1. Marginal Analysis, Optimization, and Numerical Roots

Learning goals and preparation

This chapter assumes the algebra, functions, logarithms, and rudiments of calculus appropriate to a preparatory mathematics course. We review the derivative's meaning before applying it to marginal quantities, continuous optimization, and numerical equations. You will learn to distinguish local tests from global conclusions, continuous optima from feasible integer decisions, and a numerical estimate from a verified error bound.

1.1 A derivative is a rate, not a finite difference

At an interior point $a$, the derivative is

$$f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}h,$$

if this finite limit exists. Differentiability is equivalently a local expansion

$$f(a+h)=f(a)+f'(a)h+r(h),\qquad r(h)/h\to0.$$

The term $f'(a)h$ approximates a finite change; it does not usually equal it. If $C(q)$ is cost, then $C'(q)$ is marginal cost, measured in cost units per quantity unit. The average cost $C(q)/q$, defined for $q>0$, is a different function.

Worked example 1: marginal versus next-unit cost. Let $C(q)=100+5q+0.2q^2$. Then $C'(q)=5+0.4q$. At $q=10$, marginal cost is $9$, while the exact cost of increasing output from $10$ to $11$ is

$$C(11)-C(10)=9.2.$$

The extra $0.2$ is the quadratic remainder for a unit change. For a smaller change $h$, the exact difference is $(5+0.4q)h+0.2h^2$.

We use the familiar derivative rules: $(x^n)'=nx^{n-1}$ on a suitable domain; $(e^x)'=e^x$; $(\ln x)'=1/x$ for $x>0$; the sum, product, quotient and chain rules. Differentiation must retain the original domain. For instance, differentiating a logarithmic expression does not make its nonpositive arguments admissible.

1.2 Monotonicity and the mean value theorem

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

It follows that $f'>0$ throughout an interval implies strict increase, and $f'<0$ implies strict decrease. A bound $|f'|\le M$ on a segment gives $|f(b)-f(a)|\le M|b-a|$. These are interval conclusions from local derivative information; the theorem cannot be applied across a missing domain point.

If $f'=0$ throughout an interval, $f$ is constant there.

This is a necessary condition, not a sufficient one. The function $x^3$ has derivative zero at zero but no local maximum or minimum there. Nondifferentiable points and endpoints must also be considered; neither is covered by Fermat's interior differentiable condition.

1.3 Finding and verifying optima

To find them when there are finitely many candidate points:

1. find interior points where $f'=0$ or the derivative fails to exist;
2. include both endpoints;
3. evaluate and compare all candidate values.

A first-derivative sign change from positive to negative proves a local maximum; negative to positive proves a local minimum. If $f$ has two continuous derivatives near a stationary point, $f''(a)>0$ proves a strict local minimum and $f''(a)<0$ a strict local maximum. The condition $f''(a)=0$ is inconclusive. For example, $x^4$, $-x^4$, and $x^3$ have different extremum behaviour despite the same zero first and second derivatives at zero.

Worked example 2: constrained profit. Suppose price is $p(q)=60-q$ and cost is $C(q)=100+10q+q^2$, with $0\le q\le20$. Profit is

$$\Pi(q)=qp(q)-C(q)=50q-2q^2-100.$$

Its derivative is $50-4q$, positive below $12.5$ and negative above. The stationary point is feasible, and $\Pi(12.5)=212.5$, while $\Pi(0)=-100$ and $\Pi(20)=100$. Thus the global maximum on this interval is $212.5$ at $q=12.5$. If capacity restricts $q\le10$, profit is increasing throughout the feasible interval and the maximum is instead $\Pi(10)=200$.

If quantities must be integers, the continuous optimum need not be feasible. Here completing the square gives $\Pi(q)=212.5-2(q-12.5)^2$, so both $q=12$ and $q=13$ maximize integer profit, with value $212$. Rounding is justified here by the exact quadratic shape, not by a universal rule that every continuous optimum should be rounded.

1.4 Elasticity makes a relative comparison

For positive differentiable demand $q=q(p)$ at positive price $p$, define the signed relative response $p q'(p)/q(p)$. When demand decreases, it is customary to report the nonnegative price elasticity

$$E(p)=-\frac{p q'(p)}{q(p)}.$$

For a small price change, $\Delta q/q\approx-E\,\Delta p/p$. This is a first-order statement. It does not predict exact finite percentage changes.

Revenue as a function of price is $R(p)=pq(p)$, so

$$R'(p)=q(p)+pq'(p)=q(p)(1-E(p)).$$

Under the stated positivity assumptions, revenue increases locally with price if $E<1$ and decreases locally if $E>1$.

Worked example 3: a power demand model. Let $q(p)=100p^{-2}$ for $p>0$. Then $q'=-200p^{-3}$, so $E=2$ at every price. Revenue is $100/p$, which decreases with price, exactly as the elasticity test predicts.

1.5 Bisection: an estimate with a bracket

Let real continuous $f$ satisfy $f(a)f(b)<0$ on $[a,b]$. The intermediate value theorem guarantees at least one root. Bisection evaluates the midpoint, keeps the half retaining the sign change, and repeats. After $n$ halvings the interval width is $(b-a)/2^n$; its midpoint is within $(b-a)/2^{n+1}$ of some root in that interval. Uniqueness requires an additional assumption such as strict monotonicity.

For $m=(a+b)/2$, if $f(m)=0$, stop.

Worked example 4: begin a bracketed calculation. For $f(x)=x^3+x-1$, $f(0)=-1$ and $f(1)=1$. The derivative $3x^2+1>0$ proves a unique root. At $0.5$, $f=-0.375$, so the first bracket is $[0.5,1]$. At $0.75$, $f=0.171875$, giving $[0.5,0.75]$. Its midpoint $0.625$ is already within $0.125$ of the root. Further bisections reduce the guaranteed error whether or not the exact root has a convenient formula.

The sign calculation must be reliable. With approximate arithmetic, an uncertain value near zero is not a certified positive or negative sign; increase precision or use an interval bound before claiming a rigorous bracket.

1.6 Newton's method: a fast proposal, not a guarantee

At an iterate $x_n$ where $f'(x_n)\ne0$, the tangent line is

$$y=f(x_n)+f'(x_n)(x-x_n).$$

Its zero defines Newton's proposal

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$$

This often converges rapidly near a simple root, but may fail with a poor starting point, a zero derivative, or an iterate outside the domain. A small step alone does not prove closeness to a root. Preserve a bracket or establish an error bound where a reliable numerical conclusion is required.

Worked example 5: square root and a residual bound. For $f(x)=x^2-2$ starting at $x_0=3/2$,

$$x_1=\frac12\left(\frac32+\frac{2}{3/2}\right)=\frac{17}{12}.$$

A second step gives $x_2=577/408$. The root $\alpha=\sqrt2$ and the iterate lie in $[1,2]$, where $f'(x)=2x\ge2$. By the mean value theorem,

$$|x_n-\alpha|\le\frac{|f(x_n)|}{2}.$$

This converts a residual into an error guarantee under a verified lower derivative bound. In general, $|f(x_n)|$ by itself is not an error estimate: multiplying the whole equation by a tiny constant changes residuals without moving its roots.

1.7 Graded exercises

Foundation

1. If $C(q)=80+4q+q^2/2$, find marginal and average cost for $q>0$. Find the exact change from $q$ to $q+h$.
2. Differentiate $f(x)=\ln(1+x)+x e^{-x}$ and state its real domain.
3. Find the stationary points of $f(x)=x^3-3x$ and classify them by derivative signs.

Core

4. Maximize $\Pi(q)=24q-q^2-20$ on $[0,10]$. Explain why the unconstrained stationary point does not solve this constrained problem.
5. For $q(p)=60-2p$ on $0<p<30$, calculate $E(p)$ and find where revenue is maximized.
6. Give an example of a differentiable function with $f'(0)=0$ but no local extremum at zero. Justify it.
7. For a root bracket of initial length $1$, how many bisections ensure that the midpoint error is at most $10^{-3}$?

Synthesis

8. Apply two Newton steps to $f(x)=x^2-3$ from $x_0=2$. Use a derivative bound on $[1,2]$ to give an exact error bound for the second iterate.
9. Newton's method for $f(x)=x^3-2x+2$ starts at $x_0=0$. Calculate $x_1$ and $x_2$. What does this reveal about an unconditional convergence claim?
10. Suppose $f$ is continuous on $[a,b]$, differentiable inside, has a root $\alpha$ there, and $f'(x)\ge m>0$ throughout the interior. Prove uniqueness of the root and show $|x-\alpha|\le|f(x)|/m$ for every $x\in[a,b]$.

1.8 Full solutions

1. $C'=4+q$, while $C/q=80/q+4+q/2$. Expansion gives $C(q+h)-C(q)=(4+q)h+h^2/2$. The finite difference includes a second-order term absent from the marginal prediction.
2. The domain is $x>-1$. The derivative is $1/(1+x)+e^{-x}-xe^{-x}=1/(1+x)+(1-x)e^{-x}$.
3. $f'=3(x-1)(x+1)$. It is positive for $x<-1$, negative for $-1<x<1$, and positive for $x>1$. Thus $-1$ is a local maximum and $1$ a local minimum.
4. $\Pi'=24-2q>0$ on $[0,10]$. The maximum is at $q=10$, with value $240-100-20=120$. The unconstrained stationary point $12$ lies outside the feasible interval.
5. $E=2p/(60-2p)=p/(30-p)$. Revenue is $R=60p-2p^2=450-2(p-15)^2$, so it is maximized at $p=15$, where $E=1$. The price lies in the stated domain.
6. Use $f(x)=x^3$. Its derivative is $3x^2$, giving $f'(0)=0$, but arbitrarily close negative inputs have negative values and positive inputs have positive values. Zero is neither a local maximum nor a local minimum.
7. After $n$ bisections, midpoint error is at most $1/2^{n+1}$. Since $2^9=512<1000$ and $2^{10}=1024\ge1000$, choose $n=9$. Eight bisections would only give the larger bound $1/512$.
8. The iteration is $x_{n+1}=(x_n+3/x_n)/2$. Thus $x_1=7/4$ and $x_2=97/56$. Both the root and $x_2$ lie in $[1,2]$, where $f'\ge2$. Its residual is $(97/56)^2-3=(9409-9408)/3136=1/3136$, so its error is at most $1/6272$.
9. At zero, $f=2$ and $f'=-2$, so $x_1=1$. At one, $f=1$ and $f'=1$, so $x_2=0$. The iteration cycles between $0$ and $1$ instead of converging. A nonzero derivative at each step does not ensure convergence.
10. The mean value theorem gives $f(v)-f(u)\ge m(v-u)>0$ for $u<v$, so $f$ is strictly increasing and cannot have two roots. For $x\ne\alpha$, the same theorem on the interval between them yields $f(x)-f(\alpha)=f'(c)(x-\alpha)$ for some interior $c$. Taking absolute values and using $f(\alpha)=0$ gives $|f(x)|\ge m|x-\alpha|$. At $x=\alpha$ the claimed inequality is immediate.

Summary

Marginal quantities are derivatives, not exact next-unit changes. An optimum requires domain and endpoint checks; a stationary point alone is insufficient. Elasticity compares relative rates and depends on an explicit sign convention. Bisection maintains a root bracket and a transparent error bound. Newton's method provides useful iterates but needs safeguards or analysis before convergence and accuracy can be claimed.