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Describe the domain and range of the function. $$ f(x, y)=\arcsin (x+y) $$

Short Answer

Expert verified
The domain of the given function \(f(x, y) = \arcsin (x+y)\) is the set of all \((x, y)\) such that -1 ≤ \(x+y\) ≤ 1 and its range is between -π/2 and π/2.

Step by step solution

01

Determine the domain of the function

The inverse sine function, \(\arcsin (x)\), is defined for \(x\) between -1 and 1. So, for the function \(f(x, y) = \arcsin (x+y)\), the sum of \(x\) and \(y\) must also be within -1 and 1. So the domain of \(f(x, y)\) is given by the set of all \((x, y)\) such that -1 ≤ \(x+y\) ≤ 1.
02

Determine the range of the function

The range of the inverse sine function, \(\arcsin (x)\), is between -π/2 and π/2. So, the range of the function \(f(x, y) = \arcsin (x+y)\) is also between -π/2 and π/2.

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