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I need to find the range of the function $g:(0,1)\times (0,1) \to \mathbb{R}^2$ given by

\begin{align} g_1(x,y)&= \frac{x}{x+y}\\ g_2(x,y)&= x+y \end{align}

I can see that it is $g((0,1)\times (0,1))=(0,1)\times (0,2)$ but what I need is a formal proof of this statement.

For example, for $g_2$ should be simpler, since I can see that the smallest value is reached when both $x$ and $y$ tend to 0, and the largest when they both tend to 1. However, even in this case I cannot find a formally correct "proof" for why is that. For $g_1$ I find it even more difficult.

Could someone help me?

user70645
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1 Answers1

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The range of $g$ isn’t all of $(0,1)\times(0,2)$. For example, taking $x=\frac12$, we have

$$g_1\left(\frac12,y\right)=\frac{\frac12}{\frac12+y}=\frac1{1+2y}\;,$$

and since $0<y<1$, this implies that $$\frac13<g_1\left(\frac12,y\right)<1\;.$$

In general the inequalities $0<y<1$ imply that $$1>\frac{x}{x+y}>\frac{x}{x+1}=1-\frac1{x+1}\;.$$

To show that for a fixed $x_0$ the function $g_1(x_0,y)$ takes on all values in the interval $$\left(1-\frac1{x_0+1},1\right)\;,$$ use the fact that the function $y\mapsto\frac{x_0}{x_0+y}$ is continuous and monotonically decreasing on $(0,1)$ for each $x_0\in(0,1)$.

Brian M. Scott
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