Increasing and decreasing functions — intuitive definition: Worked examples

This page is a case bank. The parent note the parent topic taught you the single rule:

Recall The one rule everything below uses

$f$ is increasing on interval $I$ if for every pair $x_{1} < x_{2}$ in $I$ we get $f (x_{1}) < f (x_{2})$ (bigger input → bigger output). It is decreasing if instead $f (x_{1}) > f (x_{2})$ (bigger input → smaller output).

Everything here is that rule, applied to every kind of function you can meet. First we lay out the full menu of cases, then we work one example per case so you never hit a scenario we didn't show.

The scenario matrix

Think of monotonicity questions as coming in a small number of flavours. Each row below is one flavour — one "cell" of the matrix. Our 8 examples fill every cell.

Cell	Case class	What makes it tricky	Example
A	Linear, positive slope	baseline, always increasing	Ex 1
B	Linear, negative slope	the sign flips — decreasing	Ex 2
C	Sign split (quadratic)	up on one side, down on the other, zero in the middle	Ex 3
D	Reciprocal / two branches	domain has a hole; each branch behaves on its own	Ex 4
E	Cube root type — degenerate slope	increasing everywhere but "flat-looking" at one point	Ex 5
F	Constant / flat (degenerate)	neither increasing nor decreasing	Ex 6
G	Word problem (real world)	translate a story into "bigger input → ?"	Ex 7
H	Exam twist — piecewise / does the join break it?	monotone on pieces ≠ monotone overall	Ex 8

The method never changes. We always compare two ordered points $x_{1} < x_{2}$ and ask: is $f (x_{2}) - f (x_{1})$ positive (increasing) or negative (decreasing)? The trick each time is turning that difference into something whose sign we can read off.

The examples

Cell A — linear, positive slope

Cell B — linear, negative slope

Cell C — sign split (quadratic)

Here the answer depends on which side of a special point you stand. See the figure: the parabola falls, touches bottom, then rises.

Figure — Increasing and decreasing functions — intuitive definition

Worked example Example 3:

$f (x) = x^{2} - 6 x + 5$ — where does it increase, where decrease? Forecast: A "smile" parabola has one lowest point. Guess whether it decreases before or after that point.

Find the turning point. Complete the square: $x^{2} - 6 x + 5 = (x - 3)^{2} - 4$ . Why this step? Writing it as $(x - 3)^{2} - 4$ shows the graph is the basic $(\cdot)^{2}$ shape shifted so its bottom sits at $x = 3$ . That $x = 3$ is the boundary between the two behaviours.
Left branch, $x < 3$ . Take $x_{1} < x_{2} < 3$ . Then $f (x_{2}) - f (x_{1}) = (x_{2} - 3)^{2} - (x_{1} - 3)^{2}$ . Factor as a difference of squares: $f (x_{2}) - f (x_{1}) = [(x_{2} - 3) - (x_{1} - 3)] [(x_{2} - 3) + (x_{1} - 3)] = (x_{2} - x_{1}) (x_{1} + x_{2} - 6) .$ Why this step? Difference of squares turns it into a product whose two factors we can sign separately.
Sign on the left. $x_{2} - x_{1} > 0$ always. But $x_{1} + x_{2} < 6$ (both below 3), so $x_{1} + x_{2} - 6 < 0$ . Product = (positive)(negative) = negative → $f (x_{2}) < f (x_{1})$ → decreasing on $(- \infty, 3)$ . Why this step? We keep the fixed $(x_{2} - x_{1})$ factor and let only the second factor decide the sign — that is how one algebraic expression covers every pair on the branch at once.
Right branch, $x > 3$ . Same factored form. Now $x_{1} + x_{2} > 6$ , so $x_{1} + x_{2} - 6 > 0$ . Product = (positive)(positive) = positive → increasing on $(3, \infty)$ . Why this step? The only thing that changed is the sign of the second factor — that single sign flip is the whole "turning point" story.
The point $x = 3$ itself is the boundary; there the graph is momentarily flat (the minimum). It belongs to neither open interval. Why this step? Monotonicity is stated on open intervals; the single turning point where the direction reverses cannot itself be labelled up or down, so we hand it to neither side.
Conclusion: decreasing on $(- \infty, 3)$ , increasing on $(3, \infty)$ . Not monotone on all of $R$ . Compare Monotone functions and Intervals and domain.

Verify: $f (0) = 5$ , $f (3) = - 4$ , $f (6) = 5$ . Left side falls $5 \to - 4$ ✓, right side rises $- 4 \to 5$ ✓.

Cell D — reciprocal / two branches

Worked example Example 4:

$f (x) = \frac{1}{x}$ on its whole domain Forecast: The parent note showed it's decreasing on $(0, \infty)$ . Careful — is it therefore decreasing "everywhere"? Guess before reading.

State the domain. $f$ is undefined at $x = 0$ , so the domain is $(- \infty, 0) \cup (0, \infty)$ — two separate pieces with a gap at $0$ . Why this step? You can only compare two points if there's an unbroken interval between them. The gap forbids comparing a negative $x_{1}$ with a positive $x_{2}$ .
Positive branch $0 < x_{1} < x_{2}$ . Form the difference our usual way: $f (x_{2}) - f (x_{1}) = \frac{1}{x _{2}} - \frac{1}{x _{1}} = \frac{x _{1} - x _{2}}{x _{1} x _{2}} .$ Numerator $x_{1} - x_{2} < 0$ (since $x_{1} < x_{2}$ ); denominator $x_{1} x_{2} > 0$ (both positive). So $f (x_{2}) - f (x_{1}) < 0$ → $f (x_{2}) < f (x_{1})$ → decreasing on $(0, \infty)$ . Why this step? We stick to the standard $f (x_{2}) - f (x_{1})$ so the sign reads directly: negative means decreasing, exactly as in every other example. The common denominator just makes that sign visible.
Negative branch $x_{1} < x_{2} < 0$ . Same formula: $f (x_{2}) - f (x_{1}) = \frac{x _{1} - x _{2}}{x _{1} x _{2}} .$ Numerator $x_{1} - x_{2} < 0$ still; denominator $x_{1} x_{2} > 0$ again, because two negatives multiply to a positive. So $f (x_{2}) - f (x_{1}) < 0$ → decreasing on $(- \infty, 0)$ too. Why this step? We must enumerate this branch on its own: the reason the denominator stays positive is different here (negative × negative), and that is precisely what a "cover every case" check demands.
The trap: crossing the gap. Take $x_{1} = - 1$ (so $f = - 1$ ) and $x_{2} = 1$ (so $f = 1$ ). Here $x_{1} < x_{2}$ but $f (x_{1}) = - 1 < 1 = f (x_{2})$ — that looks increasing! It doesn't count, because $- 1$ and $1$ are in different pieces with $0$ missing between them. Why this step? This shows why we never call $1/ x$ "decreasing on $R$ ": there's no single interval spanning the gap, so no legal pair straddles it.
Conclusion: strictly decreasing on $(- \infty, 0)$ and on $(0, \infty)$ separately — but not monotone across the whole domain.

Verify: On $(0, \infty)$ : $f (1) = 1 > f (2) = 0.5$ ✓. On $(- \infty, 0)$ : $f (- 2) = - 0.5 > f (- 1) = - 1$ ✓ (bigger $x$ , smaller $y$ ).

Cell E — increasing but with a degenerate (flat-looking) point

The cube function has a spot where it looks flat but is still climbing. See the figure — the tangent lies flat at the origin yet the curve never turns back.

Worked example Example 5:

$f (x) = x^{3}$ on $R$ Forecast: Near $x = 0$ the graph flattens out. Does it stop being increasing there?

Difference. For $x_{1} < x_{2}$ : use the identity $x_{2}^{3} - x_{1}^{3} = (x_{2} - x_{1}) (x_{2}^{2} + x_{1} x_{2} + x_{1}^{2})$ . Why this step? Factoring exposes $(x_{2} - x_{1})$ (known positive) times a second factor we must sign.
Sign of the second factor. $x_{2}^{2} + x_{1} x_{2} + x_{1}^{2}$ is a sum that is always $\geq 0$ , and it equals $0$ only when $x_{1} = x_{2} = 0$ . Since our pair has $x_{1} < x_{2}$ , the two are never both zero, so this factor is strictly positive. Why this step? Rewrite it as $(x_{1} + \frac{x _{2}}{2})^{2} + \frac{3}{4} x_{2}^{2}$ — a sum of squares, hence $\geq 0$ , and zero only when both squares vanish (both variables $0$ ).
Read the sign. (positive)(positive) $> 0$ → $f (x_{2}) > f (x_{1})$ for every legal pair. Why this step? A strictly positive difference for all ordered pairs is exactly the definition of strictly increasing — no exceptions to hunt for.
The "flat" point $x = 0$ is a red herring. The graph's steepness is zero at the origin (that's a derivative fact), but order is still preserved: any $x_{1} < 0 < x_{2}$ still gives $f (x_{1}) < 0 < f (x_{2})$ . Why this step? This is exactly Mistake 1 from the parent note — steepness $\neq =$ direction. Zero steepness at a single point does not break increasing-ness.
Conclusion: $f (x) = x^{3}$ is strictly increasing on all of $R$ . (Its being one-to-one everywhere is why it has an inverse, the cube root.)

Verify: $f (- 2) = - 8 < f (0) = 0 < f (2) = 8$ ; and even close in: $f (- 0.1) = - 0.001 < f (0.1) = 0.001$ . ✓

Cell F — constant / flat (the pure degenerate case)

Worked example Example 6:

$f (x) = 5$ on $R$ Forecast: A flat line. Increasing? Decreasing? Both? Neither?

Difference. For any $x_{1} < x_{2}$ : $f (x_{2}) - f (x_{1}) = 5 - 5 = 0$ . Why this step? The output never changes, so the difference is exactly $0$ — neither positive nor negative.
Test strict increasing. We'd need $f (x_{2}) > f (x_{1})$ , i.e. $0 > 0$ . False. Why this step? We literally check the increasing inequality against our computed difference; it fails, ruling increasing out.
Test strict decreasing. We'd need $f (x_{2}) < f (x_{1})$ , i.e. $0 < 0$ . False. Why this step? Strict definitions demand a strict inequality; equality satisfies neither, so decreasing is out too.
Conclusion: a constant function is neither strictly increasing nor strictly decreasing. (It is both "non-decreasing" and "non-increasing" under the loose definitions — see the parent note's nuance box.)

Verify: $f (1) = 5$ and $f (100) = 5$ ; $5 = 5$ , so not $<$ and not $>$ . ✓ Neither.

Cell G — real-world word problem

Worked example Example 7: Cooling coffee —

$T (t) = 20 + 70 e^{- 0.1 t}$ ( $t$ in minutes, $T$ in °C) Forecast: Coffee cools down over time. Do you expect $T$ increasing or decreasing in $t$ ?

Translate the story. "Bigger time $\to$ smaller temperature?" — that's asking whether $T$ is decreasing. Let $t_{1} < t_{2}$ . Why this step? Monotonicity questions in words always reduce to "bigger input → bigger or smaller output?"
Difference. $T (t_{2}) - T (t_{1}) = 70 (e^{- 0.1 t_{2}} - e^{- 0.1 t_{1}})$ . The $20$ cancels. Why this step? The constant offset (room temperature) never affects direction, just like the $+ c$ in a line.
Sign of $e^{- 0.1 t_{2}} - e^{- 0.1 t_{1}}$ . The exponential $e^{- 0.1 t}$ shrinks as $t$ grows (the exponent gets more negative). Since $t_{2} > t_{1}$ , we have $e^{- 0.1 t_{2}} < e^{- 0.1 t_{1}}$ , so the bracket is negative. Why this step? We use the known fact that $e^{(something decreasing)}$ decreases — the reason the tool "exponential" fits: it models decay that never turns around.
Read. $70 \times (negative) < 0$ , so $T (t_{2}) < T (t_{1})$ → temperature strictly decreasing in time. Why this step? A negative difference for every ordered time pair is the definition of decreasing — the story's "cooling" made rigorous.
Limiting behaviour. As $t \to \infty$ , $e^{- 0.1 t} \to 0$ , so $T \to 20$ °C — it cools toward room temperature, never below. It keeps decreasing but flattens near $20$ . Why this step? Checking the limit confirms the physics: coffee can't cool past the room. See Graphical analysis.
Conclusion: the coffee's temperature is strictly decreasing, approaching $20$ °C.

Verify (units + values): at $t = 0$ , $T = 20 + 70 = 90$ °C (fresh coffee ✓). At $t = 10$ , $T = 20 + 70 e^{- 1} \approx 45.75$ °C. Since $90 > 45.75$ , it cooled. ✓ Units: °C throughout.

Cell H — exam twist: piecewise, does the join break it?

Worked example Example 8: A piecewise function — monotone on each piece, but overall?

$f (x) = {x + 1, 4 - x, x < 1 x \geq 1$ Forecast: Each piece is a straight line. One goes up, one goes down. Is the whole function increasing, decreasing, or neither?

Left piece $x < 1$ : slope $+ 1 > 0$ → increasing there. Why this step? By the line rule from Cell A, a positive multiplier on $x$ means $f (x_{2}) - f (x_{1}) = + 1 \cdot (x_{2} - x_{1}) > 0$ , so this branch preserves order on its own.
Right piece $x \geq 1$ : $4 - x$ has slope $- 1 < 0$ → decreasing there. Why this step? By the Cell B line rule, a negative multiplier gives $f (x_{2}) - f (x_{1}) = - 1 \cdot (x_{2} - x_{1}) < 0$ , so this branch reverses order.
Check the join at $x = 1$ . Left approaches $1 + 1 = 2$ ; right value $4 - 1 = 3$ . So $f$ jumps up from just-below-2 to 3 at $x = 1$ . Why this step? A function's global behaviour can change exactly at the seam — you must inspect it, never assume.
Test a spanning pair. Take $x_{1} = 0.5$ (in left piece, $f = 1.5$ ) and $x_{2} = 3$ (right piece, $f = 1$ ). Here $x_{1} < x_{2}$ but $f (x_{1}) = 1.5 > 1 = f (x_{2})$ — that's decreasing behaviour. Yet another pair $x_{1} = 0, x_{2} = 0.9$ gives $f (0) = 1 < f (0.9) = 1.9$ — increasing behaviour. Why this step? Two pairs disagree, so the function preserves order for some pairs and reverses it for others.
Conclusion: $f$ is increasing on $(- \infty, 1)$ , decreasing on $[1, \infty)$ , and neither on the whole line — the classic exam trap: monotone on each piece does not make the whole thing monotone.

Verify: $f (0.5) = 1.5$ , $f (3) = 1$ , so $1.5 > 1$ ✓ (order reversed across the peak); $f (0) = 1 < f (0.9) = 1.9$ ✓ (order preserved on the left).

Wrap-up recall

Recall Which cell does each function land in?

A line with positive slope ::: Cell A — increasing everywhere A line with negative slope ::: Cell B — decreasing everywhere A parabola like $x^{2} - 6 x + 5$ ::: Cell C — sign split at the vertex The function $1/ x$ ::: Cell D — decreasing on each branch, gap at 0, not monotone overall The function $x^{3}$ ::: Cell E — increasing everywhere despite the flat spot at 0 A constant function ::: Cell F — neither increasing nor decreasing (strict) Cooling coffee $20 + 70 e^{- 0.1 t}$ ::: Cell G — decreasing, limiting to 20°C A piecewise up-then-down ::: Cell H — monotone per piece, neither overall