To determine which function has a domain of all real numbers, we need to analyze each function individually and identify any restrictions on the domain.
A. \( y = -x^{\frac{1}{2}} + 5 \)
This function includes a square root, \( x^{\frac{1}{2}} \), which is defined only for non-negative values of \( x \). Specifically, \( x \) must be \( \geq 0 \). Therefore, the domain of this function is \( x \geq 0 \), not all real numbers.
B. \( y = (x + 2)^{\frac{1}{2}} \)
Similar to the first function, this function includes a square root, \( (x + 2)^{\frac{1}{2}} \), which is defined only when the expression under the square root is non-negative. Thus, \( x + 2 \geq 0 \), or \( x \geq -2 \). Therefore, the domain of this function is \( x \geq -2 \), not all real numbers.
C. \( y = -2(3x)^{\frac{1}{6}} \)
Here, the expression involves a sixth root, \( (3x)^{\frac{1}{6}} \). Like other even roots, the sixth root is defined only for non-negative values of \( x \). Therefore, \( 3x \geq 0 \) which simplifies to \( x \geq 0 \). Hence, the domain of this function is \( x \geq 0 \), not all real numbers.
D. \( y = (2x)^{\frac{1}{3}} - 7 \)
This function involves a cube root, \( (2x)^{\frac{1}{3}} \). Unlike even roots, the cube root is defined for all real numbers \( x \); it can handle both positive and negative values as well as zero. Therefore, there are no restrictions on \( x \) and the domain of this function is all real numbers.
Upon analyzing the domains, we can conclude that the function which has a domain of all real numbers is:
[tex]\[ \text{D. } y = (2x)^{\frac{1}{3}} - 7 \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{4} \][/tex]
