To determine which of the given functions has a domain of all real numbers, we need to analyze the domain of each function.
A. [tex]\( y = -x^{\frac{1}{2}} + 5 \)[/tex]
For this function:
- [tex]\( y = -x^{\frac{1}{2}} \)[/tex] involves a square root.
- The square root function is defined only for non-negative numbers, meaning [tex]\( x \geq 0 \)[/tex].
Hence, the domain of this function is limited to non-negative real numbers, not all real numbers.
B. [tex]\( y = -2(3x)^{\frac{1}{6}} \)[/tex]
For this function:
- [tex]\( y = -2(3x)^{\frac{1}{6}} \)[/tex] involves a sixth root.
- The sixth root function is also defined only for non-negative numbers, meaning [tex]\( 3x \geq 0 \)[/tex], or [tex]\( x \geq 0 \)[/tex].
Thus, the domain of this function is also limited to non-negative real numbers, not all real numbers.
C. [tex]\( y = (x+2)^{\frac{1}{4}} \)[/tex]
For this function:
- [tex]\( y = (x+2)^{\frac{1}{4}} \)[/tex] involves a fourth root.
- The fourth root function is defined only for non-negative numbers, meaning [tex]\( x + 2 \geq 0 \)[/tex], or [tex]\( x \geq -2 \)[/tex].
So, the domain of this function is limited to [tex]\( x \geq -2 \)[/tex], not all real numbers.
D. [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
For this function:
- [tex]\( y = (2x)^{\frac{1}{3}} \)[/tex] involves a cube root.
- The cube root function is defined for all real numbers because cube roots are defined for negative, zero, and positive values of [tex]\( x \)[/tex].
Therefore, the domain of this function is all real numbers.
Given the analysis, the function that has a domain of all real numbers is:
D. [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
