Question

Where is ( f = √y-x² ln(z) ) defined and continuous?

a) ( y geq x², z > 0 )
b) ( y leq x², z > 0 )
c) ( y geq x², z neq 0 )
d) ( y leq x², z neq 0 )

Answer 1

The function f = √y-x² ln(z) is defined and continuous where y ≥ x² and z > 0, which corresponds to option (a).

Step-by-step explanation:

To determine where the function f = √y-x² ln(z) is defined and continuous, we must consider the domain of each component of the function. The square root function requires that its argument be non-negative, which implies y ≥ x². The natural logarithm function, ln(z), is defined for z > 0 because the logarithm of a non-positive number is undefined.

Therefore, the function f is defined and continuous when both of these conditions are satisfied, which corresponds to y ≥ x² and z > 0.

Comparing these conditions to the provided choices, we find that the correct answer is:

• (a) ( y ≥ x², z > 0 )