Part III · Differential Calculus and Local Approximation

Chapter 8 — Derivative Theorems and Function Behaviour

Why This Matters

The derivative becomes most useful when it reveals global behaviour: whether a function rises or falls, where a design reaches a maximum, and how a difference between two values is controlled by a slope in between. These conclusions require theorems with hypotheses, not pictures alone.

8.1 Rolle’s theorem and the Mean Value Theorem

If a smooth graph begins and ends at the same height, it must have a horizontal tangent somewhere in between. The endpoint and differentiability assumptions are essential; a sharp corner can defeat the conclusion.

The average slope of a secant line is therefore achieved by a tangent somewhere inside the interval. The MVT is the rigorous bridge from local derivative signs to interval-wide trends.

The conclusion comes after the hypotheses. Continuity on the closed interval and differentiability on its interior let the Mean Value Theorem guarantee an interior tangent parallel to the endpoint secant. Without the whole-interval conditions, the parallel tangent is not promised.

f'(c)=\frac{f(b)-f(a)}{b-a},\qquad c\in(a,b)

A smooth curve continuous on [a, b] and differentiable on (a, b) has a parallel tangent guaranteed by MVT.

8.2 Monotonicity

If $$f'(x)>0$$ throughout an interval, then $$f$$ is strictly increasing there; if $$f'(x)<0$$ , it is strictly decreasing. By the MVT,

\[f(y)-f(x)=f'(c)(y-x)\]

for some point $$c$$ between $$x$$ and $$y$$ . When $$y>x$$ , the sign of the difference is therefore the sign of $$f'(c)$$ .

Example — Sign chart

Let $$f(x)=x^3-3x$$ . Then

\[f'(x)=3x^2-3=3(x-1)(x+1).\]

The derivative is positive for $$x<-1$$ , negative for $$-1<x<1$$ , and positive for $$x>1$$ . Thus $$f$$ rises, falls, then rises. The derivative zeros identify changes to investigate; their signs explain the behaviour.

Derivative sign controls an interval through MVT. For y>x, the difference f(y)−f(x) has the sign of f′(c) at a point between them. Thus a positive derivative throughout gives increase and a negative derivative throughout gives decrease.

\[f(y)-f(x)=f'(c)(y-x)\]

A derivative sign chart translates f′ greater than zero and f′ less than zero into increasing and decreasing intervals.

8.3 Extrema and optimisation

An interior local maximum or minimum of a differentiable function must occur at a point where $$f'(c)=0$$ . Such points are critical points, but they are only candidates. A local extremum can also occur where the derivative does not exist, and an absolute extremum on a closed interval may occur at an endpoint.

Engineering Example — maximum power transfer

Let a source with internal resistance $$R_s>0$$ drive a load $$R>0$$ . Up to a positive constant, load power is

P(R)=\frac{R}{(R_s+R)^2}.

Differentiate:

P'(R)=\frac{(R_s+R)^2-2R(R_s+R)}{(R_s+R)^4} =\frac{R_s-R}{(R_s+R)^3}.

The denominator is positive. Thus $$P'(R)>0$$ for $$R<R_s$$ and $$P'(R)<0$$ for $$R>R_s$$ . Power rises then falls, so the maximum occurs at

\[R=R_s.\]

The conclusion comes from the sign change, not merely from setting a derivative to zero.

Critical points are candidates, not verdicts. On a closed interval, compare every valid interior critical point with both endpoints. A sign change from positive to negative identifies a local maximum, while the reverse identifies a local minimum; a zero derivative alone identifies neither.

P'(R)=\frac{R_s-R}{(R_s+R)^3},\qquad R=R_s\text{ gives the maximum}

List derivative-zero points and points where the derivative fails to exist.
Add both endpoints for a closed-interval absolute-extrema problem.
Compare function values or use a sign change before making the classification.

A derivative-sign chart marks rise then fall at an interior maximum while preserving endpoint candidates for absolute comparison.

8.4 L’Hôpital’s rule

The rule is not a licence to differentiate any expression in a limit. First verify the quotient form and its hypotheses. Products, differences, and powers must be transformed before the rule can be considered.

Example — repeated use with conditions checked

\lim_{x\to0}\frac{e^x-1-x}{x^2}.

Direct substitution gives $$0/0$$ . Differentiate numerator and denominator:

\lim_{x\to0}\frac{e^x-1}{2x},

which is again $$0/0$$ . Differentiate once more:

\lim_{x\to0}\frac{e^x}{2}=\frac12.

Taylor expansion in Chapter 9 will give the same result and explain the local shape behind it.

Differentiate a quotient only after its form and hypotheses qualify. L'Hôpital's rule begins with a 0/0 or ∞/∞ quotient, differentiability on a punctured interval, non-zero derivative denominator there, and a derivative-quotient limit. A product such as x sin(1/x) is not initially an eligible quotient.

\frac{e^x-1-x}{x^2}\xrightarrow{0/0}\frac{e^x-1}{2x}\xrightarrow{0/0}\frac{e^x}{2}\to\frac12

Check that the expression is an eligible quotient form.
Verify the punctured-neighbourhood derivative conditions.
Differentiate numerator and denominator together, then recheck the form before repeating.

L'Hôpital eligibility: 0/0 or ∞/∞ quotient plus the stated derivative hypotheses; x sin(1/x) is a non-example before rewriting.

Common Mistakes

Calling every derivative-zero point a maximum or minimum.
Forgetting interval endpoints in an absolute-extrema problem.
Applying MVT without checking continuity on the closed interval and differentiability on the open interval.
Applying L’Hôpital’s rule to a non-indeterminate quotient.

8.5 From derivative signs to reliable conclusions

Critical points are candidates, not automatic extrema. Make a sign chart by splitting the domain at derivative zeros and points where the derivative fails to exist; then test the sign on each interval. For absolute extrema on a closed interval, compare function values at every valid critical point and both endpoints.

Example — complete closed-interval extremum check

Find absolute extrema of $$f(x)=x^3-3x$$ on $$[-2,2]$$ . Since

\[f'(x)=3(x^2-1),\]

the interior candidates are $$x=-1,1$$ . Evaluate all required points:

f(-2)=-2,\quad f(-1)=2,\quad f(1)=-2,\quad f(2)=2.

Thus both $$2$$ and $$-2$$ occur as absolute extrema. Omitting endpoints would lose valid answers even though the derivative calculation itself is correct.

Example — when L'Hôpital may not start

For $\lim_{x\to0}x\sin(1/x)$ , direct substitution is not a quotient $$0/0$$ or $\infty/\infty$ , so L'Hôpital's rule is not the first method. Instead,

|x\sin(1/x)|\le|x|\to0,

so the squeeze theorem gives limit $$0$$ . Method selection begins with the form, not with a preferred rule.

[Figure: Secant and tangent on a smooth curve illustrating MVT; a sign chart below a function graph marking increasing/decreasing intervals and extrema.]

A sign chart becomes reliable when its domain is complete. Split at derivative zeros and derivative failures, then include both endpoints before comparing absolute values. This separates an interval-wide conclusion from the tempting but insufficient step of solving only f′=0.

f(-2)=-2,\ f(-1)=2,\ f(1)=-2,\ f(2)=2

The closed-interval check displays derivative-sign intervals together with endpoint values that can also be absolute extrema.

8.6 Theorems give conclusions only under their hypotheses

The Mean Value Theorem is a guarantee about a smooth unbroken graph on a whole interval. It does not say every secant slope occurs as a tangent slope if the graph jumps, has a corner, or is undefined at an endpoint. Derivative signs use the same hypothesis through the MVT: a positive derivative on an interval gives increase only when the function is differentiable throughout its interior.

Example — MVT applied with every condition visible

For $$f(x)=x^2$$ on $$[1,4]$$ , polynomials are continuous and differentiable everywhere. The secant slope is

\frac{f(4)-f(1)}{4-1}=\frac{16-1}{3}=5.

Set $$f'(c)=2c=5$$ , giving $$c=5/2$$ , which lies in $$(1,4)$$ as required. The theorem does not identify a unique point in general; it guarantees at least one.

Example — a hypothesis failure matters

Let $$f(x)=|x|$$ on $$[-1,1]$$ . Its endpoint values agree, and it is continuous, but it is not differentiable at $$0$$ . Rolle's theorem cannot be applied. Indeed, $$f'(x)$$ is $$-1$$ or $$1$$ where defined and is never $$0$$ . This is a counterexample to the incorrect claim that equal endpoint values alone force a horizontal tangent.

Example — L'Hôpital after form, not before

Evaluate $\lim_{x\to0}(1-\cos x)/x^2$ . It has the allowed $$0/0$$ quotient form. One application gives $\sin x/(2x)$ , still $$0/0$$ ; a second gives $\cos x/2\to1/2$ . By contrast, $\lim_{x\to0}(1-\cos x)/x$ may use the same rule once, but an expression such as $x\ln x$ is not initially a quotient and must be rewritten or handled by another argument.

[Figure: A smooth MVT curve whose secant is parallel to a tangent; beside it,

\(|x|\)

\([-1,1]\)

showing the corner that blocks Rolle's theorem.]

Equal endpoint values alone do not give Rolle's conclusion. The graph of |x| on [−1,1] is continuous but has a corner at zero, so differentiability fails and no horizontal tangent is guaranteed. The theorem's conditions are working assumptions, not formal decoration.

f(-1)=f(1),\qquad f'(x)=-1\text{ or }1\text{ for }x\ne0

The MVT contrast keeps the smooth parallel-tangent case separate from the absolute-value corner that blocks Rolle's theorem.

Exercises

Verify the hypotheses of the MVT for $$f(x)=x^2$$ on $$[1,4]$$ and find the point(s) $$c$$ it guarantees.
Find and classify the critical points of $$x^3-6x^2+9x$$ .
Optimise the area of a rectangle with perimeter $$40$$ m.
Evaluate $\lim_{x\to0}(\sin x-x)/x^3$ using L’Hôpital’s rule.
Explain why the result of Exercise 4 cannot be obtained by simply replacing $\sin x$ with $$x$$ before examining the error.
Find absolute extrema of $$x^3-3x$$ on $$[-2,2]$$ .
Evaluate $\lim_{x\to0}x\sin(1/x)$ without L'Hôpital's rule.
Apply MVT to $$x^2$$ on $$[1,4]$$ and identify every hypothesis used.
Explain precisely why Rolle's theorem cannot be used for $$|x|$$ on $$[-1,1]$$ .
Evaluate $(1-\cos x)/x^2$ as $x\to0$ using L'Hôpital's rule and state why it is permitted.