Answer :
To determine which function is continuous across its domain, we need to check each function at the boundaries where the pieces of the function change. These boundaries are at [tex]\( x = -2 \)[/tex] and [tex]\( x = 4 \)[/tex].
### Function A:
[tex]\[ f(x) = \begin{cases} x + 4, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 20 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x + 4) = -2 + 4 = 2 \)[/tex]
- Right-hand limit: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 0.5(-2)^2 = 2 \)[/tex]
- Function value at [tex]\( x = -2 \)[/tex]: [tex]\( f(-2) = 0.5 (-2)^2 = 2 \)[/tex]
- Since all these values are equal, the function is continuous at [tex]\( x = -2 \)[/tex].
2. At [tex]\( x = 4 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to 4^-} (0.5 x^2) = 0.5(4)^2 = 8 \)[/tex]
- Right-hand limit: [tex]\( \lim_{x \to 4^+} (20 - 3x) = 20 - 3(4) = 20 - 12 = 8 \)[/tex]
- Function value at [tex]\( x = 4 \)[/tex]: [tex]\( f(4) = 8 \)[/tex]
- Since all these values are equal, the function is continuous at [tex]\( x = 4 \)[/tex].
### Function B:
[tex]\[ f(x) = \begin{cases} x + 4, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 25 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x + 4) = 2 \)[/tex]
- Right-hand limit and function value at [tex]\( x = -2 \)[/tex]: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 2 \)[/tex]
- These match, so the function is continuous at [tex]\( x = -2 \)[/tex].
2. At [tex]\( x = 4 \)[/tex]:
- Left-hand limit and function value at [tex]\( x = 4 \)[/tex]: [tex]\( \lim_{x \to 4^-} (0.5 x^2) = 8 \)[/tex]
- Right-hand limit: [tex]\( \lim_{x \to 4^+} (25 - 3x) = 13 \)[/tex]
- Since these values do not match, the function is not continuous at [tex]\( x = 4 \)[/tex].
### Function C:
[tex]\[ f(x) = \begin{cases} x - 2, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 25 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x - 2) = -4 \)[/tex]
- Right-hand limit and function value at [tex]\( x = -2 \)[/tex]: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 2 \)[/tex]
- These values do not match, so the function is not continuous at [tex]\( x = -2 \)[/tex].
### Function D:
[tex]\[ f(x) = \begin{cases} x + 6, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 20 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x + 6) = 4 \)[/tex]
- Right-hand limit and function value at [tex]\( x = -2 \)[/tex]: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 2 \)[/tex]
- These values do not match, so the function is not continuous at [tex]\( x = -2 \)[/tex].
### Conclusion:
After evaluating each function for continuity at the boundaries, we observe that none of the functions is continuous across its entire domain. Thus, the answer is:
[tex]\[ \boxed{\text{None}} \][/tex]
### Function A:
[tex]\[ f(x) = \begin{cases} x + 4, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 20 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x + 4) = -2 + 4 = 2 \)[/tex]
- Right-hand limit: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 0.5(-2)^2 = 2 \)[/tex]
- Function value at [tex]\( x = -2 \)[/tex]: [tex]\( f(-2) = 0.5 (-2)^2 = 2 \)[/tex]
- Since all these values are equal, the function is continuous at [tex]\( x = -2 \)[/tex].
2. At [tex]\( x = 4 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to 4^-} (0.5 x^2) = 0.5(4)^2 = 8 \)[/tex]
- Right-hand limit: [tex]\( \lim_{x \to 4^+} (20 - 3x) = 20 - 3(4) = 20 - 12 = 8 \)[/tex]
- Function value at [tex]\( x = 4 \)[/tex]: [tex]\( f(4) = 8 \)[/tex]
- Since all these values are equal, the function is continuous at [tex]\( x = 4 \)[/tex].
### Function B:
[tex]\[ f(x) = \begin{cases} x + 4, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 25 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x + 4) = 2 \)[/tex]
- Right-hand limit and function value at [tex]\( x = -2 \)[/tex]: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 2 \)[/tex]
- These match, so the function is continuous at [tex]\( x = -2 \)[/tex].
2. At [tex]\( x = 4 \)[/tex]:
- Left-hand limit and function value at [tex]\( x = 4 \)[/tex]: [tex]\( \lim_{x \to 4^-} (0.5 x^2) = 8 \)[/tex]
- Right-hand limit: [tex]\( \lim_{x \to 4^+} (25 - 3x) = 13 \)[/tex]
- Since these values do not match, the function is not continuous at [tex]\( x = 4 \)[/tex].
### Function C:
[tex]\[ f(x) = \begin{cases} x - 2, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 25 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x - 2) = -4 \)[/tex]
- Right-hand limit and function value at [tex]\( x = -2 \)[/tex]: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 2 \)[/tex]
- These values do not match, so the function is not continuous at [tex]\( x = -2 \)[/tex].
### Function D:
[tex]\[ f(x) = \begin{cases} x + 6, & -4 \leq x < -2 \\ 0.5 x^2, & -2 \leq x < 4 \\ 20 - 3x, & 4 \leq x \leq 8 \end{cases} \][/tex]
1. At [tex]\( x = -2 \)[/tex]:
- Left-hand limit: [tex]\( \lim_{x \to -2^-} (x + 6) = 4 \)[/tex]
- Right-hand limit and function value at [tex]\( x = -2 \)[/tex]: [tex]\( \lim_{x \to -2^+} (0.5 x^2) = 2 \)[/tex]
- These values do not match, so the function is not continuous at [tex]\( x = -2 \)[/tex].
### Conclusion:
After evaluating each function for continuity at the boundaries, we observe that none of the functions is continuous across its entire domain. Thus, the answer is:
[tex]\[ \boxed{\text{None}} \][/tex]
