Monotonicity and the Sign of the Derivative

The sign-of-the-derivative test is the working engine behind curve sketching and optimization in every calculus course and in applied fields: identifying where a cost function decreases, a population model increases, or a physical potential energy has a stable region all reduce to checking the sign of a derivative on intervals between critical points. In numerical analysis and root-finding, monotonicity certificates (proving $f' > 0$ on an interval) are used to guarantee that Newton's method or bisection converges to a unique root, since a strictly monotonic continuous function crosses zero at most once. In economics and game theory, utility and production functions are routinely assumed (or shown) to be monotonic in certain variables using exactly this derivative test, which underlies comparative statics — predicting how an equilibrium shifts when a parameter increases. In probability and statistics, cumulative distribution functions are nondecreasing by definition; verifying that a proposed density integrates to a genuinely nondecreasing CDF, or that a quantile function is well-defined, leans on part (3)/(4) of this theorem (nonnegative derivative implies nondecreasing). In cryptography and computer science, monotonic functions (e.g. in the analysis of hash function avalanche properties, or in convexity/monotonicity arguments for algorithmic bounds) are frequently certified via derivative-sign arguments when the underlying functions are real-analytic approximations. Most directly within analysis itself, this theorem is the immediate stepping stone to two central results: <a class="wiki-link" href="/wiki/math-analysis-zero-derivative-constant" data-track="wiki_article.body.wiki_link">Zero Derivative Implies Constant</a>, obtained as the special case $f' \equiv 0$ (giving both $f' \ge 0$ and $f' \le 0$ simultaneously, hence $f$ both nondecreasing and nonincreasing, hence constant), and <a class="wiki-link" href="/wiki/math-analysis-inverse-function-theorem-1d" data-track="wiki_article.body.wiki_link">Derivatives of Inverse Functions</a>, where strict monotonicity (guaranteed by a strict sign condition on $f'$) is exactly the hypothesis needed to know $f$ is invertible on an interval before differentiating its inverse.

:::quiz[If $f'(x) \ge 0$ for all $x$ in an open interval $(a,b)$, and $f$ is continuous on $[a,b]$, which conclusion is justified by the theorem?]

• $f$ is strictly increasing on $[a,b]$
• $f$ is increasing (nondecreasing) on $[a,b]$ ✓
• $f$ is constant on $[a,b]$
• $f'$ must be strictly positive somewhere in $(a,b)$

:::quiz[Why does the proof of part (1) of the theorem require $f$ to be continuous on the closed interval $[x_1, x_2]$, not just differentiable on the open interval?]

• Continuity is irrelevant; only differentiability on the open interval matters
• The Mean Value Theorem's hypotheses explicitly require continuity on the closed interval to apply at the endpoints $x_1, x_2$ ✓
• Continuity is only needed to define $f'(x_1)$ and $f'(x_2)$
• It is required to make $f'$ itself continuous

Ask the tutor: Can you show me why the Mean Value Theorem itself fails without continuity on the closed interval, and how that failure would propagate into a broken monotonicity proof?

Picture the graph of $f$ as a road, and $f'(x)$ as the slope of the tangent line — the "instantaneous grade" — at each point $x$. If every tangent line along a stretch of road tilts upward, your intuition says the road itself must be climbing overall: you never actually go downhill, even for an instant, so cumulatively you must rise from any earlier point to any later one. The subtlety is that "tangent line slopes up at every point" is a statement about a limit at each individual point, while "the road climbs from $a$ to $b$" is a statement comparing the two endpoint values $f(a)$ and $f(b)$ directly. Bridging a pointwise (local) hypothesis to a two-point (global) conclusion is exactly the kind of gap the Mean Value Theorem is built to close: it guarantees that the average slope between any two points equals the instantaneous slope at some point in between. If instantaneous slopes are always positive, the average slope between any two points must be positive too — and average slope positive is precisely the statement that $f$ climbed. The picture below shows a function whose tangent slope is positive throughout an interval, illustrating how the secant line between any two points inherits its sign from some tangent line in between.

Example 1 (a rational function with an asymmetric domain split). Let $f(x) = \dfrac{x^3}{3} - 2x^2 + 3x + 1$ on $\mathbb{R}$. Determine all intervals of increase and decrease. Solution. $f$ is a polynomial, hence continuous and differentiable everywhere, with

This is a upward parabola in $x$ vanishing at $x=1$ and $x=3$, so:

• $f'(x) > 0$ for $x < 1$ (both factors negative, product positive),
• $f'(x) < 0$ for $1 < x < 3$ (factors opposite sign),
• $f'(x) > 0$ for $x > 3$ (both factors positive).

Apply the theorem on each of the three intervals $(-\infty, 1]$, $[1,3]$, $[3,\infty)$ — each is a valid interval on which $f$ is continuous and differentiable on the interior, so the theorem's hypotheses hold on each piece:

• On $(-\infty, 1]$: $f'>0$ on the interior $(-\infty,1)$, so by part (1), $f$ is strictly increasing on $(-\infty,1]$.
• On $[1,3]$: $f'<0$ on $(1,3)$, so by part (2), $f$ is strictly decreasing on $[1,3]$.
• On $[3,\infty)$: $f'>0$ on $(3,\infty)$, so by part (1), $f$ is strictly increasing on $[3,\infty)$.

(Gluing strict monotonicity across the shared endpoints $x=1,3$ is valid: e.g. for $x_1 < 1 < x_2 \le 3$ we get $f(x_1) < f(1)$ from the first piece and $f(1) \ge f(x_2)$ or $f(1) > f(x_2)$ from the second, and if $x_2=1$ trivially, so $f(x_1)<f(1)$ suffices; a full comparison across the boundary follows by chaining these strict inequalities through the shared point.) As a numerical check: $f(1) = \tfrac13 - 2 + 3 + 1 = \tfrac{7}{3} \approx 2.33$ and $f(3) = 9 - 18 + 9 + 1 = 1$, confirming $f$ decreased from $x=1$ to $x=3$, consistent with $f' < 0$ there. Also $f(0) = 1 < f(1) = 7/3$, consistent with increase on $(-\infty,1]$, and $f(4) = 64/3 - 32 + 12 + 1 = 64/3 - 19 = 7/3 \approx 2.33 > f(3) = 1$, consistent with increase on $[3,\infty)$.

Example 2 (using the corollary to handle an isolated zero of the derivative). Let $f(x) = x^5 + x$ on $\mathbb{R}$. Show $f$ is strictly increasing on all of $\mathbb{R}$, even though a naive application of part (1) alone is not immediate everywhere. Solution. Compute

Since $x^4 \ge 0$ for all real $x$, we get $f'(x) = 5x^4 + 1 \ge 1 > 0$ for every $x \in \mathbb{R}$ — in fact here $f'$ is strictly positive everywhere, with no zero at all, so part (1) of the theorem applies directly on $I = \mathbb{R}$: $f$ is strictly increasing on $\mathbb{R}$. To illustrate the corollary on a case where the derivative does vanish somewhere, modify the example slightly: let $g(x) = x^5$. Then $g'(x) = 5x^4 \ge 0$ for all $x$, with equality exactly at $x = 0$ — a single point, which contains no subinterval of $\mathbb{R}$. By part (3), $g$ is increasing on $\mathbb{R}$; by the corollary (the zero set $\{0\}$ contains no open interval), $g$ is in fact strictly increasing on $\mathbb{R}$, even though $g'(0) = 0$. Numerically: $g(-1) = -1 < g(0) = 0 < g(1) = 1$, and more finely $g(-0.1) = -0.00001 < g(0) = 0 < g(0.1) = 0.00001$, confirming strict increase straddling the point where the derivative vanishes.