# What is the range of $g(x,y)=\frac{\sqrt{5-x}}{y^2}$

How to find range of $$2$$ variable function $$\displaystyle g(x,y)=\frac{\sqrt{5-x}}{y^2}$$ for real $$x,y$$

My way

First i calculate domain of that function

Here $$5-x\geq 0\;\cap\; y\neq 0\Longrightarrow x\leq 5\;\cap\; y\in\mathbb{R}-\{0\}$$

And for calculation of range

Largest value of $$g(x,y)\rightarrow \infty$$ when $$y\rightarrow \pm \infty$$ but how i find lowest value of range

But i did not understand how to find range of function

• Are $x$ and $y$ real numbers? What is the domain of $g$? Is it the largest subset of the real numbers for which $g(x,y)$ is again a real number? Are you familiar with complex numbers? Also, there are a few speling mistakes and mathematical errors; presumably you mean $\wedge$ instead of $\cap$. Mar 30, 2020 at 19:40
• Largest value of $g(x,y)\rightarrow \infty$ when $y\rightarrow \pm \infty$ but how i find lowest value of $g(x,y)$ Mar 30, 2020 at 19:46
• If $x,y\in\mathbb R$, then the numerator is non-negative and the denominator is positive; the lowest value of $g(x,y)$ occurs when $x=5$; also, when $y\to\pm\infty,$ $g(x,y)\to0$ Mar 30, 2020 at 19:48
• @jacky: That's what I mean, so the range of $g(x,y)$ is contained in $[0,\infty)$; can you find $x$ and $y$ such that $g(x,y)=a$ for any $a\in[0,\infty)$? Mar 30, 2020 at 19:53
• Lowest value is exactly $0$ when $x=5$ Mar 30, 2020 at 19:54
Suppose $$g$$ has real inputs and outputs only. Then consider the domain: $$x \leq 5$$ and $$y \neq 0$$. The numerator can be as small as $$0$$ and as large as you'd like ($$x=-105$$ yeilds a numerator of 10). The denominator can be "very close" to $$0$$ and as large as you like. Notice that you cannot get negative results. So, the smallest the output can be is $$0$$ ($$x=5, y\neq 0$$). How big can the output get? As long as $$x$$ is not 5, we can make the number very big, for simplicity let $$x=4$$, so that the numerator takes value 1. Now, $$1/y^2$$ can be made as large as you like. Say you want to make it size M, choose $$y = 1/\sqrt{M}$$. In other words, for all $$x <5$$, $$\lim_{y\to 0}g(x,y) =\infty$$ So $$g \in [0,\infty)$$.