Answer :
To determine the domain and range of the given radical function \( f(x) = a(x + k)^{1/n} + c \), we need to understand the behavior of this function based on its mathematical form.
Firstly, let's examine the domain. The expression inside the radical, \( (x + k) \), must be defined for the function \( (x + k)^{1/n} \) to be meaningful.
Domain Analysis:
- For even values of \(n\) (e.g., \(n = 2, 4, 6, \ldots\)):
- We need \( x + k \geq 0 \) because even roots are only defined for non-negative values inside the radical.
- Therefore, \( x \geq -k \).
- For odd values of \(n\) (e.g., \(n = 3, 5, 7, \ldots\)):
- The expression \( (x + k) \) can be any real number because odd roots are defined for all real numbers.
- Therefore, \( x \) can be any real number.
However, iterating over the common notation, let's assume \(n\) is 2 for simplicity (which is the simplest case usually considered in elementary algebra):
[tex]\[ x + k \geq 0 \implies x \geq -k \][/tex]
This means the domain of \( f(x) \) is \([ -k, \infty )\).
Range Analysis:
- Consider the radical function starts for \( x = -k \):
- When \(x = -k \), \( f(x) = a(0) + c = c\).
- This gives the minimum or maximum value of the function depending on the sign and magnitude of \(a\).
- As \( x \) increases (since it can go up to \(\infty\)):
- The radical term \( (x + k)^{1/n} \) becomes larger because the inside term \( x + k \) grows, regardless of \(a\).
- Therefore, for positive \(a\), the total function value \( f(x) \rightarrow \infty \).
Hence, the function grows from \(c\) upwards, indicating that the range is \([c, \infty)\).
Based on this detailed analysis, we can say:
- Domain: \([-k, \infty)\)
- Range: \([c, \infty)\)
Thus, option D "The domain is \( (-\infty, k] \), and the range is \( [c, \infty) \)" would not be correct due to mismatched notation. Instead, what we conclude would lead us to different domain considerations if matched correctly.
However, the best fitting options considering typical expected pairs for give format should be formulated in problems.
Based on domain analysis correctly,
Correct Domain-Range Pair fit `[-k, \infty)` and `[c, \infty)` should formulate better:
D extend[-mismatched, correct conditions -> expected versatile educational concern when proper aligned).
One should consider \( [-k,\infty);[c,\infty] ` well justified typically within computed necessary truthful solutions. \]
Firstly, let's examine the domain. The expression inside the radical, \( (x + k) \), must be defined for the function \( (x + k)^{1/n} \) to be meaningful.
Domain Analysis:
- For even values of \(n\) (e.g., \(n = 2, 4, 6, \ldots\)):
- We need \( x + k \geq 0 \) because even roots are only defined for non-negative values inside the radical.
- Therefore, \( x \geq -k \).
- For odd values of \(n\) (e.g., \(n = 3, 5, 7, \ldots\)):
- The expression \( (x + k) \) can be any real number because odd roots are defined for all real numbers.
- Therefore, \( x \) can be any real number.
However, iterating over the common notation, let's assume \(n\) is 2 for simplicity (which is the simplest case usually considered in elementary algebra):
[tex]\[ x + k \geq 0 \implies x \geq -k \][/tex]
This means the domain of \( f(x) \) is \([ -k, \infty )\).
Range Analysis:
- Consider the radical function starts for \( x = -k \):
- When \(x = -k \), \( f(x) = a(0) + c = c\).
- This gives the minimum or maximum value of the function depending on the sign and magnitude of \(a\).
- As \( x \) increases (since it can go up to \(\infty\)):
- The radical term \( (x + k)^{1/n} \) becomes larger because the inside term \( x + k \) grows, regardless of \(a\).
- Therefore, for positive \(a\), the total function value \( f(x) \rightarrow \infty \).
Hence, the function grows from \(c\) upwards, indicating that the range is \([c, \infty)\).
Based on this detailed analysis, we can say:
- Domain: \([-k, \infty)\)
- Range: \([c, \infty)\)
Thus, option D "The domain is \( (-\infty, k] \), and the range is \( [c, \infty) \)" would not be correct due to mismatched notation. Instead, what we conclude would lead us to different domain considerations if matched correctly.
However, the best fitting options considering typical expected pairs for give format should be formulated in problems.
Based on domain analysis correctly,
Correct Domain-Range Pair fit `[-k, \infty)` and `[c, \infty)` should formulate better:
D extend[-mismatched, correct conditions -> expected versatile educational concern when proper aligned).
One should consider \( [-k,\infty);[c,\infty] ` well justified typically within computed necessary truthful solutions. \]
