Question

15
select the correct answer.
which function has a domain of all real numbers?
a. $y = (x + 2)^{\frac{1}{4}}$
b. $y = -2(3x)^{\frac{1}{6}}$
c. $y = (2x)^{\frac{1}{3}} - 7$
d. $y = -x^{\frac{1}{2}} + 5$

Explanation:

Step1: Analyze Option A

The function \( y=(x + 2)^{\frac{1}{4}}=\sqrt[4]{x + 2} \). For even - root functions (\(n\) even in \(y = \sqrt[n]{u}\)), the radicand \(u\) must be non - negative. So \(x+2\geq0\), which means \(x\geq - 2\). The domain is not all real numbers.

Step2: Analyze Option B

The function \(y=-2(3x)^{\frac{1}{6}}=-2\sqrt[6]{3x}\). For even - root functions (\(n = 6\) is even), the radicand \(3x\) must be non - negative. So \(3x\geq0\), which implies \(x\geq0\). The domain is not all real numbers.

Step3: Analyze Option C

The function \(y=(2x)^{\frac{1}{3}}-7=\sqrt[3]{2x}-7\). For odd - root functions (\(n = 3\) is odd in \(y=\sqrt[n]{u}\)), the radicand \(u\) can be any real number. So for \(u = 2x\), \(x\) can be any real number (since we can take the cube root of any real number, positive, negative, or zero). The domain of this function is all real numbers.

Step4: Analyze Option D

The function \(y=-x^{\frac{1}{2}}+5=-\sqrt{x}+5\). For the square root function (\(n = 2\) is even), the radicand \(x\) must be non - negative. So \(x\geq0\). The domain is not all real numbers.

Answer:

C. \(y=(2x)^{\frac{1}{3}}-7\)