Find all real m for which y = (x + 4)/(x + m) is increasing on the interval (−∞, −7).

Step-by-step solution

1. 1

   Differentiate using the Möbius formula. For $y=\dfrac{ax+b}{cx+d}$ we have $y'=\dfrac{ad-bc}{(cx+d)^{2}}$, and here $a=1$, $b=4$, $c=1$, $d=m$:

   $y'=\frac{m-4}{(x+m)^{2}}$

2. 2

   Impose the sign condition. The squared denominator is positive wherever $y$ is defined, so increasing means

   $m-4>0\quad\Longleftrightarrow\quad m>4$

3. 3

   Impose the pole condition. The function has a vertical asymptote at $x=-m$. Increasing on the whole interval $(-\infty,-7)$ requires that asymptote to lie outside it:

   $-m\notin(-\infty,-7)\quad\Longleftrightarrow\quad -m\ge -7\quad\Longleftrightarrow\quad m\le 7$

   Note the endpoint is allowed: at $m=7$ the pole sits exactly at $x=-7$, which is not an interior point of the open interval $(-\infty,-7)$.

4. 4

   Intersect the two conditions.

   $4<m\le 7$

   $\boxed{m\in(4,\,7]}$

5. 5

   Test the three boundary cases. At $m=4$: $y=\tfrac{x+4}{x+4}=1$, a constant — not increasing ✗, so $4$ is excluded. At $m=7$: $y=\tfrac{x+4}{x+7}$ with $y'=\tfrac{3}{(x+7)^{2}}>0$ on $(-\infty,-7)$ ✓, so $7$ is included. At $m=8$: the pole $x=-8$ lies inside $(-\infty,-7)$, so the function has a break there and is not increasing across the whole interval ✗.

6. 6

   Note the pattern. For $y=\dfrac{x+p}{x+m}$ increasing on $(-\infty,-q)$ with $q>p>0$, the same two conditions give $p<m\le q$ — so the answer is always the half-open interval $(p,q]$, closed at the right end.