Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

Lessons

Senior 4 · Mathematics

Differentiation of Polynomial, Rational and Irrational Functions and Its Applications

5 lessons

Lesson 4 of 5

Mean value theorems, increasing functions, turning points and L'Hôpital's rule

25 min

Caught by an average

A driver covers 120 km of road in 1 hour 30 minutes, an average speed of 80 km/h. Must the speedometer have shown exactly 80 km/h at some moment? Intuition says yes: if the car was always slower than 80, it could not keep up the average, and if always faster, it would beat it. The mean value theorem turns this intuition into a theorem about derivatives. From it follow the tests for where a function increases or decreases and where it turns, and a powerful shortcut for indeterminate limits: L'Hôpital's rule.

Important

Rolle's theorem. If $f$ is continuous on $[a, b]$ , differentiable on $(a, b)$ , and $f (a) = f (b)$ , then there is at least one $c$ in $(a, b)$ with $f^{'} (c) = 0$ .

Mean value theorem (Lagrange). If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there is at least one $c$ in $(a, b)$ with

f'(c) = \frac{f(b) - f(a)}{b - a} .

Geometrically: somewhere between $a$ and $b$ the tangent is parallel to the chord joining $(a, f (a))$ and $(b, f (b))$ . Rolle's theorem is the special case of a horizontal chord.

Mean value theorem for $f (x) = x^{3}$ on $[0, 2]$ : the chord has slope $\frac{8 - 0}{2 - 0} = 4$ , and the tangent at $c = \frac{2}{\sqrt{3}} \approx 1.15$ (red) has the same slope.

Worked example 1: finding the $c$ of each theorem

(a) Verify Rolle's theorem for $f (x) = x^{2} - 4 x + 3$ on $[1, 3]$ . (b) Find the value of $c$ given by the mean value theorem for $f (x) = x^{3}$ on $[0, 2]$ . (c) Explain why Rolle's theorem does not apply to $∣ x ∣$ on $[- 1, 1]$ .

Solution

A polynomial is continuous and differentiable everywhere, and $f (1) = f (3) = 0$ . $f^{'} (c) = 2 c - 4 = 0$ gives $c = 2$ , which lies in $(1, 3)$ ✓.
$3c^2 = \frac{8 - 0}{2 - 0} = 4$ , so $c = \frac{2}{\sqrt{3}} \approx 1.15$ (the negative root is outside $(0, 2)$ ).
$∣ - 1 ∣ = ∣ 1 ∣$ , but $∣ x ∣$ is not differentiable at 0, and indeed its derivative ( $\pm 1$ ) is never 0. A hypothesis fails, so the conclusion need not hold.

Important

Increasing and decreasing functions (a consequence of the mean value theorem). Let $f$ be differentiable on an interval $I$ .

If $f'(x) \gt 0$ on $I$ , $f$ is increasing on $I$ .
If $f'(x) \lt 0$ on $I$ , $f$ is decreasing on $I$ .
If $f^{'} (x) = 0$ on $I$ , $f$ is constant on $I$ .

Reason: for $x_1 \lt x_2$ in $I$ , $f (x_{2}) - f (x_{1}) = f^{'} (c) (x_{2} - x_{1})$ for some $c$ between them, so $f (x_{2}) - f (x_{1})$ has the sign of $f^{'} (c)$ .

Definition

Stationary point: a point where $f^{'} (x) = 0$ (horizontal tangent).

Local (relative) maximum at $c$ : $f(c) \ge f(x)$ for all $x$ near $c$ ; local minimum: $f(c) \le f(x)$ near $c$ . Both are turning points.

First derivative test: at a stationary point $c$ , if $f^{'}$ changes from $+$ to $-$ , $f (c)$ is a local maximum; from $-$ to $+$ , a local minimum; if $f^{'}$ does not change sign, there is no turning point (as for $x^{3}$ at 0).

Worked example 2: a variation table

Find the intervals on which $f (x) = x^{3} - 3 x^{2} - 9 x + 5$ increases or decreases, and its turning points.

Solution $f^{'} (x) = 3 x^{2} - 6 x - 9 = 3 (x + 1) (x - 3)$ , zero at $- 1$ and 3. $f (- 1) = - 1 - 3 + 9 + 5 = 10$ , $f (3) = 27 - 27 - 27 + 5 = - 22$ .

Variation table: f′ positive outside the roots, negative between them.
$x$	$-\infty$		$- 1$		$3$		$+\infty$
$f^{'} (x)$		+	0	−	0	+
$f (x)$		↗	10 (max)	↘	−22 (min)	↗

$f$ increases on $(-\infty, -1]$ and $[3, +\infty)$ , decreases on $[- 1, 3]$ ; local maximum $10$ at $x = - 1$ , local minimum $- 22$ at $x = 3$ .

Worked example 3: a rational function

Find the turning points of $f(x) = x + \frac{4}{x}$ , $x \ne 0$ .

Solution $f'(x) = 1 - \frac{4}{x^2} = \frac{(x - 2)(x + 2)}{x^2}$ . The denominator is positive, so $f^{'}$ has the sign of $(x - 2) (x + 2)$ : $+$ for $x \lt -2$ , $-$ on $(- 2, 0)$ and $(0, 2)$ , $+$ for $x \gt 2$ . So $f (- 2) = - 4$ is a local maximum and $f (2) = 4$ a local minimum. (The local maximum is smaller than the local minimum: "local" compares only with nearby values, and the two branches are separated by the asymptote $x = 0$ .)

Title page of Analyse des infiniment petits pour l'intelligence des lignes courbes (Paris, 1696) by Guillaume de l'Hôpital, the first textbook of differential calculus. The rule now named after him is in it; he had learned it from his teacher Johann Bernoulli. Scan: public domain, via Wikimedia Commons.

Important

L'Hôpital's rule. Suppose $f$ and $g$ are differentiable near $a$ (except possibly at $a$ ), $g'(x) \ne 0$ there, and $\frac{f(x)}{g(x)}$ has the form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$ as $x \to a$ . If $\displaystyle\lim_{x \to a} \frac{f'(x)}{g'(x)}$ exists (or is $\pm\infty$ ), then

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} .

The same holds for one-sided limits and for $x \to \pm\infty$ .

Worked example 4: L'Hôpital's rule

Find (a) $\displaystyle\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}$ (b) $\displaystyle\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}$ (c) $\displaystyle\lim_{x \to 1} \frac{x^3 - 3x + 2}{(x - 1)^2}$ (d) $\displaystyle\lim_{x \to +\infty} \frac{3x^2 + x}{5x^2 - 2}$ .

Solution

$\frac{0}{0}$ : $\lim \frac{3x^2}{2x} = \frac{12}{4} = 3$ . (Check by factorising: $\frac{x^2 + 2x + 4}{x + 2} \to \frac{12}{4}$ ✓.)
$\frac{0}{0}$ : $\lim \frac{\frac{1}{2\sqrt{1 + x}}}{1} = \frac{1}{2}$ .
$\frac{0}{0}$ : $\lim \frac{3x^2 - 3}{2(x - 1)}$ , still $\frac{0}{0}$ , so apply the rule again: $\lim \frac{6x}{2} = 3$ .
$\frac{\infty}{\infty}$ : $\lim \frac{6x + 1}{10x}$ , again $\frac{\infty}{\infty}$ : $\lim \frac{6}{10} = \frac{3}{5}$ .

Warning

Use L'Hôpital's rule only for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ . For $\displaystyle\lim_{x \to 1} \frac{x + 1}{x + 2}$ substitution gives $\frac{2}{3}$ ; applying the rule anyway gives the wrong answer $\frac{1}{1} = 1$ . And differentiate the numerator and the denominator separately: do not use the quotient rule.

Tip

For turning points, a variation table (sign of $f^{'}$ , arrows for $f$ ) is the clearest layout, and examiners expect it. Put in every value where $f^{'} = 0$ or is undefined, including the points excluded from the domain.

Professor Leonard: A brief discussion of Rolle's theorem and the mean value theorem (about 7 min). Watch how the conditions (continuous on the closed interval, differentiable inside) are checked before the theorem is used.

✦

Practice

Show a variation table in questions 4 to 6.

Verify Rolle's theorem for $f (x) = x^{3} - 4 x$ on $[- 2, 2]$ , and find every value of $c$ .
Find the value of $c$ given by the mean value theorem for $f (x) = x^{2} + 2 x$ on $[1, 4]$ .
A car's position is $s (t)$ , a differentiable function. It travels 150 km in 2 hours. Use the mean value theorem to show that at some instant its speed was 75 km/h.
Find where $f (x) = 2 x^{3} + 3 x^{2} - 12 x + 1$ increases or decreases, and its turning points.
Find the turning points of $f(x) = \frac{x^2 + 9}{x}$ .
Find the turning points of $f(x) = x\sqrt{4 - x}$ , $x \le 4$ .
Use L'Hôpital's rule: $\displaystyle\lim_{x \to 3} \frac{x^2 - 2x - 3}{x^2 - 9}$ and $\displaystyle\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ .
Use L'Hôpital's rule twice: $\displaystyle\lim_{x \to 2} \frac{x^3 - 6x^2 + 12x - 8}{x^2 - 4x + 4}$ . Then check by factorising.

Show answersHide answers

$f (- 2) = f (2) = 0$ ; $f'(c) = 3c^2 - 4 = 0 \Rightarrow c = \pm\frac{2}{\sqrt{3}} \approx \pm 1.15$ , both in $(- 2, 2)$ .
$\frac{f(4) - f(1)}{3} = \frac{24 - 3}{3} = 7 = 2c + 2 \Rightarrow c = 2.5$ .
The average speed is $\frac{150}{2} = 75$ km/h, so some $c$ has $s^{'} (c) = 75$ .
$f^{'} = 6 (x + 2) (x - 1)$ : increasing on $(-\infty, -2]$ and $[1, +\infty)$ , decreasing on $[- 2, 1]$ . Max $f (- 2) = 21$ , min $f (1) = - 6$ .
$f' = 1 - \frac{9}{x^2}$ : max $f (- 3) = - 6$ , min $f (3) = 6$ .
$f' = \sqrt{4 - x} - \frac{x}{2\sqrt{4 - x}} = \frac{8 - 3x}{2\sqrt{4 - x}}$ : maximum at $x = \frac{8}{3}$ , $f = \frac{8}{3}\sqrt{\frac{4}{3}} = \frac{16}{3\sqrt{3}} \approx 3.08$ .
$\frac{2x - 2}{2x} \to \frac{4}{6} = \frac{2}{3}$ ; $\frac{\frac{1}{2\sqrt{x}}}{1} \to \frac{1}{4}$ .
$\frac{3x^2 - 12x + 12}{2x - 4}$ is still $\frac{0}{0}$ ; then $\frac{6x - 12}{2} \to 0$ . Check: $\frac{(x - 2)^3}{(x - 2)^2} = x - 2 \to 0$ ✓.

Quick check

For f(x) = x³ on [0, 2], the mean value theorem gives a c with f′(c) equal to which number?

Key points

Rolle: $f(a) = f(b) \Rightarrow f'(c) = 0$ for some $c \in (a, b)$ . Lagrange: $f'(c) = \frac{f(b) - f(a)}{b - a}$ . Both need continuity on $[a, b]$ and differentiability on $(a, b)$ .
$f' \gt 0$ : increasing; $f' \lt 0$ : decreasing.
At a stationary point, $f^{'}$ changing $+ \to -$ gives a maximum, $- \to +$ a minimum; no change, no turning point.
L'Hôpital: for $\frac{0}{0}$ or $\frac{\infty}{\infty}$ only, $\lim \frac{f}{g} = \lim \frac{f'}{g'}$ ; repeat if needed.

Sources and credits

Written by Ganzaa: the explanations, the graph, the worked examples, the practice items and the quiz questions. The unit follows the topic list of REB's S4 Core Mathematics (2026–2027); no REB text is reproduced.
Image: L'Hospital, Analyse des infiniment petits pour l'intelligence des lignes courbes, 1696, Wikimedia Commons, public domain.
Video: Calculus 1 Lecture 3.2: A BRIEF Discussion of Rolle's Theorem and Mean-Value Theorem, Professor Leonard, embedded from YouTube.

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