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13. Law of the Mean · Increasing and Decreasing Functions

Adapted from: Classical single-variable calculus (relative extrema, Rolle, Mean Value Theorem / Law of the Mean, monotone from f′f'). Rewritten for teaching with interactive explorers — not a verbatim textbook dump.

This lesson ties geometry (horizontal tangents, parallel chords) to tests you use constantly: where can local max/min occur, and when is a function increasing or decreasing?


Relative maximum and minimum​

A function ff has a relative (local) maximum at x0x_0 if

f(x0)≥f(x)f(x_0)\ge f(x)

for every xx in some open interval around x0x_0 (where ff is defined). A relative minimum flips the inequality. Either is a relative extremum.

Local, not global

Relative extrema only compare nearby points. A local max need not be the tallest point on the whole graph.

Theorem — critical point​

If ff has a relative extremum at x0x_0 and f′(x0)f'(x_0) exists, then f′(x0)=0f'(x_0)=0.

So a differentiable local max/min has a horizontal tangent. (Nondifferentiable corners like ∣x∣|x| at 00 can still be extrema without a derivative.)

In Fig. 13-1, horizontal tangents appear at AA (relative maximum) and BB (relative minimum).

Fig. 13-1

Check: relative extrema​

Quiz

Relative extrema

1 / 2

If f is differentiable at a relative maximum x₀, then…

Pick an answer to see if you are right.


Rolle’s theorem​

Rolle. Let ff be continuous on [a,b][a,b] and differentiable on (a,b)(a,b), with f(a)=f(b)=0f(a)=f(b)=0. Then there is at least one x0∈(a,b)x_0\in(a,b) with f′(x0)=0f'(x_0)=0.

Geometrically: if the graph hits the xx-axis at both ends and is smooth in between, some interior tangent is horizontal (Fig. 13-2).

Fig. 13-2

Generalized Rolle. Same continuity/differentiability, but only g(a)=g(b)g(a)=g(b). Then g′(x0)=0g'(x_0)=0 for some interior point (Fig. 13-3). Proof idea: apply Rolle to f(x)=g(x)−g(a)f(x)=g(x)-g(a).

Fig. 13-3

Mean Value Theorem (Law of the Mean)​

MVT. If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), then some x0∈(a,b)x_0\in(a,b) satisfies

f(b)−f(a)b−a=f′(x0).\frac{f(b)-f(a)}{b-a}=f'(x_0).

The left side is the secant (chord) slope between P1=(a,f(a))P_1=(a,f(a)) and P2=(b,f(b))P_2=(b,f(b)). The right side is a tangent slope. So somewhere the tangent is parallel to P1P2P_1P_2 (Fig. 13-4).

Fig. 13-4

This is also called the Law of the Mean.

Extended / higher-order forms (preview)​

If ff and gg are continuous on [a,b][a,b], differentiable on (a,b)(a,b), and g′≠0g'\ne 0 on (a,b)(a,b), then for some x0x_0

f(b)−f(a)g(b)−g(a)=f′(x0)g′(x0).\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(x_0)}{g'(x_0)}.

(The ordinary MVT is g(x)=xg(x)=x.)

A higher-order law of the mean expands f(b)f(b) like a Taylor polynomial of degree n−1n-1 about aa, with remainder f(n)(x0)n!(b−a)n\dfrac{f^{(n)}(x_0)}{n!}(b-a)^n for some x0x_0 between aa and bb. Specializing b↦xb\mapsto x or a=0a=0 recovers the usual Taylor remainder forms.

Check: MVT​

Quiz

Mean Value Theorem

1 / 2

Geometrically, the conclusion of the MVT is that…

Pick an answer to see if you are right.

Hands On: Find an MVT point​


Increasing and decreasing functions​

On an interval, ff is increasing if u<vu<v implies f(u)<f(v)f(u)<f(v), and decreasing if u<vu<v implies f(u)>f(v)f(u)>f(v).

Theorem — sign of f′f'​

  • If f′>0f'>0 on an interval, then ff is increasing there.
  • If f′<0f'<0 on an interval, then ff is decreasing there.

Proof sketch: for u<vu<v, MVT gives f(v)−f(u)=f′(c)(v−u)f(v)-f(u)=f'(c)(v-u) with the same sign as f′(c)f'(c).

Check: monotone​

Quiz

Monotone from f′

1 / 1

If f′(x) = 2 + cos x, then on ℝ the function f is…

Pick an answer to see if you are right.

Hands On: Classify from f′​


Key takeaways
  • Differentiable relative extrema occur where f′=0f'=0 (horizontal tangent).
  • Rolle / generalized Rolle: equal endpoint values ⇒ an interior horizontal tangent.
  • MVT (Law of the Mean): average rate equals some instantaneous rate — tangent parallel to chord.
  • Sign of f′f' decides increasing vs decreasing on an interval.

Further Reading​