Monotonicity and the sign of the linear function
Learning Objectives
- Prove that \(f(x) = ax + b\) is strictly increasing when \(a > 0\) and strictly decreasing when \(a < 0\).
- Find the zero \(x = -\tfrac{b}{a}\) of a linear function and identify it as the \(x\)-intercept.
- Determine the sign of a linear function on each side of its zero.
- Build the sign table of a given linear function.
From "rises" to "increasing"
In Lesson 2 we said, looking at the picture, that a line with \(a > 0\) rises and a line with \(a < 0\) falls. We now make this precise using the language of monotonicity from the general theory of functions: a function \(f\) is strictly increasing if \(x_1 < x_2 \Rightarrow f(x_1) < f(x_2)\), and strictly decreasing if \(x_1 < x_2 \Rightarrow f(x_1) > f(x_2)\).
For a linear function the answer is decided entirely by the sign of the slope — and the one-line proof is already in our hands.
Monotonicity of the linear function
Let \(f(x) = ax + b\) with \(a \neq 0\).
If \(a > 0\), then \(f\) is strictly increasing on \(\mathbb{R}\).
If \(a < 0\), then \(f\) is strictly decreasing on \(\mathbb{R}\).
Take any two inputs with \(x_1 < x_2\). From Lesson 1,
\[f(x_2) - f(x_1) = a\,(x_2 - x_1).\]
The factor \(x_2 - x_1\) is positive, since \(x_1 < x_2\). Therefore the sign of \(f(x_2) - f(x_1)\) is exactly the sign of \(a\).
If \(a > 0\), then \(f(x_2) - f(x_1) > 0\), i.e. \(f(x_1) < f(x_2)\): strictly increasing. If \(a < 0\), then \(f(x_2) - f(x_1) < 0\), i.e. \(f(x_1) > f(x_2)\): strictly decreasing. In either case the conclusion holds for every pair \(x_1 < x_2\).
One consequence is worth naming: a strictly monotonic function is injective — it never takes the same value twice. So a linear function takes each real value exactly once, which is why its line crosses every horizontal level in a single point. In particular it crosses the level \(0\) — the \(x\)-axis — exactly once. That single crossing is the subject of the rest of the lesson.
The zero of a linear function
The zero (or root) of \(f(x) = ax + b\) is the input where the output is \(0\). Solving \(ax + b = 0\) and using \(a \neq 0\) to divide,
\[x = -\frac{b}{a}.\]
Because \(a \neq 0\), this value exists and is unique: a linear function has exactly one zero. Geometrically it is the \(x\)-intercept — the point \(\left(-\tfrac{b}{a},\, 0\right)\) where the line crosses the \(x\)-axis.
The sign of a linear function
Let \(x_0 = -\tfrac{b}{a}\) be the zero of \(f(x) = ax + b\). On the two sides of \(x_0\), the function keeps a constant sign, and the two signs are opposite:
For \(x > x_0\), \(f(x)\) has the same sign as \(a\).
For \(x < x_0\), \(f(x)\) has the opposite sign to \(a\).
At \(x = x_0\), \(f(x) = 0\).
This is recorded compactly in the sign table:
| \(x < -\tfrac{b}{a}\) | \(x = -\tfrac{b}{a}\) | \(x > -\tfrac{b}{a}\) | |
|---|---|---|---|
| \(a > 0\) | \(f(x) < 0\) | \(0\) | \(f(x) > 0\) |
| \(a < 0\) | \(f(x) > 0\) | \(0\) | \(f(x) < 0\) |
The reason is the monotonicity we just proved: the function passes through \(0\) exactly once at \(x_0\), moving steadily in one direction, so it is on one side of \(0\) before the crossing and on the other side after.
A positive slope: \(f(x) = 2x - 6\)
Here \(a = 2 > 0\), so \(f\) is increasing. Its zero is found from \(2x - 6 = 0\), giving \(x = 3\). Since \(a > 0\), the function is negative to the left of \(3\) and positive to the right:
| \(x < 3\) | \(x = 3\) | \(x > 3\) | |
|---|---|---|---|
| \(f(x) = 2x - 6\) | \(f(x) < 0\) | \(0\) | \(f(x) > 0\) |
A quick check confirms it: \(f(0) = -6 < 0\) (and \(0 < 3\)), while \(f(5) = 4 > 0\) (and \(5 > 3\)).
A negative slope: \(f(x) = -3x + 3\)
Now \(a = -3 < 0\), so \(f\) is decreasing. Its zero is found from \(-3x + 3 = 0\), giving \(x = 1\). Since \(a < 0\), the signs are reversed — positive to the left of \(1\), negative to the right:
| \(x < 1\) | \(x = 1\) | \(x > 1\) | |
|---|---|---|---|
| \(f(x) = -3x + 3\) | \(f(x) > 0\) | \(0\) | \(f(x) < 0\) |
Check: \(f(0) = 3 > 0\) (and \(0 < 1\)), while \(f(2) = -3 < 0\) (and \(2 > 1\)).
Why exactly one change of sign?
A linear function changes sign once and only once. Why can it not, say, dip back across \(0\) a second time?
Because it is strictly monotonic. A strictly increasing function that has already passed from negative to positive keeps growing; it can never return to \(0\) to cross again. The single crossing is guaranteed by \(a \neq 0\): that is exactly the condition that makes the zero \(x = -\tfrac{b}{a}\) both exist and be unique.
Contrast the excluded case \(a = 0\): a constant function \(f(x) = b\) has no change of sign at all. If \(b \neq 0\) it keeps one sign forever and has no zero; if \(b = 0\) it is zero everywhere. Neither behaviour is what we mean by a first-degree function — another reason the definition insists on \(a \neq 0\).
Exercises
Exercise 1
Consider \(f \colon \mathbb{R} \to \mathbb{R},\; f(x) = 2x - 6\).
Exercise 2
Use the sign of the slope to decide the monotonicity of each function.
Exercise 3
Find the zero \(x = -\tfrac{b}{a}\) of each function.
Exercise 4
Decide whether each statement about monotonicity and sign is true or false.
Exercise 5
For each function, compute \(f(1)\) and classify its sign.
Summary
\(f(x) = ax + b\) is strictly increasing when \(a > 0\) and strictly decreasing when \(a < 0\) — proved directly from \(f(x_2) - f(x_1) = a(x_2 - x_1)\).
Being strictly monotonic, it is injective and has exactly one zero, \(x = -\tfrac{b}{a}\), the \(x\)-intercept.
The sign table: on the far side of the zero \(f\) takes the sign of \(a\); on the near side, the opposite sign; and \(0\) at the zero itself.
The single sign change is guaranteed by \(a \neq 0\); a constant function \((a = 0)\) never changes sign. This sign analysis is the tool we use to solve first-degree inequalities in the next unit.