Question

Determine the set of points at which the function is continuous.

\(f(x,y,z) = \arcsin ({x^2} + {y^2} + {z^2})\)

Short answer

The function \(\arcsin ({x^2} + {y^2} + {z^2})\) is continuous everywhere it is defined.

Step-by-step solution

Step-1: Recognising The Form Of Function

The given function is \(\arcsin ({x^2} + {y^2} + {z^2})\)which is an arcsine function is continuous on its domain.

Step-2: The Domain Of The Function

The arcsine function is continuous on its domain, which is defined to be \(\left( {{\rm{ - 1,1}}} \right)\).

The polynomial function of three variables \({x^2} + {y^2}{\rm{ - }}z\)is continuous on , since all polynomial functions on continuous everywhere.

The composition of these functions \(\arcsin \left( {{x^2} + {y^2} + {z^2}} \right)\)is continuous everywhere it is defined.

The only restriction is that \({\rm{ - }}1 \le {x^2} + {y^2} + {z^2} \le 1\), but of course \({x^2} + {y^2} + {z^2}\)is never less than 0.

Moreover, the set can be written as \(\left\{ {\left( {x,y,z} \right)\left| {0 \le {x^2} + {y^2} + {z^2} \le 1} \right|} \right\}\)

Hence, The function\(\arcsin \left( {{x^2} + {y^2} + {z^2}} \right)\)is continuous everywhere it is Defined.

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