Let y = x/(x − k), where k is a constant and x is a real number, with k < 0.…

For a differentiable function y = f(x), y is strictly increasing on an interval where dy/dx > 0 throughout, and strictly decreasing where dy/dx < 0 throughout. If dy/dx keeps one fixed sign across an interval (never touching zero or changing sign), the function cannot switch between increasing and decreasing there; it is constant only when dy/dx is identically zero.

1. Rewrite y = x/(x − k) as y = 1 + k/(x − k), valid for x not equal to k (the domain excludes x = k).

2. Differentiate: dy/dx = −k/(x − k)2.

3. Since (x − k)2 > 0 for every x not equal to k, the sign of dy/dx is exactly the opposite of the sign of k.

4. Given k < 0 in this problem, −k is positive, so dy/dx > 0 for every x in the domain.

Because dy/dx is positive everywhere on the domain, y increases with x on each connected branch of the domain — the standard textbook sense of "increasing throughout its domain" for a function like this. The two branches (−∞, k) and (k, ∞) are separated by the excluded point x = k, so a value on one branch is never compared directly with a value on the other branch across that gap.

Check with k = −1, so y = x/(x + 1): on the left branch, y(−10) ≈ 1.11, y(−3) = 1.5, y(−2) = 2 — y rises as x rises. On the right branch, y(0) = 0, y(1) = 0.5, y(2) ≈ 0.67, y(5) ≈ 0.83 — y also rises as x rises. This confirms dy/dx > 0 throughout, and rules out a decrease-then-increase or increase-then-decrease pattern (which would need dy/dx to change sign, impossible here) and rules out a constant y (which would need k = 0).

Hence, for k < 0, y increases with an increase in x throughout its domain.