Question

which function is undefined for x = 0?
○ $y = \sqrt3{x - 2}$
○ $y = \sqrt{x - 2}$
○ $y = \sqrt3{x + 2}$
○ $y = \sqrt{x + 2}$

Explanation:

Step1: Recall Domain Rules

For cube roots ($\sqrt[3]{}$), domain is all real numbers (since we can take cube root of negative numbers). For square roots ($\sqrt{}$), the expression inside must be $\geq 0$ (non - negative) for real - valued functions.

Step2: Analyze Each Function at $x = 0$

• Function 1: $y=\sqrt[3]{x - 2}$

Substitute $x = 0$ into the expression inside the cube root: $0-2=-2$. Since we can take the cube root of $-2$, the function is defined at $x = 0$.

• Function 2: $y=\sqrt{x - 2}$

Substitute $x = 0$ into the expression inside the square root: $0 - 2=-2$. For a square root function $y=\sqrt{u}$, we need $u\geq0$. Here, $u=-2<0$, so the square root of a negative number is not a real number. Thus, the function $y = \sqrt{x - 2}$ is undefined at $x = 0$.

• Function 3: $y=\sqrt[3]{x + 2}$

Substitute $x = 0$ into the expression inside the cube root: $0 + 2=2$. We can take the cube root of $2$, so the function is defined at $x = 0$.

• Function 4: $y=\sqrt{x + 2}$

Substitute $x = 0$ into the expression inside the square root: $0+2 = 2$. Since $2\geq0$, we can take the square root of $2$, so the function is defined at $x = 0$.

Answer:

The function $y=\sqrt{x - 2}$ (the second option) is undefined for $x = 0$.