Let's analyze the function [tex]\( f(x) = a(x+k)^{1/n} + c \)[/tex] in terms of its domain and range step-by-step.
### Domain:
1. The domain of a function is the set of all possible input values (x-values) that the function can accept.
2. Here, we have a radical function with a fractional exponent [tex]\( (1/n) \)[/tex].
3. For even [tex]\( n \)[/tex], the expression inside the radical, [tex]\( (x+k) \)[/tex], must be non-negative because the even root of a negative number is not defined in the set of real numbers.
4. Therefore, for even [tex]\( n \)[/tex], the value of [tex]\( (x+k) \)[/tex] must be greater than or equal to 0:
[tex]\[
x + k \geq 0 \implies x \geq -k
\][/tex]
5. Hence, for even [tex]\( n \)[/tex], the domain is [tex]\([-k, \infty)\)[/tex].
### Range:
1. The range of the function is the set of all possible output values (y-values) the function can produce.
2. Evaluate the function at the endpoint of the domain. When [tex]\( x = -k \)[/tex], the expression inside the radical becomes zero:
[tex]\[
f(-k) = a(0)^{1/n} + c = a \cdot 0 + c = c
\][/tex]
3. As [tex]\( x \)[/tex] increases from [tex]\(-k\)[/tex] to [tex]\(\infty\)[/tex], [tex]\( (x+k)^{1/n} \)[/tex] produces values starting from 0 and increasing to infinity.
4. Because the radical function produces non-negative values, multiplying by [tex]\( a \)[/tex] keeps it non-negative if [tex]\( a \)[/tex] is positive. Adding [tex]\( c \)[/tex] shifts the entire range upwards by [tex]\( c \)[/tex]:
[tex]\[
f(x) = a(x+k)^{1/n} + c \quad \text{starts from} \quad c \quad \text{and goes to} \quad \infty
\][/tex]
5. Thus, the range of the function is [tex]\([c, \infty)\)[/tex].
By analyzing both the domain and range, the correct answer is:
A. The domain is [tex]\([-k, \infty)\)[/tex], and the range is [tex]\([c, \infty)\)[/tex].
