To determine which functions have a domain of all real numbers, we need to analyze the domain of each function provided.
1. Function A: [tex]\( y = -x^{\frac{1}{2}} + 5 \)[/tex]
- This function includes the term [tex]\( x^{\frac{1}{2}} \)[/tex], which is the square root of [tex]\( x \)[/tex].
- The square root function [tex]\( x^{\frac{1}{2}} \)[/tex] is only defined for [tex]\( x \geq 0 \)[/tex].
- Therefore, Function A is only defined for non-negative real numbers.
- Domain: [tex]\( x \geq 0 \)[/tex].
2. Function B: [tex]\( y = -2(3x)^{\frac{1}{d}} \)[/tex]
- The term [tex]\( (3x)^{\frac{1}{d}} \)[/tex] involves taking the [tex]\( 1/d \)[/tex]-th power of [tex]\( 3x \)[/tex].
- The domain of this function depends on the value of [tex]\( d \)[/tex].
- If [tex]\( d \)[/tex] is a positive integer, [tex]\( (3x)^{\frac{1}{d}} \)[/tex] will be defined for all real numbers [tex]\( x \)[/tex].
- If [tex]\( d \)[/tex] is an even integer, [tex]\( 3x \)[/tex] must be non-negative for the expression to be real, hence [tex]\( x \geq 0 \)[/tex].
- Without knowing the value of [tex]\( d \)[/tex], we cannot definitively say that the domain is all real numbers.
- Domain: Depends on [tex]\( d \)[/tex].
3. Function C: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
- The term [tex]\( (2x)^{\frac{1}{3}} \)[/tex] is the cube root of [tex]\( 2x \)[/tex].
- The cube root function is defined for all real numbers [tex]\( x \)[/tex], because any real number can be cubed.
- Domain: All real numbers [tex]\( x \)[/tex].
4. Function D: [tex]\( y = (x + 2) \cdot \frac{1}{1} \)[/tex]
- Simplifying the expression, we get [tex]\( y = x + 2 \)[/tex].
- This is a linear function, which is defined for all real numbers [tex]\( x \)[/tex].
- Domain: All real numbers [tex]\( x \)[/tex].
Thus, the functions that have a domain of all real numbers are:
- Function C: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
- Function D: [tex]\( y = (x + 2) \cdot \frac{1}{1} \)[/tex]
Given the choices, the correct answers are:
- C and D
If we refer to the problem's format, the choices corresponding to these functions are:
- 3 and 4
Therefore, the correct selection is:
- [3, 4]
