Answer :
To determine which function has a domain of all real numbers, we need to analyze each given function in terms of its domain, i.e., the set of all possible input values \(x\) for which the function is defined.
Option A: \(y = -x^{\frac{1}{2}} + 5\)
This function contains the term \(x^{\frac{1}{2}}\), which represents the square root of \(x\). The square root function is only defined for non-negative values of \(x\):
[tex]\[ x^{\frac{1}{2}} \text{ is defined if and only if } x \geq 0 \][/tex]
Thus, the domain of this function is \(x \geq 0\). This function does not have a domain of all real numbers.
Option B: \(y = (2x)^{\frac{1}{3}} - 7\)
This function contains the term \((2x)^{\frac{1}{3}}\), which represents the cube root of \(2x\). The cube root function is defined for all real numbers, whether positive, negative, or zero:
[tex]\[ (2x)^{\frac{1}{3}} \text{ is defined for all } x \in \mathbb{R} \][/tex]
Thus, the domain of this function is all real numbers.
Option C: \(y = (x + 2)^{\frac{1}{4}}\)
This function involves the term \((x + 2)^{\frac{1}{4}}\), which represents the fourth root of \((x + 2)\). Similar to the square root, the fourth root function is only defined for non-negative values of the argument:
[tex]\[ (x + 2)^{\frac{1}{4}} \text{ is defined if and only if } x + 2 \geq 0 \implies x \geq -2 \][/tex]
Thus, the domain of this function is \(x \geq -2\). This function does not have a domain of all real numbers.
Option D: \(y = -2(3x)^{\frac{1}{6}}\)
This function includes the term \((3x)^{\frac{1}{6}}\), which represents the sixth root of \(3x\). Like the square root and fourth root, the sixth root function is only defined for non-negative values of the argument:
[tex]\[ (3x)^{\frac{1}{6}} \text{ is defined if and only if } 3x \geq 0 \implies x \geq 0 \][/tex]
Thus, the domain of this function is \(x \geq 0\). This function does not have a domain of all real numbers.
After analyzing all the options, we conclude that the only function with a domain of all real numbers is:
B. [tex]\(y = (2x)^{\frac{1}{3}} - 7\)[/tex]
Option A: \(y = -x^{\frac{1}{2}} + 5\)
This function contains the term \(x^{\frac{1}{2}}\), which represents the square root of \(x\). The square root function is only defined for non-negative values of \(x\):
[tex]\[ x^{\frac{1}{2}} \text{ is defined if and only if } x \geq 0 \][/tex]
Thus, the domain of this function is \(x \geq 0\). This function does not have a domain of all real numbers.
Option B: \(y = (2x)^{\frac{1}{3}} - 7\)
This function contains the term \((2x)^{\frac{1}{3}}\), which represents the cube root of \(2x\). The cube root function is defined for all real numbers, whether positive, negative, or zero:
[tex]\[ (2x)^{\frac{1}{3}} \text{ is defined for all } x \in \mathbb{R} \][/tex]
Thus, the domain of this function is all real numbers.
Option C: \(y = (x + 2)^{\frac{1}{4}}\)
This function involves the term \((x + 2)^{\frac{1}{4}}\), which represents the fourth root of \((x + 2)\). Similar to the square root, the fourth root function is only defined for non-negative values of the argument:
[tex]\[ (x + 2)^{\frac{1}{4}} \text{ is defined if and only if } x + 2 \geq 0 \implies x \geq -2 \][/tex]
Thus, the domain of this function is \(x \geq -2\). This function does not have a domain of all real numbers.
Option D: \(y = -2(3x)^{\frac{1}{6}}\)
This function includes the term \((3x)^{\frac{1}{6}}\), which represents the sixth root of \(3x\). Like the square root and fourth root, the sixth root function is only defined for non-negative values of the argument:
[tex]\[ (3x)^{\frac{1}{6}} \text{ is defined if and only if } 3x \geq 0 \implies x \geq 0 \][/tex]
Thus, the domain of this function is \(x \geq 0\). This function does not have a domain of all real numbers.
After analyzing all the options, we conclude that the only function with a domain of all real numbers is:
B. [tex]\(y = (2x)^{\frac{1}{3}} - 7\)[/tex]