Math Analysis 1

Curve sketching in depth: monotonicity, extrema, convexity and asymptotes proved

Mean value theorem and monotonicity, Fermat's theorem and the higher derivative test, convexity and inflection points, corner points and cusps with the limit of the derivative, oblique asymptote, summary sheet with mistakes to avoid.

ielettronica 25 October 2026 8 min read 0 comments

This page goes deeper into the lesson on curve sketching (studying the graph of a function). There the criteria are stated and applied; here they are proved, because understanding where they come from avoids the typical mistakes: concluding that a function with positive derivative is increasing even where the domain is split, mistaking a point with zero derivative for an extremum, looking for inflection points only where $f'' = 0$ . At the end there is a summary sheet to keep beside you during the exercises.

The mean value theorem, the key to everything

Theorem (Lagrange, or mean value theorem). If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , there is at least one point $\xi \in (a, b)$ with

$f'(\xi) = \frac{f(b) - f(a)}{b - a}$

Geometrically: there is a point where the tangent is parallel to the chord joining the ends of the graph. The proof, from Rolle’s theorem, is in the lesson on the theorems of differential calculus. Here we are interested in what follows from it.

The mean value theorem — The orange chord joins the points A and B of the graph; at the point xi the tangent, dashed in green, has the same slope as the chord.

Monotonicity and the sign of the derivative

Theorem. Let $f$ be differentiable on an interval $I$ . If $f'(x) \gt 0$ for every $x \in I$ , then $f$ is strictly increasing on $I$ ; if $f'(x) = 0$ for every $x$ , $f$ is constant on $I$ .

Proof. Let $x_1 \lt x_2$ be in $I$ . Since $I$ is an interval, all of $[x_1, x_2]$ lies in $I$ , and the mean value theorem gives a $\xi$ with $f(x_2) - f(x_1) = f'(\xi)(x_2 - x_1)$ . If $f' \gt 0$ , the right-hand side is positive, so $f(x_2) \gt f(x_1)$ . If $f' = 0$ , $f(x_2) = f(x_1)$ . ∎

The “interval” hypothesis is essential. $f(x) = -1/x$ has $f'(x) = 1/x^2 \gt 0$ on its whole domain, but $f(-1) = 1 \gt f(1) = -1$ : it is not increasing on the domain, because this is the union of two intervals separated by $x = 0$ . One says “increasing on $(-\infty, 0)$ and on $(0, +\infty)$ “, never “increasing on $\mathbb R \setminus \{0\}$ “.

The converse holds in weak form: if $f$ is increasing, $f' \ge 0$ , but it can vanish at isolated points, like $x^3$ at 0.

Stationary points and extrema

Theorem (Fermat). If $x_0$ is an interior point of the domain, is a local maximum or minimum and $f$ is differentiable at $x_0$ , then $f'(x_0) = 0$ .

Proof. Let $x_0$ be a local maximum: $f(x_0 + h) \le f(x_0)$ for small $h$ . For $h \gt 0$ the difference quotient $\frac{f(x_0 + h) - f(x_0)}{h}$ is $\le 0$ , for $h \lt 0$ it is $\ge 0$ . Both limits equal $f'(x_0)$ , which is therefore both $\le 0$ and $\ge 0$ : it is 0. ∎

Three practical consequences. Extrema must be looked for in three places: stationary points, points where $f$ is not differentiable, endpoints of the domain. The converse is false: $x^3$ has $f'(0) = 0$ but 0 is not an extremum. And a stationary point must always be classified, with the sign of $f'$ or with higher derivatives.

Theorem (higher derivative test). If $f'(x_0) = \cdots = f^{(n-1)}(x_0) = 0$ and $f^{(n)}(x_0) \ne 0$ , then: with $n$ even, $x_0$ is a minimum if $f^{(n)}(x_0) \gt 0$ and a maximum if $f^{(n)}(x_0) \lt 0$ ; with $n$ odd, $x_0$ is not an extremum (it is an inflection point with horizontal tangent).

Proof. By Taylor’s formula with Peano remainder, $f(x_0 + h) - f(x_0) = \frac{f^{(n)}(x_0)}{n!}h^n + o(h^n)$ . For small $h$ the sign of the difference is that of the first term. With $n$ even $h^n \gt 0$ on both sides, and the sign is that of $f^{(n)}(x_0)$ ; with $n$ odd $h^n$ changes sign with $h$ , and the function lies above on one side and below on the other. ∎

The case $n = 2$ is the second derivative test. Example: $x^4$ has its first three derivatives zero at 0 and $f^{(4)}(0) = 24 \gt 0$ : minimum.

Convexity

Definition. $f$ is convex on an interval if every chord lies above the graph: for $a \lt x \lt b$ , $f(x) \le f(a) + \frac{f(b) - f(a)}{b - a}(x - a)$ . It is concave if $-f$ is convex.

Theorem. If $f$ is twice differentiable on an interval, the following are equivalent: (i) $f'' \ge 0$ ; (ii) $f'$ is increasing; (iii) the graph lies above each of its tangent lines; (iv) $f$ is convex.

Proof. (i) ⇒ (ii) is the monotonicity theorem applied to $f'$ .

(ii) ⇒ (iii): let $x_0$ be a point and $x \gt x_0$ . By the mean value theorem $f(x) - f(x_0) = f'(\xi)(x - x_0)$ with $\xi \gt x_0$ , and since $f'$ increases $f'(\xi) \ge f'(x_0)$ : so $f(x) \ge f(x_0) + f'(x_0)(x - x_0)$ . For $x \lt x_0$ we have $\xi \lt x_0$ , $f'(\xi) \le f'(x_0)$ , but also $x - x_0 \lt 0$ , and the inequality is the same.

(ii) ⇒ (iv): with $a \lt x \lt b$ , the mean value theorem on $[a, x]$ and on $[x, b]$ gives two points $\xi_1 \lt \xi_2$ with slopes $\frac{f(x) - f(a)}{x - a} = f'(\xi_1) \le f'(\xi_2) = \frac{f(b) - f(x)}{b - x}$ : the slope of the chords increases, which is an equivalent way of saying that the point of the graph at $x$ lies below the chord from $a$ to $b$ .

The reverse implications are proved in a similar way; they are not needed in the exercises. ∎

A convex function with a chord and a tangent — The orange chord between the points with abscissas a and b lies above the graph; the green tangent at the point with abscissa x zero lies below it. These are the two equivalent characterisations of convexity.

Inflection points. A point $x_0$ is an inflection point if the concavity changes across $x_0$ (and the graph has a tangent there, possibly vertical). If $f$ is twice differentiable, at an inflection point $f''(x_0) = 0$ : indeed $f''$ changes sign. But the converse is false: $x^4$ has $f''(0) = 0$ and at 0 it has a minimum, not an inflection point. And an inflection point can also lie where $f''$ does not exist: $\sqrt[3]{x}$ has at 0 an inflection point with vertical tangent. The correct method is always to study the sign of $f''$ .

Points where the function is not differentiable

Where $f$ is continuous but not differentiable, compute the right and left derivatives (or their limits):

corner point: right and left derivatives finite and different, like $|x|$ at 0 ( $-1$ and $+1$ );
cusp: right and left derivatives infinite with opposite signs, like $x^{2/3}$ at 0;
inflection point with vertical tangent: right and left derivatives infinite with the same sign, like $x^{1/3}$ at 0.

Three kinds of points of non-differentiability — From the left: the corner point of the absolute value of x, the cusp of x to the two thirds and the inflection point with vertical tangent of x to the one third, all at the origin.

Theorem (limit of the derivative). If $f$ is continuous at $x_0$ , differentiable near $x_0$ and $\lim_{x\to x_0^+}f'(x) = \ell$ , then the right derivative at $x_0$ exists and equals $\ell$ (even if $\ell$ is infinite).

Proof. For $h \gt 0$ , the mean value theorem on $[x_0, x_0 + h]$ gives $\frac{f(x_0 + h) - f(x_0)}{h} = f'(\xi_h)$ with $x_0 \lt \xi_h \lt x_0 + h$ . As $h \to 0^+$ also $\xi_h \to x_0^+$ , so the difference quotient tends to $\ell$ . ∎

This is the theorem that justifies the method used in all the exercises: to know what kind of point it is, it is enough to compute the limit of the derivative from the right and from the left. Beware of the continuity hypothesis: without it the theorem is false (a step function has zero derivative everywhere except at the jump).

The oblique asymptote, proved

Theorem. The line $y = mx + q$ is an oblique asymptote of $f$ as $x \to +\infty$ , i.e. $f(x) - (mx + q) \to 0$ , if and only if

$m = \lim_{x\to+\infty}\frac{f(x)}{x}\ \ (\text{finite and} \ne 0),\qquad q = \lim_{x\to+\infty}\big(f(x) - mx\big)\ \ (\text{finite})$

Proof. If the line is an asymptote, $f(x) = mx + q + d(x)$ with $d(x) \to 0$ . Dividing by $x$ : $\frac{f(x)}{x} = m + \frac{q + d(x)}{x} \to m$ ; and $f(x) - mx = q + d(x) \to q$ . Conversely, if both limits exist and are finite, $f(x) - mx - q \to q - q = 0$ . ∎

Consequences: if $f(x)/x \to \pm\infty$ (like $e^x$ ) or $f(x)/x \to m$ but $f(x) - mx \to \infty$ (like $x + \sqrt x$ ) there is no oblique asymptote. And the sign of $d(x) = f(x) - (mx + q)$ tells whether the graph lies above or below the asymptote.

A function with an oblique asymptote — The vertical distance d of x between the graph and the line y equal to m x plus q tends to zero: this is the definition of an oblique asymptote. Here d of x is positive, so the graph lies above the line.

The curve-sketching sheet

Step	What to compute	Mistakes to avoid
Domain	conditions of existence: denominators $\ne 0$ , even roots $\ge 0$ , arguments of logarithms $\gt 0$	forgetting combined conditions, such as $\ln$ of a fraction
Symmetry	$f(-x) = f(x)$ even, $f(-x) = -f(x)$ odd, period	looking for symmetry when the domain is not symmetric
Sign and zeros	$f(x) \gt 0$ , $f(x) = 0$	getting the sign of denominator factors wrong
Limits	at the ends of the domain and at excluded points, right and left	forgetting the one-sided limit
Asymptotes	vertical, horizontal, oblique with $m$ and $q$	looking for an oblique one where there is already a horizontal one (on the same side)
First derivative	$f'$ , its domain, zeros and sign	saying “increasing on the domain” when it is split
Non-differentiable points	limits of $f'$ from the right and from the left	applying the theorem without continuity
Second derivative	$f''$ , zeros and sign	looking for inflection points only where $f'' = 0$
Graph	put everything together, check consistency	inconsistencies between the sign of $f'$ and the limits, such as a function that tends to $+\infty$ on the right but turns out decreasing to the end

The complete exercises, with the graph of each function, are on the exercise page linked at the top.

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The mean value theorem, the key to everything

Monotonicity and the sign of the derivative

Stationary points and extrema

Convexity

Points where the function is not differentiable

The oblique asymptote, proved

The curve-sketching sheet

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