Bounding of $\frac{x-y}{x+y}$ over $x \in [-5,-3]$ and $y \in [2,4]$

Question

Suppose $x \in [-5,-3]$ and $y \in [2,4]$. Find all possible values of $\dfrac{x-y}{x+y}$.

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I wrote $\dfrac{x-y}{x+y}$ as $\dfrac{x}{x+y}-\dfrac{y}{x+y}$, but I don't see how this would help, and I'm not sure how to start this problem.

Thanks in advance

Answer 1

HINT

Make the substitution $u = x/y$ so that we get \begin{align*} f(u) = \frac{u - 1}{u + 1} = 1 - \frac{2}{u + 1} \end{align*} whose derivative is given by \begin{align*} f'(u) = \frac{2}{(u+1)^{2}} > 0 \end{align*}

whence it can be concluded that $f$ is strictly increasing.

Now it remains to study the possible values of $u$.

Can you take it from here?

Answer 2

For each particular $y$, it looks like you know how to solve the problem.

• In a case like $y=2$, $x+y$ is always negative. Then $\frac{x-y}{x+y}$ is steadily increasing from $-5$ to $-3$, and we get the interval $[\frac{-5-y}{-5+y}, \frac{-3-y}{-3+y}]$. All of these values are positive.
• In a case like $y=4$, it's possible to have $y+x=0$. Then $\frac{x-y}{x+y}$ has an asymptote at $x=-y$. It is increasing from its value at $x=-5$ to $+\infty$ at the asymptote, and then increasing from $-\infty$ at the asymptote to its value at $x=-3$. We get the interval $[\frac{-5-y}{-5+y}, \infty) \cup (-\infty, \frac{-3-y}{-3+y}]$. The first interval is always positive, the second interval is always negative.

You should ask yourself:

• For which values of $y$ are we in which case?
• For which $y$ do we get the smallest positive value we can? Can we get all values higher than this value?
• For which $y$ do we get the largest negative value we can? Can we get all values lower than this value?