Question

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Answer 1

For this case, we must indicate which of the given functions is not defined for$x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$.

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$, your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.

$y = \sqrt [3] {x + 2}$with x = 0: $y = \sqrt [3] {2}$in the same way is defined.

Answer:

$y = \sqrt {x-2}$

Option b

Answer 2

The function that are undefined for (x=0) is  $y = \sqrt[3]{x-2}$ and $y = \sqrt{x-2}$ and this can be determined by putting (x=0) in all the given expressions.

Given :

Function 1 - $y = \sqrt[3]{x-2}$

Function 2 - $y = \sqrt{x-2}$

Function 3 - $y = \sqrt[3]{x+2}$

Function 4 - $y = \sqrt{x+2}$

Evaluate each function at (x = 0) to determine which function is undefined for (x = 0).

Function 1 - $y = \sqrt[3]{x-2}$

put (x = 0) in above function:

$y = \sqrt[3]{-2}$

Therefore, it can be concluded that the above function is not defined for (x=0).

Function 2 - $y = \sqrt{x-2}$

put (x = 0) in above function:

$y = \sqrt{-2}$

Therefore, it can be concluded that the above function is not defined for (x=0).

Function 3 - $y = \sqrt[3]{x+2}$

put (x = 0) in above function:

$y = \sqrt[3]{2}$

Therefore, it can be concluded that the above function is defined for (x=0).

Function 4 - $y = \sqrt{x+2}$

put (x = 0) in above function:

$y = \sqrt{2}$

Therefore, it can be concluded that the above function is defined for (x=0).

For more information, refer to the link given below:

https://brainly.com/question/2263981