# Consider the function f(x, y):= xy^3/x^4 + y^4. (a) Find the maximal domain D so that f mapping D...

## Question:

Consider the function {eq}\displaystyle f(x, y):= \frac{xy^3}{x^4 + y^4} {/eq}

(a) Find the maximal domain {eq}D {/eq} so that {eq}f:D \rightarrow \mathbb{R} {/eq} is a function (i.e. find all points {eq}(x, y) {/eq} at which {eq}f {/eq} is defined).

(b) At which points of {eq}D {/eq} from part (a) is {eq}f {/eq} continuous?

(c) If {eq}f {/eq} has singularities, identify them, and determine if they are continuously removable.

## Domain and Continuity

A function without gaps, holes or interruptions, is a continuous function, also, some times, a discontinuity is removable if and only if the function has a simplification that clear discontinuity factor. Therefore, with a factoring procedure, we can determine if a function has a removable discontinuity.

## Answer and Explanation:

The function is:

{eq}\displaystyle f(x,y)= \frac{xy^3}{x^4 + y^4} {/eq}

(a)

Denominator can't be zero, then, function domain is:

{eq}\displaystyle (x,y) \in \mathbb{R} \; \text{ / } \; x^4+y^4 \neq 0 \\ \; \text{ or } \; \\ \displaystyle \mathbb{R}^2 - \{ (0,0) \} {/eq}

(b)

The function is continuous for all area given by {eq}\displaystyle (x,y) \; \text{ with } \; x \; \text{ , } \; y \in \mathbb{R} {/eq} except the point {eq}\displaystyle (0,0) {/eq}

(c)

The function hasn't singularities, also its discontinuity removable if and only if factoring we cancel the factor that determines the discontinuity. But {eq}\displaystyle x^4+y^4 {/eq} can't be simplified.