Question

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

Explanation:

Step1: Analyze square root functions

For a square root function \( y = \sqrt{f(x)} \), the expression inside the square root (the radicand) must be non - negative (\( f(x)\geq0 \)) for the function to be defined in the set of real numbers. For cube root functions \( y=\sqrt[3]{f(x)} \), the radicand can be any real number (positive, negative, or zero) because we can take the cube root of any real number.

Step2: Check \( y = \sqrt{x + 2} \) at \( x = 0 \)

Substitute \( x = 0 \) into the radicand: \( 0+2=2\geq0 \). So the function \( y=\sqrt{x + 2} \) is defined at \( x = 0 \).

Step3: Check \( y=\sqrt[3]{x + 2} \) at \( x = 0 \)

Since it is a cube root function, the radicand \( 0 + 2=2 \) is a real number. Cube root of a real number is defined, so \( y=\sqrt[3]{x + 2} \) is defined at \( x = 0 \).

Step4: Check \( y=\sqrt[3]{x-2} \) at \( x = 0 \)

The radicand is \( 0-2=- 2 \), and since it is a cube root function, the cube root of - 2 is defined. So \( y=\sqrt[3]{x - 2} \) is defined at \( x = 0 \).

Step5: Check \( y=\sqrt{x-2} \) at \( x = 0 \)

Substitute \( x = 0 \) into the radicand: \( 0 - 2=-2<0 \). Since the radicand of a square root function must be non - negative, the function \( y=\sqrt{x - 2} \) is undefined at \( x = 0 \).

Answer:

\( y=\sqrt{x - 2} \)