Answer :
To determine all the discontinuities of the given function \( f(x) \), we need to analyze the behavior of the function at the boundaries where the definition changes. The critical points to check are \( x = -4 \), \( x = -2 \), and \( x = 4 \).
1. At \( x = -4 \):
- For \( x < -4 \), \( f(x) = 4 \).
- For \( -4 \leq x \leq -2 \), \( f(x) = (x + 2)^2 \).
To check if \( f(x) \) is continuous at \( x = -4 \), we need to compare the limits approaching from both sides:
- As \( x \) approaches \(-4\) from the right:
[tex]\[ \lim_{x \to -4^+} f(x) = (-4 + 2)^2 = 4 \][/tex]
- As \( x \) approaches \(-4\) from the left:
[tex]\[ \lim_{x \to -4^-} f(x) = 4 \][/tex]
The values match, thus \( f(x) \) is continuous at \( x = -4 \).
2. At \( x = -2 \):
- For \( -4 \leq x \leq -2 \), \( f(x) = (x + 2)^2 \).
- For \( -2 < x < 4 \), \( f(x) = -\frac{1}{2}x + 1 \).
To check if \( f(x) \) is continuous at \( x = -2 \), we need to compare the limits approaching from both sides:
- As \( x \) approaches \(-2\) from the right:
[tex]\[ \lim_{x \to -2^+} f(x) = -\frac{1}{2}(-2) + 1 = 2 \][/tex]
- As \( x \) approaches \(-2\) from the left:
[tex]\[ \lim_{x \to -2^-} f(x) = (-2 + 2)^2 = 0 \][/tex]
Since \( 0 \neq 2 \), \( f(x) \) has a jump discontinuity at \( x = -2 \).
3. At \( x = 4 \):
- For \( -2 < x < 4 \), \( f(x) = -\frac{1}{2}x + 1 \).
- For \( x > 4 \), \( f(x) = -1 \).
To check if \( f(x) \) is continuous at \( x = 4 \), we need to compare the limits approaching from both sides:
- As \( x \) approaches \( 4 \) from the right:
[tex]\[ \lim_{x \to 4^+} f(x) = -1 \][/tex]
- As \( x \) approaches \( 4 \) from the left:
[tex]\[ \lim_{x \to 4^-} f(x) = -\frac{1}{2}(4) + 1 = -1 \][/tex]
The values match, thus \( f(x) \) is continuous at \( x = 4 \).
Considering all necessary points, the function \( f(x) \) has a jump discontinuity at \( x = -2 \) and no discontinuities at \( x = -4 \) and \( x = 4 \).
Thus, the most accurate list of discontinuities for the function \( f(x) \) would be:
- Jump discontinuity at \( x = -2 \).
Therefore, the correct option is:
```
jump discontinuity at x=-2
```
1. At \( x = -4 \):
- For \( x < -4 \), \( f(x) = 4 \).
- For \( -4 \leq x \leq -2 \), \( f(x) = (x + 2)^2 \).
To check if \( f(x) \) is continuous at \( x = -4 \), we need to compare the limits approaching from both sides:
- As \( x \) approaches \(-4\) from the right:
[tex]\[ \lim_{x \to -4^+} f(x) = (-4 + 2)^2 = 4 \][/tex]
- As \( x \) approaches \(-4\) from the left:
[tex]\[ \lim_{x \to -4^-} f(x) = 4 \][/tex]
The values match, thus \( f(x) \) is continuous at \( x = -4 \).
2. At \( x = -2 \):
- For \( -4 \leq x \leq -2 \), \( f(x) = (x + 2)^2 \).
- For \( -2 < x < 4 \), \( f(x) = -\frac{1}{2}x + 1 \).
To check if \( f(x) \) is continuous at \( x = -2 \), we need to compare the limits approaching from both sides:
- As \( x \) approaches \(-2\) from the right:
[tex]\[ \lim_{x \to -2^+} f(x) = -\frac{1}{2}(-2) + 1 = 2 \][/tex]
- As \( x \) approaches \(-2\) from the left:
[tex]\[ \lim_{x \to -2^-} f(x) = (-2 + 2)^2 = 0 \][/tex]
Since \( 0 \neq 2 \), \( f(x) \) has a jump discontinuity at \( x = -2 \).
3. At \( x = 4 \):
- For \( -2 < x < 4 \), \( f(x) = -\frac{1}{2}x + 1 \).
- For \( x > 4 \), \( f(x) = -1 \).
To check if \( f(x) \) is continuous at \( x = 4 \), we need to compare the limits approaching from both sides:
- As \( x \) approaches \( 4 \) from the right:
[tex]\[ \lim_{x \to 4^+} f(x) = -1 \][/tex]
- As \( x \) approaches \( 4 \) from the left:
[tex]\[ \lim_{x \to 4^-} f(x) = -\frac{1}{2}(4) + 1 = -1 \][/tex]
The values match, thus \( f(x) \) is continuous at \( x = 4 \).
Considering all necessary points, the function \( f(x) \) has a jump discontinuity at \( x = -2 \) and no discontinuities at \( x = -4 \) and \( x = 4 \).
Thus, the most accurate list of discontinuities for the function \( f(x) \) would be:
- Jump discontinuity at \( x = -2 \).
Therefore, the correct option is:
```
jump discontinuity at x=-2
```
