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

Explanation:

Step1: Analyze the domain of cube - root functions

For a cube - root function (y = \sqrt[3]{u}), the domain is all real numbers because for any real number (u), the cube root (\sqrt[3]{u}) is defined. So for (y=\sqrt[3]{x - 2}) and (y=\sqrt[3]{x + 2}), when (x = 0):

• For (y=\sqrt[3]{x - 2}), (y=\sqrt[3]{0-2}=\sqrt[3]{-2})
• For (y=\sqrt[3]{x + 2}), (y=\sqrt[3]{0 + 2}=\sqrt[3]{2})

Step2: Analyze the domain of square - root functions

For a square - root function (y=\sqrt{u}), the domain is (u\geq0).

• For (y=\sqrt{x - 2}), when (x = 0), (u=x - 2=0 - 2=-2). Since (-2<0), the function (y = \sqrt{x - 2}) is undefined at (x = 0).
• For (y=\sqrt{x + 2}), when (x = 0), (u=x + 2=0+2 = 2). Since (2\geq0), (y=\sqrt{0 + 2}=\sqrt{2})

Answer:

(y=\sqrt{x - 2})