Increasing and decreasing functions — intuitive definition: Exercises

The core move in almost every solution is the difference trick: to compare two outputs, look at the sign of $f (x_{2}) - f (x_{1})$ . If this difference is greater than zero (written $f (x_{2}) - f (x_{1}) > 0$ ) the function rose between the two points; if it is $< 0$ the function fell. Figure s01 draws exactly this: pick a left point and a right point, then read whether the vertical gap goes up or down.

Figure — Increasing and decreasing functions — intuitive definition

Figure s01 — The difference trick. The blue dashed segment is the horizontal step $x_{2} - x_{1} > 0$ (we always move rightward). The pink dashed segment is the vertical change $f (x_{2}) - f (x_{1})$ . When the pink segment points up (positive), the function increased on that step; when it points down, it decreased. Every solution below just computes the sign of this pink gap in symbols.

Level 1 — Recognition

L1.1

Problem. From the graph rule of thumb (a rising left-to-right curve is increasing), classify each on all of $R$ : (a) $f (x) = 7 x - 2$ , (b) $g (x) = - 3 x + 1$ , (c) $h (x) = 5$ .

Recall Solution

(a) Slope $+ 7 > 0$ , so as $x$ grows $y$ grows → strictly increasing. (b) Slope $- 3 < 0$ , so as $x$ grows $y$ shrinks → strictly decreasing. (c) Output never changes: $h (x_{1}) = h (x_{2}) = 5$ . Since $5 \neq < 5$ and $5 \neq > 5$ , it is neither — it is constant. (This is the flat path from the parent note.)

L1.2

Problem. Look at figure s02. On which of the marked intervals $A = (- 4, - 1)$ , $B = (- 1, 2)$ , $C = (2, 5)$ is the drawn curve increasing?

Recall Solution

Walk left to right. On $A$ the curve climbs → increasing. On $B$ it drops → decreasing. On $C$ it climbs again → increasing. Answer: increasing on $A$ and $C$ .

Level 2 — Application

L2.1

Problem. Using the difference trick, prove $f (x) = - 5 x + 8$ is strictly decreasing on $R$ .

Recall Solution

Take any $x_{1} < x_{2}$ . $f (x_{2}) - f (x_{1}) = (- 5 x_{2} + 8) - (- 5 x_{1} + 8) = - 5 (x_{2} - x_{1}) .$ Since $x_{2} > x_{1}$ , we have $x_{2} - x_{1} > 0$ , so $- 5 (x_{2} - x_{1}) < 0$ . Thus $f (x_{2}) < f (x_{1})$ for every such pair → strictly decreasing. ∎

L2.2

Problem. Show $f (x) = x^{3}$ is strictly increasing on $R$ .

Recall Solution

Take $x_{1} < x_{2}$ . We need the sign of $x_{2}^{3} - x_{1}^{3}$ . Factor: $x_{2}^{3} - x_{1}^{3} = (x_{2} - x_{1}) (x_{2}^{2} + x_{1} x_{2} + x_{1}^{2}) .$ Why factor? It isolates the always-positive part $(x_{2} - x_{1})$ from the bracket. The bracket is a sum of squares in disguise: $x_{2}^{2} + x_{1} x_{2} + x_{1}^{2} = (x_{1} + \frac{x _{2}}{2})^{2} + \frac{3}{4} x_{2}^{2} \geq 0$ , and it is $0$ only if $x_{1} = x_{2} = 0$ , which is excluded since $x_{1} < x_{2}$ . So the bracket is $> 0$ . Then $(x_{2} - x_{1}) > 0$ times a positive bracket gives $x_{2}^{3} - x_{1}^{3} > 0$ , i.e. $f (x_{2}) > f (x_{1})$ → strictly increasing. ∎

L2.3

Problem. Prove $f (x) = \frac{1}{x ^{2}}$ is strictly decreasing on $(0, \infty)$ .

Recall Solution

Recall the criterion. Decreasing means: for every $x_{1} < x_{2}$ we need $f (x_{1}) > f (x_{2})$ , i.e. the left output sits above the right output. So instead of $f (x_{2}) - f (x_{1})$ , it is cleaner here to compute $f (x_{1}) - f (x_{2})$ and show it is positive — that is the very same difference trick, just written as (left) minus (right); a positive answer says "left is higher," which is exactly decreasing. Take $0 < x_{1} < x_{2}$ . $f (x_{1}) - f (x_{2}) = \frac{1}{x _{1}^{2}} - \frac{1}{x _{2}^{2}} = \frac{x _{2}^{2} - x _{1}^{2}}{x _{1}^{2} x _{2}^{2}} .$ Numerator: since $0 < x_{1} < x_{2}$ , we get $x_{2}^{2} > x_{1}^{2}$ , so $x_{2}^{2} - x_{1}^{2} > 0$ . Denominator $x_{1}^{2} x_{2}^{2} > 0$ . So $f (x_{1}) - f (x_{2}) > 0$ , meaning $f (x_{1}) > f (x_{2})$ → strictly decreasing. ∎

Level 3 — Analysis

L3.1

Problem. Find all intervals on which $f (x) = x^{2} - 6 x + 5$ is increasing and where it is decreasing, using only the difference trick (no calculus).

Recall Solution

Take $x_{1} < x_{2}$ (for now, without assuming which side of anything they lie on). $f (x_{2}) - f (x_{1}) = (x_{2}^{2} - 6 x_{2}) - (x_{1}^{2} - 6 x_{1}) = (x_{2}^{2} - x_{1}^{2}) - 6 (x_{2} - x_{1}) .$ Why factor out $(x_{2} - x_{1})$ ? Because we already know its sign ( $x_{2} - x_{1} > 0$ ), so pulling it out leaves a smaller expression whose sign fully decides the answer: $= (x_{2} - x_{1}) [(x_{2} + x_{1}) - 6] .$ Since $x_{2} - x_{1} > 0$ always, the sign is decided entirely by the bracket $(x_{1} + x_{2}) - 6$ . Why split into two sides of $x = 3$ ? The bracket $(x_{1} + x_{2}) - 6$ can be positive for some pairs and negative for others, so $f$ cannot be one single type on all of $R$ (this is why we must restrict to sub-intervals — monotonicity is an interval property). The bracket changes sign around where $x_{1} + x_{2} = 6$ , i.e. when both points hug $x = 3$ . So we test the two regions cut by $3$ :

Both points in $(3, \infty)$ : then $x_{1} > 3$ and $x_{2} > 3$ , so $x_{1} + x_{2} > 6$ , bracket $> 0$ → difference $> 0$ → increasing on $(3, \infty)$ .
Both points in $(- \infty, 3)$ : then $x_{1} < 3$ and $x_{2} < 3$ , so $x_{1} + x_{2} < 6$ , bracket $< 0$ → difference $< 0$ → decreasing on $(- \infty, 3)$ . What about a straddling pair (one point below $3$ , one above)? Such a pair mixes the two behaviours and is not covered by either interval statement — that is precisely why we never claim a single behaviour across $x = 3$ . The point $x = 3$ itself is the single turning point (the vertex); it is the boundary shared by both intervals and belongs to neither "increasing" nor "decreasing" as an interior point. Answer: decreasing on $(- \infty, 3)$ , increasing on $(3, \infty)$ ; turning point at $x = 3$ .

L3.2

Problem. Is $f (x) = ∣ x ∣$ monotonic on $[- 2, 2]$ ? Identify each piece.

Recall Solution

Split at $0$ where the definition of absolute value changes.

On $[- 2, 0]$ , $∣ x ∣ = - x$ , slope $- 1 < 0$ → decreasing.
On $[0, 2]$ , $∣ x ∣ = x$ , slope $+ 1 > 0$ → increasing. Because it falls then rises, $f$ is not monotonic on the whole $[- 2, 2]$ — a V shape. (See Monotone functions.)

L3.3

Problem. For which real values of $k$ is $f (x) = k x + 3$ strictly increasing on $R$ ? Cover every case of $k$ .

Recall Solution

$f (x_{2}) - f (x_{1}) = k (x_{2} - x_{1})$ with $x_{2} - x_{1} > 0$ .

$k > 0$ : difference $> 0$ → increasing.
$k < 0$ : difference $< 0$ → decreasing.
$k = 0$ : $f (x) = 3$ , constant → neither. Answer: strictly increasing exactly when $k > 0$ .

Level 4 — Synthesis

L4.1

Problem. Let $g$ be strictly increasing on $R$ . Prove that $h (x) = - g (x)$ is strictly decreasing.

Recall Solution

Take $x_{1} < x_{2}$ . Because $g$ is increasing, $g (x_{1}) < g (x_{2})$ . Multiply the inequality by $- 1$ — this flips the sign: $- g (x_{1}) > - g (x_{2}) ⟹ h (x_{1}) > h (x_{2}) .$ That is exactly the decreasing condition for every pair → $h$ is strictly decreasing. ∎ (Geometrically: negating $g$ reflects the graph over the $x$ -axis, turning every uphill into a downhill.)

L4.2

Problem. If $f$ is strictly increasing on an interval $I$ , argue that $f$ is one-to-one on $I$ (hence invertible there — link to Inverse functions).

Recall Solution

One-to-one means: different inputs give different outputs. Take any $x_{1} \neq = x_{2}$ in $I$ . One is smaller; say $x_{1} < x_{2}$ . Since $f$ is strictly increasing, $f (x_{1}) < f (x_{2})$ , so in particular $f (x_{1}) \neq = f (x_{2})$ . Because any two distinct inputs produce distinct outputs, $f$ is one-to-one on $I$ , so its inverse exists on $I$ . ∎

L4.3

Problem. Suppose $f$ and $g$ are both strictly increasing on $R$ . Is the composition $f \circ g$ (meaning $f (g (x))$ ) strictly increasing? Prove or give a counterexample.

Recall Solution

Claim: yes. Take $x_{1} < x_{2}$ . Step 1: $g$ increasing → $g (x_{1}) < g (x_{2})$ . Step 2: feed those two ordered outputs into $f$ . Since $f$ is increasing and $g (x_{1}) < g (x_{2})$ , $f (g (x_{1})) < f (g (x_{2})) .$ That is $(f \circ g) (x_{1}) < (f \circ g) (x_{2})$ for every pair → strictly increasing. ∎ (Intuition: uphill-then-uphill is still uphill.)

Level 5 — Mastery

L5.1

Problem. Prove from the definition that $f (x) = x$ is strictly increasing on $[0, \infty)$ , including the degenerate endpoint $x = 0$ .

Recall Solution

Take $0 \leq x_{1} < x_{2}$ . Both square roots are defined and non-negative. Consider $x_{2} - x_{1} = \frac{( x _{2} - x _{1} ) ( x _{2} + x _{1} )}{x _{2} + x _{1}} = \frac{x _{2} - x _{1}}{x _{2} + x _{1}} .$ Why rationalize? It turns the hard root difference into $x_{2} - x_{1}$ , whose sign we know. Numerator $x_{2} - x_{1} > 0$ . Denominator $x_{2} + x_{1} > 0$ (it is zero only if both are $0$ , impossible since $x_{1} < x_{2}$ ). So the whole thing is $> 0$ : $x_{2} > x_{1}$ . Endpoint check: at $x_{1} = 0$ , $x_{2} - 0 = x_{2} > 0$ for any $x_{2} > 0$ — still increasing. → strictly increasing on $[0, \infty)$ . ∎

L5.2

Problem. Determine every interval of monotonicity of $f (x) = x + \frac{1}{x}$ on its domain (all $x \neq = 0$ ). Handle both signs of $x$ .

Recall Solution

Domain excludes $0$ . Take two points $x_{1} < x_{2}$ in the same sign region (we do not mix a negative point with a positive one, because the graph is broken at $0$ — a straddling pair is never covered, exactly the L5.3 warning). Compute $f (x_{2}) - f (x_{1}) = (x_{2} - x_{1}) + (\frac{1}{x _{2}} - \frac{1}{x _{1}}) = (x_{2} - x_{1}) + \frac{x _{1} - x _{2}}{x _{1} x _{2}} = (x_{2} - x_{1}) (1 - \frac{1}{x _{1} x _{2}}) .$ Why factor out $(x_{2} - x_{1})$ again? Same reason as L3.1: its sign is known ( $> 0$ ), so the leftover bracket $(1 - \frac{1}{x _{1} x _{2}})$ decides everything. Its sign equals the sign of $\frac{x _{1} x _{2} - 1}{x _{1} x _{2}}$ .

Region $x > 0$ : here $x_{1} > 0$ and $x_{2} > 0$ , so their product $x_{1} x_{2} > 0$ (positive × positive). The sign therefore follows the numerator $x_{1} x_{2} - 1$ .

Both in $(1, \infty)$ : $x_{1} x_{2} > 1$ → positive → increasing.
Both in $(0, 1)$ : $x_{1} x_{2} < 1$ → negative → decreasing.

Region $x < 0$ : here $x_{1} < 0$ and $x_{2} < 0$ . Why is $x_{1} x_{2} > 0$ ? Because a negative number times a negative number is positive (two minus signs cancel), so the product of two negatives is positive. Hence $x_{1} x_{2} > 0$ again, and the sign once more follows $x_{1} x_{2} - 1$ .

Both in $(- \infty, - 1)$ : here $∣ x_{1} ∣, ∣ x_{2} ∣ > 1$ so $x_{1} x_{2} > 1$ → positive → increasing.
Both in $(- 1, 0)$ : here $∣ x_{1} ∣, ∣ x_{2} ∣ < 1$ so $x_{1} x_{2} < 1$ → negative → decreasing.

Summary: increasing on $(- \infty, - 1)$ and $(1, \infty)$ ; decreasing on $(- 1, 0)$ and $(0, 1)$ . The turning points sit at $x = - 1$ (local max, $f = - 2$ ) and $x = 1$ (local min, $f = 2$ ), matching figure s03.

L5.3

Problem. True or false, with proof: "If $f$ is strictly increasing on $(- \infty, 0)$ and strictly increasing on $(0, \infty)$ , then $f$ is strictly increasing on all of $R ∖ {0}$ treated as one set." Use $f (x) = 1/ x$ to guide you.

Recall Solution

False. Increasing on each piece separately does not force increasing across the gap. Counterexample $f (x) = \frac{1}{x}$ : it is strictly decreasing on each side (a cleaner illustration of the same danger). Take a stronger example, $f (x) = - \frac{1}{x}$ — check: on $(0, \infty)$ , using L2.3 logic $- 1/ x$ is increasing; on $(- \infty, 0)$ it is also increasing. Yet pick $x_{1} = - 1 < x_{2} = 1$ : $f (- 1) = 1$ , $f (1) = - 1$ , so $f (x_{1}) = 1 > - 1 = f (x_{2})$ — the order is broken across $0$ . Lesson: "increasing on $A$ " and "increasing on $B$ " only combine into "increasing on $A \cup B$ " when the pieces don't jump down as you cross the gap. Always re-check pairs that straddle the excluded point. ∎

Recall Quick self-check (reason it out before revealing)

For a function to be increasing on an interval, the difference $f (x_{2}) - f (x_{1})$ (right output minus left output, with $x_{1} < x_{2}$ ) must have which sign for every such pair? ::: It must be $> 0$ (positive) for all pairs — the right output always sits above the left. Using the difference-trick factoring $(x_{2} - x_{1}) [(x_{1} + x_{2}) - 6]$ , on which interval is $f (x) = x^{2} - 6 x + 5$ decreasing? ::: $(- \infty, 3)$ — there $x_{1} + x_{2} < 6$ makes the bracket negative. For $f (x) = k x + 3$ , name every $k$ giving a strictly increasing function, and say what happens otherwise. ::: Increasing exactly when $k > 0$ ; $k < 0$ gives decreasing; $k = 0$ gives a constant (neither). If $g$ is strictly increasing, what is $- g$ , and which inequality rule makes it so? ::: $- g$ is strictly decreasing — multiplying $g (x_{1}) < g (x_{2})$ by the negative number $- 1$ flips $<$ into $>$ . On which intervals is $f (x) = x + \frac{1}{x}$ increasing, and what algebraic fact about products of negatives did the negative- $x$ region rely on? ::: Increasing on $(- \infty, - 1)$ and $(1, \infty)$ ; the negative region used that (negative)×(negative)=positive, so $x_{1} x_{2} > 0$ .