True or false; a function with a square root cannot have a domain that is the set of all real numbers

The answer is false, if the domain is the set of all inputs for which the function is defined then logically it is then an example function which breaks for certain input values. A function is needed for certain inputs that does not produce a valid output for example the function is undefined for the input. y = 3/x-1 this function is defined for almost any real x but what is the value of y when x is equals to 1 then, it will be 3/0 which is undefined. Division by zero is undefined therefore 1 is not in the domain of the function and 1 cannot be used as an input because it breaks the function. All other real number are valid inputs so the domain is all real numbers except for 1.

What is the diagram u didn't post that