Which function is undefined for [tex]$x=0$[/tex]?

A. [tex]$y=\sqrt[3]{x-2}$[/tex]
B. [tex]$y=\sqrt{x-2}$[/tex]
C. [tex]$y=\sqrt[3]{x+2}$[/tex]
D. [tex]$y=\sqrt{x+2}$[/tex]



Answer :

To determine which function is undefined at \( x = 0 \), let's analyze each function one by one:

1. Function: \( y = \sqrt[3]{x - 2} \)

- Substitute \( x = 0 \):
[tex]\[ y = \sqrt[3]{0 - 2} = \sqrt[3]{-2} \][/tex]
- The cube root of \(-2\) is defined for all real numbers.
- Therefore, this function is defined at \( x = 0 \).

2. Function: \( y = \sqrt{x - 2} \)

- Substitute \( x = 0 \):
[tex]\[ y = \sqrt{0 - 2} = \sqrt{-2} \][/tex]
- The square root of a negative number is not defined in the set of real numbers.
- Therefore, this function is undefined at \( x = 0 \).

3. Function: \( y = \sqrt[3]{x + 2} \)

- Substitute \( x = 0 \):
[tex]\[ y = \sqrt[3]{0 + 2} = \sqrt[3]{2} \][/tex]
- The cube root of 2 is defined for all real numbers.
- Therefore, this function is defined at \( x = 0 \).

4. Function: \( y = \sqrt{x + 2} \)

- Substitute \( x = 0 \):
[tex]\[ y = \sqrt{0 + 2} = \sqrt{2} \][/tex]
- The square root of 2 is defined in the set of real numbers.
- Therefore, this function is defined at \( x = 0 \).

By analyzing all four functions, we can conclude that the function [tex]\( y = \sqrt{x - 2} \)[/tex] is the only one that is undefined at [tex]\( x = 0 \)[/tex].