f(x,y)=(1)/(\sqrt(x^(2)+y^(2)+16)), find domain and range

asked by guest
on Jan 14, 2025 at 3:40 pm



You asked:

Find the domain and the range of f(x,y)=1x2+y2+16f\left( x, y \right) = \frac{1}{\sqrt{{x}^{2} + {y}^{2} + 16}}

MathBot Answer:

Domain

{(x,y)∈R2:x2+y2+16≥0∧x≠−−y2−16∧x≠−y2−16∧y≠−−x2−16∧y≠−x2−16}\{(x, y) \in \mathbb{R}^{2}:x^{2} + y^{2} + 16 \geq 0\wedge x \neq - \sqrt{- y^{2} - 16} \wedge x \neq \sqrt{- y^{2} - 16} \wedge y \neq - \sqrt{- x^{2} - 16} \wedge y \neq \sqrt{- x^{2} - 16}\}

Range

Set notation:

{f∈R:0≤f<4}\displaystyle\left\{f \in \mathbb{R}:0 \le f \lt 4\right\}

Interval notation:

[0,4)\displaystyle\left[0, 4\right)

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