Answer :
To determine which function has a domain of all real numbers, we need to examine each given option and determine the set of [tex]\(x\)[/tex]-values for which the function is defined.
Option A: [tex]\( y = (x + 2)^{\frac{1}{4}} \)[/tex]
- This function involves taking the fourth root of [tex]\(x + 2\)[/tex]. For the fourth root to be defined as a real number, the expression inside the root, [tex]\(x + 2\)[/tex], must be non-negative.
- Therefore, [tex]\(x + 2 \geq 0\)[/tex], which gives [tex]\(x \geq -2\)[/tex].
- Domain of Option A: [tex]\( x \geq -2 \)[/tex]
Option B: [tex]\( y = -2(3x)^{\frac{1}{6}} \)[/tex]
- This function involves taking the sixth root of [tex]\(3x\)[/tex]. For the sixth root to be defined as a real number, the expression inside the root, [tex]\(3x\)[/tex], must be non-negative.
- Hence, [tex]\(3x \geq 0\)[/tex], which simplifies to [tex]\(x \geq 0\)[/tex].
- Domain of Option B: [tex]\( x \geq 0 \)[/tex]
Option C: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
- This function involves taking the cube root of [tex]\(2x\)[/tex]. The cube root is defined for all real numbers since any real number has a real cube root.
- Therefore, there are no restrictions on [tex]\(x\)[/tex].
- Domain of Option C: All real numbers
Option D: [tex]\( y = -x^{\frac{1}{2}} + 5 \)[/tex]
- This function involves taking the square root of [tex]\(x\)[/tex]. For the square root to be defined as a real number, the expression inside the root, [tex]\(x\)[/tex], must be non-negative.
- Hence, [tex]\(x \geq 0\)[/tex].
- Domain of Option D: [tex]\( x \geq 0 \)[/tex]
Given the above analysis, the function that is defined for all real numbers is:
Option C: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
Therefore, the answer is:
[tex]\( C\)[/tex]
Option A: [tex]\( y = (x + 2)^{\frac{1}{4}} \)[/tex]
- This function involves taking the fourth root of [tex]\(x + 2\)[/tex]. For the fourth root to be defined as a real number, the expression inside the root, [tex]\(x + 2\)[/tex], must be non-negative.
- Therefore, [tex]\(x + 2 \geq 0\)[/tex], which gives [tex]\(x \geq -2\)[/tex].
- Domain of Option A: [tex]\( x \geq -2 \)[/tex]
Option B: [tex]\( y = -2(3x)^{\frac{1}{6}} \)[/tex]
- This function involves taking the sixth root of [tex]\(3x\)[/tex]. For the sixth root to be defined as a real number, the expression inside the root, [tex]\(3x\)[/tex], must be non-negative.
- Hence, [tex]\(3x \geq 0\)[/tex], which simplifies to [tex]\(x \geq 0\)[/tex].
- Domain of Option B: [tex]\( x \geq 0 \)[/tex]
Option C: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
- This function involves taking the cube root of [tex]\(2x\)[/tex]. The cube root is defined for all real numbers since any real number has a real cube root.
- Therefore, there are no restrictions on [tex]\(x\)[/tex].
- Domain of Option C: All real numbers
Option D: [tex]\( y = -x^{\frac{1}{2}} + 5 \)[/tex]
- This function involves taking the square root of [tex]\(x\)[/tex]. For the square root to be defined as a real number, the expression inside the root, [tex]\(x\)[/tex], must be non-negative.
- Hence, [tex]\(x \geq 0\)[/tex].
- Domain of Option D: [tex]\( x \geq 0 \)[/tex]
Given the above analysis, the function that is defined for all real numbers is:
Option C: [tex]\( y = (2x)^{\frac{1}{3}} - 7 \)[/tex]
Therefore, the answer is:
[tex]\( C\)[/tex]
