from the left must exist and equal the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 1 from the right.
3. The value of the function at [tex]\( x = 1 \)[/tex] must equal this common limit.
Let's check these conditions step by step:
### 1. The function [tex]\( f(x) \)[/tex] at [tex]\( x = 1 \)[/tex]:
For [tex]\( x \leq 1 \)[/tex], the function is given by [tex]\( f(x) = -4x + 3 \)[/tex]. Thus, at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -4(1) + 3 = -4 + 3 = -1 \][/tex]
So, [tex]\( f(x) \)[/tex] is defined at [tex]\( x = 1 \)[/tex] and [tex]\( f(1) = -1 \)[/tex].
### 2. The limits from the left and right of [tex]\( x = 1 \)[/tex]:
- Left-hand limit (as [tex]\( x \)[/tex] approaches 1 from the left, [tex]\( x \to 1^- \)[/tex]):
For [tex]\( x \leq 1 \)[/tex], [tex]\( f(x) = -4x + 3 \)[/tex]. So,
[tex]\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-4x + 3) = -4(1) + 3 = -1 \][/tex]
- Right-hand limit (as [tex]\( x \)[/tex] approaches 1 from the right, [tex]\( x \to 1^+ \)[/tex]):
For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = x - 3 \)[/tex]. So,
[tex]\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 3) = 1 - 3 = -2 \][/tex]
### 3. Comparison of the limits and the function value at [tex]\( x = 1 \)[/tex]:
Comparing the limits
[tex]\[ \lim_{x \to 1^-} f(x) = -1 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = -2 \][/tex]
Since the left-hand limit (-1) is not equal to the right-hand limit (-2), the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 1 does not exist. Therefore, the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = 1 \)[/tex] because the limits from either side of 1 are not equal.
In conclusion, the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = 1 \)[/tex] because there is a value [tex]\( a \)[/tex] for which [tex]\( \lim_{x \to a} f(x) \)[/tex] does not exist.
Thus, the correct choice is:
"It is discontinuous because there is a value [tex]\( a \)[/tex] such that [tex]\( \lim_{x \rightarrow a} f(x) \)[/tex] does not exist."
[tex]\[ f(x) = \begin{cases} -4x + 3, & \text{if } x \leq 1 \\ x - 3, & \text{if } x > 1 \end{cases} \][/tex]
is continuous, we need to examine the behavior of the function around the point where the definition of the function changes, which is at [tex]\( x = 1 \)[/tex].
The criteria for continuity at [tex]\( x = 1 \)[/tex] require three conditions to be satisfied:
1. The function [tex]\( f(x) \)[/tex] must be defined at [tex]\( x = 1 \)[/tex].
2. The limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 1 from the left must exist and equal the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 1 from the right.
3. The value of the function at [tex]\( x = 1 \)[/tex] must equal this common limit.
Let's check these conditions step by step:
### 1. The function [tex]\( f(x) \)[/tex] at [tex]\( x = 1 \)[/tex]:
For [tex]\( x \leq 1 \)[/tex], the function is given by [tex]\( f(x) = -4x + 3 \)[/tex]. Thus, at [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -4(1) + 3 = -4 + 3 = -1 \][/tex]
So, [tex]\( f(x) \)[/tex] is defined at [tex]\( x = 1 \)[/tex] and [tex]\( f(1) = -1 \)[/tex].
### 2. The limits from the left and right of [tex]\( x = 1 \)[/tex]:
- Left-hand limit (as [tex]\( x \)[/tex] approaches 1 from the left, [tex]\( x \to 1^- \)[/tex]):
For [tex]\( x \leq 1 \)[/tex], [tex]\( f(x) = -4x + 3 \)[/tex]. So,
[tex]\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (-4x + 3) = -4(1) + 3 = -1 \][/tex]
- Right-hand limit (as [tex]\( x \)[/tex] approaches 1 from the right, [tex]\( x \to 1^+ \)[/tex]):
For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = x - 3 \)[/tex]. So,
[tex]\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x - 3) = 1 - 3 = -2 \][/tex]
### 3. Comparison of the limits and the function value at [tex]\( x = 1 \)[/tex]:
Comparing the limits
[tex]\[ \lim_{x \to 1^-} f(x) = -1 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = -2 \][/tex]
Since the left-hand limit (-1) is not equal to the right-hand limit (-2), the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 1 does not exist. Therefore, the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = 1 \)[/tex] because the limits from either side of 1 are not equal.
In conclusion, the function [tex]\( f(x) \)[/tex] is discontinuous at [tex]\( x = 1 \)[/tex] because there is a value [tex]\( a \)[/tex] for which [tex]\( \lim_{x \to a} f(x) \)[/tex] does not exist.
Thus, the correct choice is:
"It is discontinuous because there is a value [tex]\( a \)[/tex] such that [tex]\( \lim_{x \rightarrow a} f(x) \)[/tex] does not exist."
