To determine which function has a restricted domain, we need to analyze each function and see if there are any values of [tex]\( x \)[/tex] that cause the function to be undefined or invalid.
Function A: [tex]\( k(x) = (-x + 3)^{\frac{1}{2}} \)[/tex]
- The expression [tex]\((-x + 3)\)[/tex] is under a square root. A square root is only defined for non-negative values, meaning [tex]\(-x + 3 \geq 0\)[/tex].
- Solving the inequality:
[tex]\[
-x + 3 \geq 0
\][/tex]
[tex]\[
-x \geq -3
\][/tex]
[tex]\[
x \leq 3
\][/tex]
- Therefore, the domain of [tex]\( k(x) \)[/tex] is all [tex]\( x \)[/tex] such that [tex]\( x \leq 3 \)[/tex].
Function B: [tex]\( h(x) = (4x)^2 - 5 \)[/tex]
- This is a polynomial function, and polynomials are defined for all real numbers. Therefore, there are no restrictions on [tex]\( x \)[/tex].
- The domain of [tex]\( h(x) \)[/tex] is all real numbers.
Function C: [tex]\( g(x) = -(x + 8)^3 \)[/tex]
- This is also a polynomial function in the form of a cubic function. Cubic functions are defined for all real numbers. Therefore, there are no restrictions on [tex]\( x \)[/tex].
- The domain of [tex]\( g(x) \)[/tex] is all real numbers.
Function D: [tex]\( j(x) = (3x)^{\frac{1}{5}} - 4 \)[/tex]
- This function involves taking the fifth root of [tex]\( 3x \)[/tex]. The fifth root (or any odd root) is defined for all real numbers, which means there are no restrictions on [tex]\( x \)[/tex].
- The domain of [tex]\( j(x) \)[/tex] is all real numbers.
After analyzing all four functions, we see that the only function with a restricted domain is Function A, [tex]\( k(x) = (-x + 3)^{\frac{1}{2}} \)[/tex].
Thus, the correct answer is:
[tex]\[
\boxed{A}
\][/tex]
