To determine which function has a domain of all real numbers, we need to analyze each function individually and identify the domain for each one.
Option A: [tex]\( y = -2(3x)^{\frac{1}{6}} \)[/tex]
This function involves a sixth root, [tex]\((3x)^{\frac{1}{6}}\)[/tex]. The expression inside the root, [tex]\(3x\)[/tex], must be non-negative for the root to be defined.
[tex]\[ 3x \geq 0 \implies x \geq 0 \][/tex]
Thus, the domain of this function is [tex]\( x \geq 0 \)[/tex]. This means Option A does not have a domain of all real numbers.
Option B: [tex]\( y = -x^{\frac{1}{2}} + 5 \)[/tex]
This function involves a square root, [tex]\(x^{\frac{1}{2}}\)[/tex]. The expression inside the root, [tex]\(x\)[/tex], must be non-negative for the root to be defined.
[tex]\[ x \geq 0 \][/tex]
Thus, the domain of this function is [tex]\( x \geq 0 \)[/tex]. This means Option B does not have a domain of all real numbers.
Option C: [tex]\( y = (x + 2)^{\frac{1}{4}} \)[/tex]
This function involves a fourth root, [tex]\((x+2)^{\frac{1}{4}}\)[/tex]. The expression inside the root, [tex]\(x + 2\)[/tex], must be non-negative for the root to be defined.
[tex]\[ x + 2 \geq 0 \implies x \geq -2 \][/tex]
Thus, the domain of this function is [tex]\( x \geq -2 \)[/tex]. This means Option C does not have a domain of all real numbers.
Option D: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
This function involves a cube root, [tex]\((2x)^{\frac{1}{3}}\)[/tex]. Cube roots are defined for all real numbers.
There are no restrictions on the value of [tex]\(x\)[/tex]. This means the domain of this function is all real numbers.
Conclusion:
From the analysis, it is clear that Option D, [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex], is the only function that has a domain of all real numbers.
So, the correct answer is:
[tex]\[ \boxed{4} \][/tex]
