## Calculus: Early Transcendentals 8th Edition

${(x,y)| {1+x-y}\geq 0}$
Given: $f(x,y)=cos\sqrt {1+x-y}$ The given function is defined for all values of x and y except at $\sqrt {1+x-y}\geq 0$ Since, $cos(x,y)$ is continuous at $R^{2}$ and square root of the function does not exist at $R$ when it contains non-negative value. Square both sides to obtain an inequality to represent the domain. ${1+x-y}\geq 0$ Hence, the function $f(x,y)=cos\sqrt {1+x-y}$ is continuous on ${(x,y)| {1+x-y}\geq 0}$.