Applications — increasing - decreasing, local extrema (first derivative test): Visual walkthrough

We will use these words. Let us pin them down before anything else.

Definition The words we must earn

$f (x)$ — the height of a curve above the horizontal position $x$ . Think: how high the road is when you have walked $x$ steps to the right.
$f^{'} (x)$ — the slope at position $x$ : the steepness of the road right there. Positive = tilts up-to-the-right, negative = tilts down-to-the-right, zero = momentarily flat.
Interval $I$ — a stretch of $x$ -values, like all $x$ between $- 2$ and $5$ .
The arrow " $\to$ " in patterns like " $+ \to -$ " means "as we walk left-to-right through the critical point $c$ , the slope sign changes from the first to the second." It is not multiplication; it is a before/after snapshot.
"Positive" and "negative" mean greater than zero and less than zero; whenever we say a run or a quantity is $> 0$ we mean strictly greater than zero.

That is the entire vocabulary. Everything below is drawn from these.

Step 1 — What does "slope" actually measure?

WHAT. Pick two points on the curve, a left one at position $a$ and a right one at position $b$ , with $a < b$ . The average slope between them is the rise divided by the run: $average slope \frac{f ( b ) - f ( a )}{b - a} .$

$f (b) - f (a)$ — the rise: how much higher (or lower) the road is at $b$ than at $a$ .
$b - a$ — the run: how far right we walked. Because $a < b$ , this is always $> 0$ (strictly greater than zero).

WHY. Slope is the tool for "rising or falling", because a fraction's sign is decided by the signs of its top and bottom. The bottom $b - a$ is locked strictly positive, so the sign of the rise alone tells us up or down. This is the seed of the whole theorem.

PICTURE. The right triangle formed by the horizontal run and the vertical rise — the hypotenuse is the straight line joining the two points.

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

Step 2 — Shrink the run: from average slope to $f^{'}$

WHAT. Slide $b$ toward $a$ . As the run $b - a$ gets tiny, the joining line stops being a far-apart chord and hugs the curve — it becomes the tangent line, and its slope is exactly $f^{'} (a)$ : $f^{'} (a) = lim_{b \to a} \frac{f ( b ) - f ( a )}{b - a} .$

$lim_{b \to a}$ — "watch what the fraction approaches as $b$ creeps up to $a$ ".
$f^{'} (a)$ — the result: the slope of the road at the single point $a$ .

WHY. We want steepness at a point, not averaged over a stretch. The limit is the only tool that turns a two-point ratio into a one-point steepness — it answers "what does the average slope settle down to as the two points merge?"

PICTURE. Three chords of shrinking run, flattening onto the tangent.

Step 3 — The Mean Value Theorem: the bridge back

WHAT. The Mean Value Theorem says: on any stretch $[a, b]$ where $f$ is continuous (no breaks) and differentiable inside (no corners), there is at least one interior point $c$ where the tangent slope equals the average slope: $f^{'} (c) = \frac{f ( b ) - f ( a )}{b - a} .$

$c$ — a mystery point strictly between $a$ and $b$ ; we don't know where, only that it exists.
Left side $f^{'} (c)$ — an instant slope (one point).
Right side — the average slope (two points) from Step 1.

WHY. Step 1 gave us average slope; Step 2 gave us instant slope. MVT is the bridge that says these are equal somewhere. That equality is what lets us convert a fact about $f^{'}$ (instant) into a fact about $f (b)$ vs $f (a)$ (heights).

PICTURE. The chord and a parallel tangent kissing the curve at $c$ .

Step 4 — Rearrange: heights obey the slope's sign

WHAT. Multiply MVT by the run $b - a$ : $f (b) - f (a) = sign we control f^{'} (c) \cdot > 0 (b - a) .$

WHY. Now the rise $f (b) - f (a)$ is written as a product. One factor $(b - a)$ is always $> 0$ . So the sign of the rise is entirely decided by the sign of $f^{'} (c)$ . This is the payoff: control the slope's sign, control whether the road rose or fell.

If $f^{'} > 0$ everywhere on $I$ : then $f^{'} (c) > 0$ , product is positive $\Rightarrow f (b) > f (a)$ → increasing.
If $f^{'} < 0$ everywhere: product is negative $\Rightarrow f (b) < f (a)$ → decreasing.

PICTURE. A sign multiplication table drawn as a ledger: (slope sign) $\times$ (positive run) = (rise sign).

Step 5 — Where can the road turn around? Critical points

WHAT. A peak or valley is where the road stops rising and starts falling (or vice versa). At the very tip, the tangent is either flat ( $f^{'} (c) = 0$ ) or broken ( $f^{'} (c)$ does not exist — a corner). An interior point of either kind (one that has curve on both sides of it) is a critical point.

WHY. For the height to switch from rising to falling, the slope must pass through — or jump across — zero. It cannot go from $+$ to $-$ while staying strictly positive. So turnarounds are trapped at critical points. Everywhere else, Step 4 forbids a turn.

PICTURE. A smooth hilltop (flat tangent) beside a sharp corner (no tangent) — both are turnarounds.

Step 6 — The First Derivative Test: reading the turn

WHAT. Sit at a critical point $c$ . Look at the slope sign just left and just right of it (recall $+ \to -$ means "sign was $+$ before $c$ , becomes $-$ after $c$ "): $up then down + \to - \Rightarrow local max, down then up - \to + \Rightarrow local min .$

WHY. By Step 4, " $+$ on the left" means the road climbed up to $c$ ; " $-$ on the right" means it descended away from $c$ . Climb-then-descend is exactly a hilltop. The mirror pattern is a valley. We only ever needed the sign of $f^{'}$ on each side — nothing else.

PICTURE. A hill labelled $+ \to -$ next to a valley labelled $- \to +$ , arrows showing walking direction.

Step 7 — The degenerate cases: same sign, and true plateaus

WHAT. What if $f^{'} = 0$ at $c$ but the road rises on both sides? Example: $f (x) = x^{3}$ at $x = 0$ , where $f^{'} (x) = 3 x^{2} \geq 0$ with the only zero at $x = 0$ . $+ \to + (or - \to -) \Rightarrow no extremum .$ There is also a third possibility: $f^{'} = 0$ on a whole sub-interval (a genuine flat plateau, e.g. a constant stretch). There the slope isn't " $+$ " or " $-$ " on that side at all — it is exactly $0$ across the stretch, so the road is level, and again there is no strict peak or valley.

WHY. In the $x^{3}$ case the road never turned around — it merely paused, like a flat landing on an upward staircase: rising before, rising after, no peak or trough. On a true plateau the road is flat over a whole stretch, so no single point is strictly highest or lowest. In both, the sign of $f^{'}$ never actually flipped, which is the only thing that would signal an extremum. This is the trap summarised as Mistake A below: $f^{'} (c) = 0$ alone is not enough — you must see the sign change.

PICTURE. The graph of $x^{3}$ flattening at the origin but continuing upward — a horizontal tangent with no turn.

Step 8 — Worked landscape: $f (x) = x^{3} - 3 x$

Before we differentiate, we need one earned tool.

WHAT. Apply everything to a real cubic.

$f^{'} (x) = 3 x^{2} - 3$ . (Power rule on $x^{3}$ gives $3 x^{2}$ ; on $- 3 x = - 3 x^{1}$ gives $- 3$ ; constant terms would give $0$ .)
$3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ — two critical points.
Test $x = - 2$ : $f^{'} = 3 (4) - 3 = 9 > 0$ (increasing). Test $x = 0$ : $f^{'} = - 3 < 0$ (decreasing). Test $x = 2$ : $f^{'} = 9 > 0$ (increasing).
At $x = - 1$ : $+ \to -$ → local max, height $f (- 1) = (- 1)^{3} - 3 (- 1) = 2$ .
At $x = + 1$ : $- \to +$ → local min, height $f (1) = 1 - 3 = - 2$ .

WHY. This shows the full pipeline: differentiate → find critical points → number-line the signs → classify. It also drills Mistake D: report the value ( $2$ and $- 2$ ), not just the location.

PICTURE. The curve $x^{3} - 3 x$ with the sign-of- $f^{'}$ number line beneath, extrema marked.

The one-picture summary

Everything above collapses into one diagram: heights on top, the slope sign map on the number line below, turnarounds exactly where the sign flips.

Recall Feynman retelling — the whole walkthrough in plain words

Imagine walking a hilly road left to right. Slope is just how steep the ground is under your feet: uphill is "positive", downhill is "negative". If I want the steepness at one exact spot, I take two nearby footprints, measure rise-over-run, then slide them together until they're the same spot — that limiting steepness is $f^{'}$ . The Mean Value Theorem is a promise: over any stretch, somewhere the ground's exact steepness equals the average steepness of the whole stretch. Multiply that by the (always-positive) length walked, and you learn whether you ended higher or lower — so the sign of the slope alone decides rising or falling; it's even enough for the slope to just touch zero at scattered points, as long as it's never negative. To turn around — hilltop or valley — the ground must go flat or kink; those are critical points, and the far ends of a fenced-off road (endpoints) are candidates too. Stand on one and peek: the little arrow " $\to$ " means "before then after". Uphill-before, downhill-after ( $+ \to -$ ) means you're on a peak; downhill-before, uphill-after ( $- \to +$ ) means a valley. But if it's uphill on both sides, or dead flat over a whole stretch, you only stepped over a landing — no peak at all. That's why $f^{'} = 0$ by itself (Mistake A) never proves a summit; you must watch the sign change. And when you report it, give the height $f (c)$ , not just the spot $c$ (Mistake D). Test it all on $x^{3} - 3 x$ : flat at $x = - 1$ (peak, height $2$ ) and $x = 1$ (valley, height $- 2$ ), and you've mapped the entire landscape.

Connections

Mean Value Theorem — Steps 3–4 are its direct consequence.
Fermat's Theorem on Stationary Points — why turnarounds are trapped at interior critical points (Step 5).
Second Derivative Test — classifies the same critical points using concavity instead of sign changes.
Concavity and Inflection Points — the $x^{3}$ pause of Step 7 is an inflection.
Optimization (Closed Interval Method) — extends this local picture to global extrema, including endpoints.
Curve Sketching — combines this walkthrough with $f^{''}$ analysis.