Applications — increasing - decreasing, local extrema (first derivative test): Worked examples

The scenario matrix

Every problem this topic can throw belongs to one of these cells. The examples that follow are labelled with the cell they cover, so by the end no scenario is unseen.

Cell	Situation	What makes it tricky	Example
C1	Smooth polynomial, sign flips $+ \to -$ and $- \to +$	standard peak and valley	Ex 1
C2	Stationary point with no sign change	$f^{'} (c) = 0$ but same sign both sides	Ex 2
C3	Derivative undefined (corner)	test still works on a kink	Ex 3
C4	Derivative undefined (cusp, vertical tangent)	$f^{'} \to \pm \infty$ , still a critical point	Ex 4
C5	Monotone everywhere (no critical points)	must report "no extrema" confidently	Ex 5
C6	Rational function — $f^{'}$ undefined at a point not in the domain	that point is NOT critical	Ex 6
C7	Trig / periodic — infinitely many critical points	classify a whole family	Ex 7
C8	Word problem (real-world max)	translate words → $f$ , then test	Ex 8
C9	Exam twist — parameter $a$ decides the answer	case-split on a constant	Ex 9

Worked example Example 1 — cell C1: smooth quartic, a valley–peak–valley

Classify every local extremum of $f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2}$ .

Forecast: A quartic opening upward — guess: how many turning points, and are they max or min?

Step 1 — differentiate. $f^{'} (x) = 12 x^{3} - 24 x^{2} + 12 x$ . Why this step? The sign of $f^{'}$ is our entire map of rising/falling.

Step 2 — factor to find critical points. $f^{'} (x) = 12 x (x^{2} - 2 x + 1) = 12 x (x - 1)^{2} .$ So $f^{'} (x) = 0$ at $x = 0$ and $x = 1$ . Both are in the domain (all reals), so both are critical. Why this step? Factoring exposes the roots and their multiplicity — the squared factor $(x - 1)^{2}$ is a warning that $f^{'}$ may touch zero without crossing.

Step 3 — sign of each factor. $12 x$ is negative for $x < 0$ , positive for $x > 0$ . The factor $(x - 1)^{2} \geq 0$ always (a square is never negative). So the sign of $f^{'}$ is just the sign of $x$ , except it dips to $0$ at $x = 1$ without changing sign. Why this step? Reading factor-by-factor beats blind test points when squares are present.

Step 4 — number line (look at figure).

Interval	test $x$	$12 x$	$(x - 1)^{2}$	$f^{'}$	behaviour
$(- \infty, 0)$	$- 1$	$-$	$+$	$-$	decreasing
$(0, 1)$	$0.5$	$+$	$+$	$+$	increasing
$(1, \infty)$	$2$	$+$	$+$	$+$	increasing

Step 5 — classify.

At $x = 0$ : $- \to +$ → local minimum, value $f (0) = 0$ .
At $x = 1$ : $+ \to +$ → no extremum (a horizontal-tangent inflection, exactly the C2 warning).

Verify: $f (0) = 0$ . Test a point either side of $0$ in the original $f$ : $f (- 1) = 3 + 8 + 6 = 17 > 0$ and $f (0.5) = 3 (0.0625) - 8 (0.125) + 6 (0.25) = 0.1875 - 1 + 1.5 = 0.6875 > 0$ . Both above $f (0) = 0$ ✓ — it really is the bottom of a valley.

Worked example Example 2 — cell C2: stationary but NOT an extremum (isolated)

Where, if anywhere, does $f (x) = x^{5}$ have a local extremum?

Forecast: $f^{'} (0) = 0$ , so a beginner shouts "extremum!" Is it?

Step 1 — differentiate. $f^{'} (x) = 5 x^{4}$ . Why this step? We need the sign map around the only candidate.

Step 2 — critical point. $5 x^{4} = 0 \Rightarrow x = 0$ (only root). Why this step? $x^{4}$ is a fourth power — it is $\geq 0$ everywhere and hits $0$ only at $x = 0$ .

Step 3 — sign check. $5 x^{4} > 0$ for every $x \neq = 0$ . So $f^{'}$ is $+ \to +$ across $0$ . Why this step? Same sign both sides is the parent's "neither" case.

Step 4 — conclude. $x = 0$ is a critical point but not an extremum (horizontal tangent inflection, like $x^{3}$ ).

Verify: $f (- 0.1) = - 1 0^{- 5} < 0 < 1 0^{- 5} = f (0.1)$ while $f (0) = 0$ sits between — the function keeps climbing through $0$ , no peak or valley ✓.

Worked example Example 3 — cell C3: derivative undefined at a corner

Classify extrema of $f (x) = ∣ x - 2∣ + 1$ .

Forecast: A "V" shifted right and up. Where is its tip, and is it a max or min?

Step 1 — piecewise derivative. $f (x) = {(x - 2) + 1 = x - 1, - (x - 2) + 1 = 3 - x, x \geq 2 x < 2$ so $f^{'} (x) = + 1$ for $x > 2$ and $f^{'} (x) = - 1$ for $x < 2$ ; at $x = 2$ the left slope $- 1$ and right slope $+ 1$ disagree, so $f^{'} (2)$ does not exist. Why this step? A corner has no single tangent — that's precisely a critical point of the "undefined" kind (parent's Mistake B).

Step 2 — the only critical point is $x = 2$ .

Step 3 — sign change. $- \to +$ across $x = 2$ . Why this step? Down then up = valley.

Step 4 — conclude. Local (and global) minimum at $x = 2$ , value $f (2) = 0 + 1 = 1$ .

Verify: $f (1) = ∣ - 1∣ + 1 = 2$ and $f (3) = ∣1∣ + 1 = 2$ ; both exceed $f (2) = 1$ ✓ — the tip is the lowest point.

Worked example Example 4 — cell C4: cusp with a vertical tangent

Classify extrema of $f (x) = x^{2/3}$ .

Forecast: Even root of a square-ish power — sketch it, where's the sharp point?

Step 1 — differentiate. $f^{'} (x) = \frac{2}{3} x^{- 1/3} = \frac{2}{3 3 x}$ . Why this step? Power rule; keep the negative exponent visible so we can see where it blows up.

Step 2 — where is $f^{'}$ zero or undefined? The fraction $\frac{2}{3 3 x}$ is never zero (numerator is $2$ ). It is undefined at $x = 0$ (division by $30 = 0$ ). Since $x = 0$ is in the domain of $f$ , it is a critical point. Why this step? "Undefined derivative but point in domain" = critical (contrast with Ex 6, where the bad point is NOT in the domain).

Step 3 — sign of $f^{'}$ . $3 x$ is negative for $x < 0$ , positive for $x > 0$ . So

$x < 0$ : $f^{'} = \frac{+}{neg} < 0$ → decreasing.
$x > 0$ : $f^{'} = \frac{+}{pos} > 0$ → increasing.

Step 4 — conclude. $- \to +$ at $x = 0$ → local minimum, value $f (0) = 0$ . Note $f^{'} \to - \infty$ as $x \to 0^{-}$ and $+ \infty$ as $x \to 0^{+}$ : the two branches shoot up into a cusp, yet the sign test still classifies it perfectly.

Verify: $f (- 8) = (- 8)^{2/3} = (3 - 8)^{2} = (- 2)^{2} = 4$ and $f (8) = 4$ ; both above $f (0) = 0$ ✓.

Worked example Example 5 — cell C5: monotone everywhere, no extrema

Show $f (x) = x^{3} + x$ has no local extrema, and state its behaviour.

Forecast: A cubic — does it have to have turning points?

Step 1 — differentiate. $f^{'} (x) = 3 x^{2} + 1$ . Why this step? We hunt for sign changes.

Step 2 — solve $f^{'} = 0$ . $3 x^{2} + 1 = 0 \Rightarrow x^{2} = - \frac{1}{3}$ , which has no real solution (a square can't be negative). And $f^{'}$ is defined everywhere. So there are no critical points at all. Why this step? No critical point ⇒ nowhere for the sign to flip.

Step 3 — sign of $f^{'}$ . $3 x^{2} \geq 0$ , so $3 x^{2} + 1 \geq 1 > 0$ for every $x$ . Thus $f^{'} > 0$ everywhere. Why this step? Confirms one unbroken sign.

Step 4 — conclude. $f$ is strictly increasing on all of $R$ ; no local maximum or minimum exists. This is a clean use of the parent's Mean Value Theorem argument: $f^{'} > 0$ on the whole line ⇒ increasing throughout.

Verify: pick any $a < b$ , say $0 < 1$ : $f (0) = 0 < 2 = f (1)$ ✓; and $- 1 < 0$ : $f (- 1) = - 2 < 0 = f (0)$ ✓ — never turns.

Worked example Example 6 — cell C6: rational function, "bad" point NOT in domain

Classify the extrema of $f (x) = \frac{x ^{2}}{x - 1}$ .

Forecast: There's trouble at $x = 1$ . Is $x = 1$ a critical point? (Careful!)

Step 1 — domain first. $x = 1$ makes the denominator $0$ , so $x = 1$ is not in the domain. It can never be a critical point or an extremum — extrema must be domain points. Why this step? Skipping this is the classic trap; a point off the domain is a vertical asymptote, not a peak.

Step 2 — differentiate (quotient rule). $f^{'} (x) = \frac{( 2 x ) ( x - 1 ) - x ^{2} ( 1 )}{( x - 1 ) ^{2}} = \frac{2 x ^{2} - 2 x - x ^{2}}{( x - 1 ) ^{2}} = \frac{x ^{2} - 2 x}{( x - 1 ) ^{2}} = \frac{x ( x - 2 )}{( x - 1 ) ^{2}} .$ Why this step? Quotient rule: $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ .

Step 3 — critical points. Numerator zero: $x (x - 2) = 0 \Rightarrow x = 0$ or $x = 2$ (both in domain). The denominator $(x - 1)^{2} > 0$ except at $x = 1$ , which is already excluded. So critical points are $x = 0, x = 2$ .

Step 4 — sign of $f^{'}$ (denominator always $+$ , so sign $=$ sign of $x (x - 2)$ ):

Interval	$x (x - 2)$	$f^{'}$	behaviour
$(- \infty, 0)$	$(-) (-) = +$	$+$	increasing
$(0, 1)$	$(+) (-) = -$	$-$	decreasing
$(1, 2)$	$(+) (-) = -$	$-$	decreasing
$(2, \infty)$	$(+) (+) = +$	$+$	increasing

(The sign does not flip across $x = 1$ ; we just skip that excluded point.)

Step 5 — classify.

$x = 0$ : $+ \to -$ → local maximum, value $f (0) = \frac{0}{- 1} = 0$ .
$x = 2$ : $- \to +$ → local minimum, value $f (2) = \frac{4}{1} = 4$ .

Verify: $f (- 1) = \frac{1}{- 2} = - 0.5 < 0 = f (0)$ and $f (0.5) = \frac{0.25}{- 0.5} = - 0.5 < 0$ , so $0$ tops its neighbours ✓. $f (1.5) = \frac{2.25}{0.5} = 4.5 > 4 = f (2)$ and $f (3) = \frac{9}{2} = 4.5 > 4$ , so $4$ is the valley floor ✓.

Worked example Example 7 — cell C7: periodic function, infinitely many critical points

Find and classify the local extrema of $f (x) = sin x + cos x$ on all of $R$ .

Forecast: Waves repeat — expect a pattern of alternating peaks and troughs, not just one.

Step 1 — differentiate. $f^{'} (x) = cos x - sin x$ . Why this step? Sign of $f^{'}$ still rules, even for periodic functions.

Step 2 — solve $f^{'} = 0$ . $cos x = sin x \Rightarrow tan x = 1 \Rightarrow x = \frac{π}{4} + nπ$ for every integer $n$ . Why this step? $tan x = 1$ repeats every $π$ , giving the infinite family. $f^{'}$ is defined everywhere, so these are the only critical points.

Step 3 — sign test on one period. Take $n = 0$ , $x = \frac{π}{4}$ . Just left ( $x = 0$ ): $f^{'} = cos 0 - sin 0 = 1 > 0$ . Just right ( $x = \frac{π}{2}$ ): $f^{'} = 0 - 1 = - 1 < 0$ . So $+ \to -$ → local maximum at $x = \frac{π}{4}$ , value $f (\frac{π}{4}) = \frac{2}{2} + \frac{2}{2} = 2$ . Take $n = 1$ , $x = \frac{5 π}{4}$ : left ( $x = π$ ) $f^{'} = - 1 - 0 = - 1 < 0$ , right ( $x = \frac{3 π}{2}$ ) $f^{'} = 0 + 1 = 1 > 0$ , so $- \to +$ → local minimum, value $f (\frac{5 π}{4}) = - \frac{2}{2} - \frac{2}{2} = - 2$ . Why this step? One period settles the whole pattern by periodicity.

Step 4 — the rule. Maxima at $x = \frac{π}{4} + 2 nπ$ (value $2$ ), minima at $x = \frac{5 π}{4} + 2 nπ$ (value $- 2$ ). They alternate forever.

Verify: amplitude of $sin x + cos x = 2 sin (x + \frac{π}{4})$ is $2$ , so the true max is $2$ and min $- 2$ ✓ — matches.

Worked example Example 8 — cell C8: word problem (real-world maximum)

A farmer has $200$ m of fence to build a rectangular pen against a straight river (no fence needed on the river side). What dimensions give the largest area?

Forecast: With a fixed perimeter, guess: is a long-thin pen or a squarish pen bigger?

Step 1 — translate to a function. Let $x$ = length of each of the two sides perpendicular to the river, and $y$ = the single side parallel to it. Fence used: $2 x + y = 200 \Rightarrow y = 200 - 2 x$ . Area $A = x y = x (200 - 2 x) = 200 x - 2 x^{2}$ , valid for $0 \leq x \leq 100$ . Why this step? We must get a single-variable function before the derivative test can run.

Step 2 — differentiate. $A^{'} (x) = 200 - 4 x$ . Why this step? The sign of $A^{'}$ tells us where area rises/falls.

Step 3 — critical point. $200 - 4 x = 0 \Rightarrow x = 50$ . Defined everywhere, one critical point.

Step 4 — sign test. $A^{'} (40) = 200 - 160 = 40 > 0$ (increasing), $A^{'} (60) = 200 - 240 = - 40 < 0$ (decreasing). $+ \to -$ → local maximum at $x = 50$ . Why this step? First derivative test proves it's a genuine peak, not a min or inflection.

Step 5 — dimensions & area. $x = 50$ m, $y = 200 - 2 (50) = 100$ m, area $A = 50 \cdot 100 = 5000 m^{2}$ . Report both: location $x = 50$ and value $5000 m^{2}$ (parent's Mistake D).

Verify: endpoints give $A (0) = 0$ and $A (100) = 100 (0) = 0$ ; the interior point $A (50) = 5000$ beats both, and units are m·m $= m^{2}$ ✓.

Worked example Example 9 — cell C9: exam twist, answer depends on a parameter

For which values of the constant $a$ does $f (x) = x^{3} + a x$ have local extrema, and where?

Forecast: Guess: does every choice of $a$ give turning points, or only some?

Step 1 — differentiate. $f^{'} (x) = 3 x^{2} + a$ . Why this step? Critical points depend on $a$ , so we keep it symbolic.

Step 2 — solve $f^{'} = 0$ . $3 x^{2} + a = 0 \Rightarrow x^{2} = - \frac{a}{3}$ . A real solution exists only when $- \frac{a}{3} \geq 0$ , i.e. $a \leq 0$ . Why this step? $x^{2} \geq 0$ , so the right side must be non-negative — this splits the problem into cases.

Step 3 — case split.

$a > 0$ : no real critical point ⇒ $f^{'} = 3 x^{2} + a > 0$ everywhere ⇒ strictly increasing, no extrema (like Ex 5).
$a = 0$ : $f = x^{3}$ , single critical point $x = 0$ with $f^{'} = 3 x^{2} \geq 0$ , same sign ⇒ no extremum (like Ex 2).
$a < 0$ : two critical points $x = \pm - a /3$ . Since $3 x^{2} + a$ is an upward parabola in $x$ , it is $+$ outside the roots and $-$ between them. So at the left root $+ \to -$ = local max, at the right root $- \to +$ = local min. Why this step? Each cell of the matrix reappears as a case controlled by $a$ .

Step 4 — concrete check. For $a = - 3$ : roots $x = \pm 1 = \pm 1$ ; this is exactly the parent's Example 1 ( $x^{3} - 3 x$ ): max at $- 1$ , min at $+ 1$ .

Verify: with $a = - 3$ , $f (- 1) = - 1 + 3 = 2$ (max value) and $f (1) = 1 - 3 = - 2$ (min value) ✓ — matches the parent note. With $a = 3$ : $f^{'} = 3 x^{2} + 3 \geq 3 > 0$ , monotone, no turning point ✓.

Recall Active recall — cover the answers

In Ex 1, why is $x = 1$ not an extremum? ::: $f^{'} = 12 x (x - 1)^{2}$ ; the $(x - 1)^{2}$ factor keeps $f^{'}$ positive both sides — same sign.
In Ex 6, why is $x = 1$ NOT a critical point? ::: $x = 1$ is not in the domain (denominator zero) — extrema must be domain points.
Ex 4: what kind of point is $x = 0$ for $x^{2/3}$ ? ::: A cusp / local minimum; $f^{'}$ is undefined there but $- \to +$ .
Ex 9: for which $a$ does $x^{3} + a x$ have extrema? ::: Only $a < 0$ ; then max at $- - a /3$ , min at $+ - a /3$ .
Ex 8: max area with 200 m against a river? ::: $x = 50$ , $y = 100$ , area $5000 m^{2}$ .
Ex 7: extrema of $sin x + cos x$ ? ::: Max $2$ at $\frac{π}{4} + 2 nπ$ , min $- 2$ at $\frac{5 π}{4} + 2 nπ$ .

Connections

Fermat's Theorem on Stationary Points — why we only ever search critical points (Ex 6's domain caveat).
Mean Value Theorem — engine behind the monotone conclusion in Ex 5.
Second Derivative Test — faster classifier when it applies; fails at cusps like Ex 4 and at $x^{5}$ in Ex 2.
Optimization (Closed Interval Method) — Ex 8 done with endpoints for a global answer.
Curve Sketching — assembling all these sign charts into full graphs.
Concavity and Inflection Points — the inflection nature of Ex 1's $x = 1$ and Ex 2's $x = 0$ .

Worked examples — Applications — increasing - decreasing, local extrema (first derivative test)

The scenario matrix

Connections

Concept Map