How many integer values of m make the function f(x) = (mx - 4)/(x - m) increasing on the interval (0, infinity)?

Step-by-step solution

1. 1

   Differentiate with the quotient rule and watch the numerator collapse.

   $f'(x)=\frac{m(x-m)-(mx-4)\cdot 1}{(x-m)^2}=\frac{mx-m^2-mx+4}{(x-m)^2}=\frac{4-m^2}{(x-m)^2}$

   The $x$ terms cancel, so $f'$ has a constant numerator. This is a general feature of $\frac{ax+b}{cx+d}$: its derivative sign is fixed by $ad-bc$ and never changes within a branch.

2. 2

   Turn the sign of $f'$ into a condition on $m$. The denominator $(x-m)^2$ is positive wherever $f$ is defined, so

   $f'(x)>0\iff 4-m^2>0\iff -2<m<2.$

   The endpoints must be excluded, not included: $m=\pm2$ gives $f'\equiv0$, and indeed $f$ collapses to the constant $2$ or $-2$ (for example $m=2$ gives $\frac{2x-4}{x-2}=2$), which is not increasing.

3. 3

   Add the condition that the interval must sit inside a single branch. A positive derivative only proves the function rises on each connected piece of its domain. The point $x=m$ is a vertical asymptote, and if it falls inside $(0,+\infty)$ the function jumps from $+\infty$ to $-\infty$ there, so it is not increasing on the whole interval. Hence we also need

   $m\notin(0,+\infty)\quad\Longleftrightarrow\quad m\le 0.$

4. 4

   See the failure concretely for $m=1$. Then $f(x)=\frac{x-4}{x-1}$, and $f(0.5)=7$ while $f(2)=-2$. The derivative is positive on both sides of $x=1$, yet the function decreases across the asymptote — exactly the trap the extra condition is there to catch.

5. 5

   Intersect the two conditions.

   $-2<m<2\quad\text{and}\quad m\le 0\qquad\Longrightarrow\qquad -2<m\le 0$

6. 6

   Count the integers. The integers in $(-2,\,0]$ are $m=-1$ and $m=0$, so there are $\mathbf{2}$ of them. Quick check for $m=0$: $f(x)=-\frac{4}{x}$, which rises from $-\infty$ toward $0$ on $(0,\infty)$ — increasing, as claimed.