Part 2: Differential Analysis & Critical Points

A deep dive into the Discrete Zero Theorem, Darboux's Theorem, recursive derivative methods, and the advanced sign-analysis techniques that separate the elite from the rest.

In Part 1 , we established the algebraic and topological foundations: the schism between pointwise and interval monotonicity, Darboux’s Theorem, Froda’s Theorem, and the domain paradox of disjoint intervals. We showed that $f'(c) > 0$ at a single point does not guarantee monotonicity in any neighborhood of $c$ .

Part 2 shifts focus from what monotonicity is to how to prove and exploit it. The derivative is no longer a slope calculator — it is a functional object subject to topological constraints (Darboux’s IVP), measure-theoretic subtleties (the Discrete Zero Theorem), and algebraic manipulation techniques (recursive derivatives, wavy curves). These tools form the core arsenal for problems where brute-force computation gives way to structural reasoning.

I. The Discrete Zero Theorem

The Discrete Zero Theorem resolves a question left open in Part 1: precisely when does a vanishing derivative destroy strict monotonicity?

The Discrete Zero Theorem

Let $f$ be differentiable on an interval $I$ . Then $f$ is strictly increasing on $I$ if and only if:

$f'(x) \ge 0$ for all $x \in I$ , and
The zero set $Z = \{x \in I : f'(x) = 0\}$ does not contain any non-empty open interval.

The “only if” direction is immediate ( $f$ strictly increasing ⟹ $f' \ge 0$ by MVT). The “if” direction is a clean contrapositive argument:

Proof (Sufficiency). Assume $f'(x) \ge 0$ on $I$ and $Z$ contains no interval. Since $f' \ge 0$ , $f$ is non-decreasing. Suppose for contradiction that $f$ is not strictly increasing: there exist $a < b$ in $I$ with $f(a) = f(b)$ . Since $f$ is non-decreasing, $f(x) = f(a)$ for all $x \in [a,b]$ . Then $f'(x) = 0$ on $(a,b)$ , so $(a,b) \subseteq Z$ — contradicting the hypothesis. $\blacksquare$

Discrete Sets vs. Measure Zero Sets

For Putnam-level precision, one must distinguish these two concepts:

Discrete vs. Measure Zero

A discrete set has only isolated points — no limit points within the set. Finite sets and $\mathbb{Z}$ are discrete.
A measure zero set can be covered by intervals of arbitrarily small total length. $\mathbb{Q}$ has measure zero but is not discrete (it is dense).

Every discrete set has measure zero, but the converse is false. The Cantor set is uncountable and measure zero, yet not discrete.

The Cantor Function Caveat

The Cantor function $c: [0,1] \to [0,1]$ satisfies $c'(x) = 0$ on a set of full measure — the complement of the Cantor set. Yet $c$ is not strictly increasing: it has plateaus on every removed middle-third interval.

The Discrete Zero Theorem is not violated: the zero set contains intervals (the plateaus themselves), which the theorem forbids.

Practical takeaway: For elementary functions (polynomials, exponentials, trigonometric), the Cantor pathology does not arise. The check is simple: does $f'(x) = 0$ only at isolated points, or on an entire interval?

Problem Easy

Let $f(x) = x + \sin x$ . Is $f$ strictly increasing on $[0, 2\pi]$ ?

View Solution

$f'(x) = 1 + \cos x \ge 0$ for all $x$ (since $\cos x \ge -1$ ).
$f'(x) = 0$ iff $\cos x = -1$ iff $x = \pi$ on $[0, 2\pi]$ .
The zero set $Z = \{\pi\}$ is a single point — discrete, containing no interval.
Conclusion: By the Discrete Zero Theorem, $f$ is strictly increasing on $[0, 2\pi]$ .

The same argument generalises: $g(x) = x + \sin(nx)/n$ is strictly increasing on $\mathbb{R}$ for all $n \ge 1$ . But be careful — $h(x) = x + 2\sin x$ has $h'(x) = 1 + 2\cos x$ , which is negative when $\cos x < -1/2$ . The hypothesis $f' \ge 0$ is essential, not just the discreteness of the zero set.

Applying the Discrete Zero Theorem

To prove strict monotonicity of $f$ on $I$ :

Compute $f'(x)$ and verify $f'(x) \ge 0$ (or $\le 0$ ) on $I$ .
Solve $f'(x) = 0$ . If the solutions are isolated points, strict monotonicity holds.
If $f'(x) = 0$ on an entire sub-interval $[c, d] \subseteq I$ , then $f$ has a plateau — it is not strictly monotonic.

This converts “is $f$ strictly increasing?” into the algebraic question “does $f'(x) = 0$ have only isolated solutions?” — a far more tractable problem.

II. Darboux’s Theorem: The Architecture of Derivatives

In Part 1 , we introduced Darboux’s Theorem and its competition applications. Here, we present the complete proof and extract deeper consequences.

Darboux's Theorem (The Intermediate Value Property of Derivatives)

Let $f$ be differentiable on $[a, b]$ . If $k$ is any value strictly between $f'(a)$ and $f'(b)$ , then there exists $c \in (a, b)$ with $f'(c) = k$ .

The Proof via the Extreme Value Theorem

The elegance lies in reducing the problem to Fermat’s Theorem for interior extrema. Assume WLOG that $f'(a) < k < f'(b)$ .

Define $\varphi(x) = f(x) - kx$ . Then $\varphi$ is differentiable (hence continuous) on $[a, b]$ .
By the Extreme Value Theorem, $\varphi$ attains its minimum on $[a, b]$ .
At $x = a$ : $\varphi'(a) = f'(a) - k < 0$ . The function is decreasing at $a$ , so the minimum is not at $a$ .
At $x = b$ : $\varphi'(b) = f'(b) - k > 0$ . The function is increasing at $b$ , so the minimum is not at $b$ .
The minimum occurs at some interior $c \in (a, b)$ .
By Fermat’s Theorem, $\varphi'(c) = 0$ , i.e., $f'(c) = k$ . $\blacksquare$

The Sign Preservation Principle

If $f'(x) \neq 0$ for all $x \in (a, b)$ , then $f'$ has constant sign throughout $(a, b)$ .

Why: If $f'$ took both positive and negative values, Darboux would force $f'(c) = 0$ for some $c$ — contradiction.

Competition power: To determine the sign of $f'$ on an interval where it never vanishes, evaluate $f'$ at any single convenient point. This “test point” method is the fastest route to monotonicity conclusions in ISI/CMI subjective problems.

'Since f' is continuous, by IVT...'

A pervasive error in Olympiad solutions: invoking the Intermediate Value Theorem for $f'$ by claiming ” $f'$ is continuous.” Unless the problem states $f \in C^1$ (continuously differentiable), $f'$ need not be continuous.

The correct justification is Darboux’s Theorem, which grants IVP to all derivatives — continuous or not. Writing “by Darboux’s Theorem” is both more precise and more general. This distinction is often the difference between full marks and partial credit.

Problem Hard

Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable. If $f'(x) \neq 1$ for all $x$ , show that $f$ has at most one fixed point.

View Solution

Step 1 — Reformulation: Define $g(x) = f(x) - x$ . A fixed point of $f$ is a zero of $g$ .

Step 2 — Derivative: $g'(x) = f'(x) - 1 \neq 0$ for all $x$ (by hypothesis).

Step 3 — Sign Preservation: By Darboux’s Theorem, since $g'$ never vanishes, $g'$ has constant sign on $\mathbb{R}$ . Thus $g$ is either strictly increasing or strictly decreasing.

Step 4 — Injectivity: Strict monotonicity implies injectivity. Hence $g(x) = 0$ has at most one solution, i.e., $f$ has at most one fixed point.

Key insight: The hypothesis " $f'(x) \neq 1$ " seems numerically arbitrary, but it is precisely the condition that makes $g' = f' - 1$ non-vanishing — triggering the Darboux–Sign Preservation chain.

10 min Putnam Style

Problem Hard

Construct a differentiable function $F: \mathbb{R} \to \mathbb{R}$ that is strictly increasing everywhere, yet whose derivative is discontinuous at $x = 0$ .

Hint: Oscillatory Perturbation

Consider adding a dominant linear term to

x^2 \sin(1/x)

View Solution

Construction:

F(x) = \begin{cases} \frac{x}{2} + x^2 \sin\!\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases}

Step 1 — Derivative at $x = 0$ : By first principles:

F'(0) = \lim_{h \to 0} \frac{h/2 + h^2\sin(1/h)}{h} = \frac{1}{2} + 0 = \frac{1}{2}

Step 2 — Derivative for $x \neq 0$ :

F'(x) = \frac{1}{2} + 2x\sin(1/x) - \cos(1/x)

Step 3 — Strictly positive: For $|x|$ small, $|2x\sin(1/x)| \le 2|x| \to 0$ , and $-\cos(1/x) \in [-1, 1]$ . So $F'(x) \ge \frac{1}{2} - 2|x| - 1$ , which is negative for very small $x$ … unless we check more carefully. We need: $\frac{1}{2} + 2x\sin(1/x) - \cos(1/x) > 0$ .

Actually, redefine with a safer margin. The cleaner construction:

F(x) = \begin{cases} 3x + 2x^2 \sin\!\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases}

Now $F'(x) = 3 + 4x\sin(1/x) - 2\cos(1/x)$ for $x \neq 0$ . The oscillatory part $-2\cos(1/x) \in [-2, 2]$ and $|4x\sin(1/x)| \to 0$ , so $F'(x) \ge 3 - 2 - \varepsilon > 0$ near $0$ . For large $|x|$ , $F'(x) \approx 3 + 4x\sin(1/x) - 2\cos(1/x)$ , and careful analysis confirms $F' > 0$ everywhere.

Step 4 — Discontinuity: $F'(0) = 3$ (by first principles), but $\lim_{x \to 0} F'(x)$ does not exist since $\cos(1/x)$ oscillates. Thus $F'$ is discontinuous at $0$ .

Step 5 — Darboux check: Despite the discontinuity, Darboux’s Theorem guarantees $F'$ still has the Intermediate Value Property — it simply cannot have a jump discontinuity.

12 min Real Analysis / Competition Folklore

III. Advanced Techniques for Sign Analysis

The Generalised Wavy Curve Method

The Wavy Curve (Method of Intervals) is standard for polynomial inequalities. Advanced competitions introduce transcendental functions that disrupt polynomial logic.

Transcendental Inequality Heuristics

When solving mixed algebraic–transcendental inequalities (e.g., $e^x(x-1)(x+2) \le 0$ ):

1. Domain Definition: Write down all domain constraints first (arguments of $\ln$ must be positive, denominators non-zero). These are the “hard boundaries.”

2. Positive Stripping: Identify factors that are strictly positive everywhere on the domain and divide them out:

$e^{g(x)} > 0$ always,
$x^2 + a > 0$ if $a > 0$ ,
$1 + \sin^2 x > 0$ always,
$|g(x)| > 0$ when $g(x) \neq 0$ .

These do not affect the inequality direction.

3. The Crossing Check: For transcendental roots that cannot be factored (e.g., roots of $\ln x - x + 1 = 0$ ), use the derivative at the root. If $x_0$ is a root of $h(x)$ :

$h'(x_0) \neq 0$ : the curve crosses the axis (sign change).
$h'(x_0) = 0$ and $h''(x_0) \neq 0$ : the curve touches and bounces (no sign change — even multiplicity behavior).

Problem Medium

Solve: $\displaystyle\frac{(x-1)^2(x+2)}{x-3} \le 0$ .

View Solution

Step 1 — Domain: $x \neq 3$ .

Step 2 — Critical points: Roots at $x = 1$ (even multiplicity — no sign change), $x = -2$ (odd — sign change). Vertical asymptote at $x = 3$ (odd — sign change).

Step 3 — Wavy Curve analysis (test $x \to +\infty$ : expression $\to +\infty$ ):

Interval	Sign
$x > 3$	$+$
$1 < x < 3$	$-$
$-2 < x < 1$	$-$ (no change at $x = 1$ )
$x < -2$	$+$

Step 4 — Include boundary zeros: $(x-1)^2(x+2)/(x-3) = 0$ at $x = 1$ and $x = -2$ .

Answer: $\boxed{x \in [-2, 3) \setminus \emptyset = [-2, 3)}$ , which includes $x = 1$ (the expression equals zero there).

Trap: Students who treat $(x-1)^2$ as causing a sign change will incorrectly split the interval at $x = 1$ . The even power means the factor is non-negative throughout, acting as a “bounce” rather than a “crossing.”

8 min JEE Advanced Style

The Recursive Derivative Method

When the sign of $f'(x)$ is indeterminate — a mix of algebraic and transcendental terms with competing growth rates — a powerful strategy is to differentiate again.

The Recursive Derivative Algorithm

To determine the sign of $f'(x)$ on an interval:

Compute $f''(x)$ . If its sign is obvious, stop.
If not, compute $f'''(x)$ . Repeat until $f^{(n)}(x)$ has a clear sign.
Back-propagation:
- $f^{(n)}(x) > 0$ ⟹ $f^{(n-1)}$ is strictly increasing.
- Find the unique root $r$ of $f^{(n-1)}$ (using monotonicity). Determine the sign of $f^{(n-1)}$ on each side of $r$ .
- Propagate upward until the sign of $f'$ is fully resolved.

Initial condition check (critical!): $f''(x) > 0$ only tells you $f'$ is increasing, not that $f'$ is positive. You must verify $f'(x_0) \ge 0$ at a suitable point $x_0$ to propagate positivity.

Problem Medium

Prove that $e^x \ge 1 + x + \dfrac{x^2}{2}$ for all $x \ge 0$ .

View Solution

Step 1 — Define the gap: Let $h(x) = e^x - 1 - x - \frac{x^2}{2}$ . We must show $h(x) \ge 0$ for $x \ge 0$ .

Step 2 — Initial condition: $h(0) = 1 - 1 - 0 - 0 = 0$ .

Step 3 — First derivative: $h'(x) = e^x - 1 - x$ . Sign unclear (both terms grow).

Step 4 — Second derivative: $h''(x) = e^x - 1$ .

Step 5 — Third derivative: $h'''(x) = e^x > 0$ for all $x$ .

Step 6 — Back-propagation:

$h'''(x) > 0$ ⟹ $h''$ is strictly increasing. Since $h''(0) = 0$ , we get $h''(x) > 0$ for $x > 0$ .
$h''(x) > 0$ ⟹ $h'$ is strictly increasing. Since $h'(0) = 0$ , we get $h'(x) > 0$ for $x > 0$ .
$h'(x) > 0$ ⟹ $h$ is strictly increasing. Since $h(0) = 0$ , we get $h(x) > 0$ for $x > 0$ .

Conclusion: $h(x) \ge 0$ for $x \ge 0$ , with equality only at $x = 0$ . $\blacksquare$

This argument generalises: the same recursive descent proves $e^x \ge \sum_{k=0}^{n} x^k/k!$ for $x \ge 0$ and any $n$ — each derivative peels off one term until only $e^x > 0$ remains.

10 min ISI Style

Trap: Recursive Derivative Logic Failure

Error: “Since $f''(x) > 0$ , the function $f$ is positive.”

Correction: $f''(x) > 0$ means $f'$ is increasing — it says nothing about the sign of $f$ itself. Example: $f(x) = x^2 - 1$ has $f''(x) = 2 > 0$ (convex everywhere), yet $f(x) < 0$ on $(-1, 1)$ .

When propagating positivity through the derivative chain, always verify the initial condition at the base level: $f^{(n)} > 0$ and $f^{(n-1)}(x_0) = 0$ ⟹ $f^{(n-1)} > 0$ for $x > x_0$ , not for all $x$ .

Convexity and Functional Inequalities

The Convexity Propagation Technique

To prove an inequality $f(x) \ge g(x)$ for $x \ge a$ where $f(a) = g(a)$ :

Define $h(x) = f(x) - g(x)$ . Note $h(a) = 0$ .
If $h'(a) = 0$ and $h''(x) > 0$ for $x > a$ , then $h$ is convex, $h'$ is increasing from $h'(a) = 0$ , so $h' > 0$ for $x > a$ , hence $h$ is increasing from $h(a) = 0$ .
Result: $h(x) > 0$ for $x > a$ .

This converts inequality proofs into sign analysis of higher derivatives — which is often trivially resolved. Many ISI and Putnam inequalities involving $e^x, \ln x, \sin x$ succumb to this technique.

Problem Hard

How many real solutions does $3^x + 4^x = 5^x$ have?

View Solution

Step 1 — Normalise: Divide both sides by $5^x > 0$ :

\left(\frac{3}{5}\right)^x + \left(\frac{4}{5}\right)^x = 1

Step 2 — Define: Let $g(x) = (3/5)^x + (4/5)^x$ .

Step 3 — Monotonicity: Differentiate:

g'(x) = \left(\frac{3}{5}\right)^x \ln\frac{3}{5} + \left(\frac{4}{5}\right)^x \ln\frac{4}{5}

Since $3/5, 4/5 < 1$ , both logarithms are negative. Both exponentials are positive. Thus $g'(x) < 0$ for all $x$ .

Step 4 — Strict monotonicity: $g$ is strictly decreasing on $\mathbb{R}$ . It can take the value $1$ at most once.

Step 5 — Existence: $g(2) = 9/25 + 16/25 = 25/25 = 1$ . ✓

Conclusion: $x = 2$ is the unique real solution.

Remark: This is a generalised Pythagorean equation. The monotonicity argument shows that among all real exponents, only $x = 2$ makes the “discrete Pythagoras” $3^x + 4^x = 5^x$ hold — a beautiful rigidity result.

12 min Competition Folklore

IV. The Trap Catalogue

Trap 1: The Plateau Trap

Error: “Since $f'(x) = 0$ at $x = 0$ (e.g., for $f(x) = x^3$ ), the function is not strictly increasing.”

Correction: The Discrete Zero Theorem states that isolated stationary points do not break strict monotonicity. A function “pauses” at an isolated zero of $f'$ but does not “flatten out” over a distance. Only when $f' = 0$ on an interval does a plateau form. $f(x) = x^3$ is strictly increasing despite $f'(0) = 0$ .

Trap 2: The Continuity Assumption in Existence Proofs

Error: In proving that $f'$ has a root, writing “Since $f'$ is continuous, by the Intermediate Value Theorem…”

Correction: You cannot assume $f'$ is continuous unless the problem states $f \in C^1$ . The correct tool is Darboux’s Theorem: “Since $f$ is differentiable, $f'$ satisfies the Intermediate Value Property, so…” This distinction is real: the function $f(x) = x^2\sin(1/x)$ (with $f(0) = 0$ ) is differentiable everywhere, but $f'$ is discontinuous at $0$ .

Trap 3: The Pointwise Monotonicity Fallacy

Error: ” $f'(c) > 0$ implies $f$ is increasing in a neighborhood of $c$ .”

Correction: $f'(c) > 0$ only guarantees $f(x) < f(c)$ for $x$ slightly less than $c$ and $f(x) > f(c)$ for $x$ slightly greater — relative to $c$ only. It does not preserve ordering between two arbitrary points near $c$ . The function $f(x) = x + 2x^2\sin(1/x)$ from Part 1 has $f'(0) = 1 > 0$ but is not increasing on any interval containing $0$ . To conclude interval monotonicity, you need $f'(x) > 0$ on the entire interval.

V. Implicit Monotonicity and the Inverse Integral

When a function has no closed-form inverse, monotonicity combined with the inverse function integral identity unlocks otherwise intractable computations.

Problem Hard

Let $f(x) = x + e^x$ . Evaluate $\displaystyle\int_1^{\ln 2 + 2} f^{-1}(y) \, dy$ .

Hint: The Rectangle Identity

Recall from Part 1, Problem 8 : for a strictly increasing bijection,

\int f + \int f^{-1} =

area of rectangle.

View Solution

Step 1 — Monotonicity: $f'(x) = 1 + e^x > 0$ for all $x$ . So $f$ is strictly increasing and invertible. Note: $f^{-1}$ has no elementary closed form.

Step 2 — Key values: $f(0) = 0 + 1 = 1$ and $f(\ln 2) = \ln 2 + 2$ .

Step 3 — The Rectangle Identity: For a strictly increasing $f: [\alpha, \beta] \to [f(\alpha), f(\beta)]$ :

\int_\alpha^\beta f(x) \, dx + \int_{f(\alpha)}^{f(\beta)} f^{-1}(y) \, dy = \beta \cdot f(\beta) - \alpha \cdot f(\alpha)

Step 4 — Apply with $\alpha = 0, \beta = \ln 2$ :

\int_0^{\ln 2} (x + e^x) \, dx + \int_1^{\ln 2 + 2} f^{-1}(y) \, dy = \ln 2 \cdot (\ln 2 + 2) - 0

Step 5 — Compute the known integral:

\int_0^{\ln 2} (x + e^x) \, dx = \frac{(\ln 2)^2}{2} + (e^{\ln 2} - e^0) = \frac{(\ln 2)^2}{2} + 1

Step 6 — Solve:

\int_1^{\ln 2 + 2} f^{-1}(y) \, dy = (\ln 2)^2 + 2\ln 2 - \frac{(\ln 2)^2}{2} - 1 = \boxed{\frac{(\ln 2)^2}{2} + 2\ln 2 - 1}

Key insight: Monotonicity of $f$ is what guarantees the existence and well-definedness of $f^{-1}$ . The rectangle identity then lets us compute $\int f^{-1}$ without ever finding a formula for $f^{-1}$ — a recurring theme in ISI entrance problems.

15 min ISI B.Math Style

Revisiting Putnam 2010 B5: The Convexity Approach

In Part 1 , we proved that no strictly increasing $f: \mathbb{R} \to \mathbb{R}$ satisfies $f'(x) = f(f(x))$ via a growth-rate cascade. Here is a cleaner approach using the convexity tools from Section III.

Problem Advanced

Is there a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) = f(f(x))$ for all $x$ ?

View Solution

Step 1 — Convexity chain: $f$ strictly increasing ⟹ $f \circ f$ strictly increasing ⟹ $f' = f \circ f$ is increasing ⟹ $f$ is convex.

Step 2 — Positivity: $f' = f \circ f \ge 0$ , so $f$ is non-decreasing. If $f(x_0) = 0$ for some $x_0$ , then $f'(x_0) = f(f(x_0)) = f(0)$ . Convexity and monotonicity together force $f$ to eventually dominate any polynomial.

Step 3 — The inverse function argument: Since $f$ is strictly increasing and convex, $f^{-1}$ exists. Substituting $x = f^{-1}(t)$ :

f'(f^{-1}(t)) = f(t) \quad \text{for all } t \in \text{range}(f)

But $f'(f^{-1}(t)) = \frac{1}{(f^{-1})'(t)}$ by the inverse function theorem, giving $(f^{-1})'(t) = \frac{1}{f(t)}$ .

Step 4 — Growth contradiction: Integrating: $f^{-1}(T) - f^{-1}(T_0) = \int_{T_0}^{T} \frac{dt}{f(t)}$ . As $T \to \infty$ , the LHS $\to \infty$ (since $f^{-1}$ is unbounded). But convexity forces $f(t) \ge ct$ for large $t$ (some $c > 0$ ), making $\int dt/f(t)$ diverge only logarithmically — while convexity simultaneously forces $f^{-1}$ to grow faster than logarithmically.

Rigorous bounding produces a contradiction: no such function exists.

15 min Putnam 2010, Problem B5

VI. Beyond the Syllabus

Combinatorics: Stirling Numbers and Unimodality

Many combinatorial sequences are unimodal (increase then decrease) — a discrete analog of “concave function.” The Stirling numbers of the second kind $S(n, k)$ form a unimodal sequence in $k$ for fixed $n$ .

The tool: a sequence $\{a_k\}$ is log-concave if $a_k^2 \ge a_{k-1} \cdot a_{k+1}$ for all $k$ . Log-concavity implies unimodality and is the discrete equivalent of the second derivative test ( $f'' \le 0$ ). This powerful principle reduces many “find the maximum term” competition problems to verifying a single quadratic inequality.

Differential Equations: Monotone Dynamical Systems

In systems of ODEs $\mathbf{x}' = F(\mathbf{x})$ , if the Jacobian satisfies $\partial F_i / \partial x_j \ge 0$ for $i \neq j$ (a cooperative system), then the flow preserves partial orderings.

Hirsch’s Theorem states that bounded trajectories of cooperative systems converge to equilibria — they cannot exhibit chaos. This is the $n$ -dimensional generalisation of the principle that “bounded monotone sequences converge,” connecting the Monotone Convergence Theorem to the geometry of dynamical systems.

Selected Problems

Problem Hard

(The Oscillating Neighborhood): Let $f$ be differentiable on $\mathbb{R}$ with $f'(0) > 0$ . True or false: there exists $\varepsilon > 0$ such that $f$ is increasing on $(-\varepsilon, \varepsilon)$ .

Hint

Consider the pathological function from Part 1:

f(x) = x + 2x^2\sin(1/x)

. What does its derivative do near

0

Problem Hard

(Darboux Integration): Let $f$ be differentiable on $[a, b]$ with $f'(x) \neq 0$ for all rational $x \in [a, b]$ . Prove that $f$ is strictly monotonic on $[a, b]$ .

Hint

f'

changes sign, Darboux gives a zero of

f'

. Can a zero of

f'

avoid all rationals? Consider the density of

\mathbb{Q}

… but be careful — the zero might be irrational. What constraint does the problem actually give you?

Problem Medium

(Composite Monotonicity): Let $f, g: \mathbb{R} \to \mathbb{R}$ be differentiable with $f'$ strictly decreasing and $g'$ strictly decreasing. Determine the monotonicity of $f \circ g$ and $g \circ f$ .

Hint

(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

. What does “strictly decreasing

f'

” tell you about convexity?

Problem Medium

(The Sinc Function): Prove that $\dfrac{\sin x}{x}$ is strictly decreasing on $(0, \pi)$ .

Hint

Let

h(x) = \sin x - x\cos x

. Show

h > 0

(0, \pi)

by analyzing

h'

Problem Advanced

(Matrix Monotonicity): Let $A$ be an $n \times n$ symmetric positive definite matrix. Define $f(t) = \det(A + tI)$ for $t > 0$ . Show that $f'(t) > 0$ for all $t > 0$ .

Hint

Express

\det(A + tI)

in terms of the eigenvalues

\lambda_1, \ldots, \lambda_n

A

. Then

f(t) = \prod(\lambda_i + t)

, and differentiate.

Problem Hard

(Recursive Roots): Define $p_1(x) = x$ and $p_{n+1}(x) = p_n(x) - p_n'(x)$ for $n \ge 1$ . Determine the monotonicity of the sequence of real roots of $p_n$ .

Hint

Compute the first few:

p_2(x) = x - 1

p_3(x) = x - 2

p_4(x) = x - 3

. Prove the pattern.

Problem Hard

(Integral Inequality): Let $f: [0,1] \to [0,1]$ be continuous, strictly increasing, with $f(0) = 0$ and $f(1) = 1$ . Prove:

\int_0^1 f(x) \, dx \cdot \int_0^1 f^{-1}(y) \, dy \ge \frac{1}{4}

Hint

You know

\int f + \int f^{-1} = 1

(the rectangle identity). Now use the AM-GM inequality on the two integrals.

Problem Hard

(Functional Equation): Find all strictly monotonic functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ .

Hint

Show

f > 0

everywhere. Then

\ln f(x+y) = \ln f(x) + \ln f(y)

, i.e.,

g = \ln \circ f

is additive. Monotonicity + additivity forces

g(x) = cx

Problem Medium

(The Cubic Trap): Let $p(x) = x^3 + bx^2 + cx + d$ . Find necessary and sufficient conditions on $b, c$ such that $p$ is strictly increasing on $\mathbb{R}$ but $p'(x)$ is not strictly positive everywhere.

Hint

p'(x) = 3x^2 + 2bx + c \ge 0

with equality at isolated points. What discriminant condition gives a non-negative quadratic with real roots?

Problem Advanced

(Limit via Monotonicity): Evaluate $\displaystyle\lim_{n \to \infty} \frac{(1 + 1/n)^{n^2}}{e^n}$ using monotonicity properties of the sequence $a_n = n^2 \ln(1 + 1/n) - n$ .

Hint

Show

\{a_n\}

is monotone by analysing

g(x) = x^2\ln(1 + 1/x) - x

for large

x

. The limit is related to

e^{-1/2}

Challenge Problem

Problem Advanced

The Threshold of Monotonicity

Let $h(x) = 2x^2 \sin(1/x)$ for $x \neq 0$ , $h(0) = 0$ , and define $F_c(x) = h(x) + cx$ for a constant $c > 0$ (with $F_c(0) = 0$ ).

Part (a): Determine all values of $c$ for which $F_c$ is strictly increasing on some interval containing $0$ .

Part (b): For the critical threshold value $c^*$ , is $F_{c^*}$ strictly increasing near $0$ , merely non-decreasing, or neither?

Part (c): Prove your answer to (a) rigorously. You will need both the Discrete Zero Theorem and Darboux’s Theorem.

Hint: Compute the Derivative

$F_c'(x) = 4x\sin(1/x) - 2\cos(1/x) + c$ for $x \neq 0$ . Near $0$ , the dominant oscillatory term is $-2\cos(1/x) \in [-2, 2]$ . What value of $c$ ensures $F_c' \ge 0$ despite this oscillation?

Hint: The Critical Threshold

The critical value is $c^* = 2$ . For $c > 2$ , the constant term dominates the oscillation. For $c < 2$ , at points $x_k = 1/(2k\pi)$ where $\cos(1/x_k) = 1$ , the derivative becomes $F_c'(x_k) = c - 2 < 0$ . What happens at $c = 2$ exactly?