CommeUnJeu · L1 PCSI

Real functions: recap

⌚ ~99 min ▢ 12 blocks ✓ 36 exercises ➣ Prerequisites : Maps, Functions

A real function attaches one real number to each input. From this single idea grow all of analysis: limits, continuity, derivatives, integration. This chapter is a recap chapter: it cleans up the lycée vocabulary about real functions --- domain, parity, periodicity, monotonicity, derivative rules --- with explicit quantifiers and rigorous statements, and adds the mature tools (class $C^k$, inverse function differentiation) that the next chapters will rely on.
\medskip
Convention. Throughout the chapter, $\ln$, $\exp$, $\sin$, $\cos$, $\tan$ and $\sqrt{\cdot}$ are used as known lycée functions; their systematic study is postponed to the chapters Trigonometry and Standard functions. Derivability and class-$C^k$ statements are made on an interval $I \subset \mathbb{R}$ with non-empty interior; for statements about general subsets (domain, parity, periodicity, monotonicity, boundedness), an arbitrary subset $E \subset \mathbb{R}$ is allowed.

I Domain$\virgule$ image$\virgule$ graph

A real function attaches one real number to each input --- but only when the formula makes sense. Before any study of $f$, nail down where $f$ is defined. The domain is read off the formula by intersecting all the conditions each elementary block demands ($\ln$ wants a positive argument, $\sqrt{\cdot}$ wants a non-negative one, division wants a non-zero denominator, etc.).

Definition — Domain$\virgule$ image$\virgule$ graph

Let $E \subset \mathbb{R}$ and $f : E \to \mathbb{R}$ a function.

The domain of definition of $f$, denoted $\textcolor{colordef}{D_f}$, is the set $E$ on which $f$ is defined. When $f$ is given by a formula without an explicit $E$, $D_f$ is the largest subset of $\mathbb{R}$ on which the formula makes sense.
For $x \in D_f$, $f(x)$ is the image of $x$ by $f$ ; $x$ is an antecedent of $f(x)$.
The image set of $f$ is $\textcolor{colordef}{f(D_f) = \{f(x) \mid x \in D_f\}} \subset \mathbb{R}$.
In a fixed orthonormal frame $(O, \vec{\imath}, \vec{\jmath})$, the graph of $f$ is $\textcolor{colordef}{\Gamma_f = \{(x, f(x)) \mid x \in D_f\}} \subset \mathbb{R}^2$. The equation of the graph is $y = f(x)$.

Example

Determine the domain of $f : x \mapsto \sqrt{x+3}$.

Answer

The square root requires a non-negative argument: $x + 3 \ge 0$, i.e., $x \ge -3$. So $D_f = [-3, +\infty[$.

Example

Determine the domain of $g : x \mapsto \ln(x^2 - 3x + 2)$.

Answer

The logarithm requires a strictly positive argument: $x^2 - 3x + 2 > 0$. The polynomial factorizes as $(x-1)(x-2)$ ; with leading coefficient $+1$, it is strictly positive outside its roots. So $D_g = ]-\infty, 1[ \,\cup\, ]2, +\infty[$.

Example

Determine the domain of $h : x \mapsto \dfrac{1 + \mathrm{e}^{\sqrt{x}}}{x \sqrt{2 - x}}$.

Answer

Three conditions must hold simultaneously : $\sqrt{x}$ requires $x \ge 0$ ; $\sqrt{2-x}$ requires $2 - x \ge 0$, i.e., $x \le 2$ ; the denominator $x \sqrt{2-x}$ must be non-zero, hence $x \ne 0$ and $x \ne 2$. Intersecting: $D_h = ]0, 2[$.

Method — Finding the domain of a real function

Algorithm.

List every elementary block in the formula : $\sqrt{u}$, $\ln u$, $1/u$, $u^\alpha$ (for $\alpha \notin \mathbb{N}$), $\arccos u$, $\arcsin u$, $\tan u$, etc.
Write the native condition for each block ($u \ge 0$ for $\sqrt{u}$, $u > 0$ for $\ln u$, $u \ne 0$ for $1/u$, $u \ne \pi/2 + k\pi$ for $\tan u$, \dots ; for $\arccos u$ and $\arcsin u$ the condition $u \in [-1, 1]$ applies once those functions are introduced in the next chapter).
Take the intersection of all these conditions. The result is $D_f$.

Trap. The block $1/(x \sqrt{2-x})$ requires $\sqrt{2-x}$ to be defined and non-zero --- two conditions, not one.

Skills to practice

Finding the domain of a real function

II Operations on real functions

The set $\mathcal{F}(E, \mathbb{R})$ of real functions defined on $E$ inherits an algebraic structure: you can add functions, multiply them by scalars or pointwise, and apply $|\cdot|$. A separate operation, composition, builds new functions by feeding the output of one into the input of another --- and breaks naïve intuition (composition is not commutative; the domain of $g \circ f$ is delicate).

Definition — Sum$\virgule$ product$\virgule$ scalar multiple

Let $E \subset \mathbb{R}$ and $f, g : E \to \mathbb{R}$ two functions, $\lambda \in \mathbb{R}$. We define on $E$:

$\textcolor{colordef}{(f + g)(x) = f(x) + g(x)}$ (sum) ;
$\textcolor{colordef}{(\lambda f)(x) = \lambda \cdot f(x)}$ (scalar multiple) ;
$\textcolor{colordef}{(fg)(x) = f(x) \cdot g(x)}$ (pointwise product) ;
$\textcolor{colordef}{|f|(x) = |f(x)|}$ (absolute value).

The set $\mathcal{F}(E, \mathbb{R})$, equipped with $+$ and the scalar multiplication, is a $\mathbb{R}$-vector space.

Example

For $f(x) = x^2$ and $g(x) = x + 1$ on $\mathbb{R}$, compute $(f + g)(x)$, $(fg)(x)$, $(3f)(x)$, $|f - g|(x)$.

Answer

$(f+g)(x) = x^2 + x + 1$. $(fg)(x) = x^2(x+1) = x^3 + x^2$. $(3f)(x) = 3x^2$. $|f-g|(x) = |x^2 - x - 1|$ ; the sign of $x^2 - x - 1$ depends on $x$ relative to its roots $(1 \pm \sqrt{5})/2$.

Definition — Composition of functions

Strict form. Let $f : X \to Y$ and $g : Y \to Z$ where $X, Y, Z$ are subsets of $\mathbb{R}$. The composition $\textcolor{colordef}{g \circ f : X \to Z}$ is defined by $(g \circ f)(x) = g(f(x))$ for $x \in X$.
Permissive form (used with formulas). Let $f : X \to \mathbb{R}$ and $g : Y \to \mathbb{R}$. Then $g \circ f$ is defined on $$ \textcolor{colordef}{\{x \in X \mid f(x) \in Y\}}. $$ This set may be smaller than $X$ (when $f$ takes some values outside $Y$) or may equal $X$ (when $f(X) \subset Y$); only computation reveals which.

Example

Let $f(x) = x^2$ and $g(x) = x + 1$, both defined on $\mathbb{R}$. Compute $f \circ g$ and $g \circ f$, and check that they differ.

Answer

$(f \circ g)(x) = f(g(x)) = (x+1)^2 = x^2 + 2x + 1$. $(g \circ f)(x) = g(f(x)) = x^2 + 1$. The two functions differ : composition is not commutative.

Example

Let $g(y) = \sqrt{y}$ defined on $\mathbb{R}_+$ and $f(x) = x^2 + 1$. Determine the domain of $g \circ f$.

Answer

$g \circ f$ is defined on $\{x \in \mathbb{R} \mid f(x) \ge 0\}$. Since $f(x) = x^2 + 1 \ge 1 > 0$ for all $x \in \mathbb{R}$, the condition is satisfied everywhere : $g \circ f$ is defined on $\mathbb{R}$. We have $(g \circ f)(x) = \sqrt{x^2 + 1}$.

Definition — Positive and negative parts

For $f : E \to \mathbb{R}$, define on $E$ : $$ \textcolor{colordef}{f^+ = \frac{|f| + f}{2}}, \qquad \textcolor{colordef}{f^- = \frac{|f| - f}{2}}. $$ $f^+$ is the positive part of $f$ ; $f^-$ is the negative part. Both take non-negative values.

Example

For $f : x \mapsto x - 1$ on $\mathbb{R}$, compute $f^+(x)$ and $f^-(x)$, and check $f = f^+ - f^-$.

Answer

$|f|(x) = |x - 1| = \begin{cases} x - 1 & \text{if } x \ge 1 \\ 1 - x & \text{if } x < 1 \end{cases}$. Hence $f^+(x) = (|f|+f)/2 = \begin{cases} x - 1 & \text{if } x \ge 1 \\ 0 & \text{if } x < 1 \end{cases}$ and $f^-(x) = (|f|-f)/2 = \begin{cases} 0 & \text{if } x \ge 1 \\ 1 - x & \text{if } x < 1 \end{cases}$. Both are non-negative. Check : for $x \ge 1$, $f^+ - f^- = (x-1) - 0 = x - 1 = f(x)$ ; for $x < 1$, $f^+ - f^- = 0 - (1-x) = x - 1 = f(x)$. $\checkmark$

Proposition — Decomposition $f \equal f^+ - f^-$

For every function $f : E \to \mathbb{R}$ : $$ \textcolor{colorprop}{f = f^+ - f^-} \qquad \text{and} \qquad \textcolor{colorprop}{|f| = f^+ + f^-}. $$

Proof

Direct expansion : $$ f^+ - f^- = \frac{|f| + f}{2} - \frac{|f| - f}{2} = \frac{2f}{2} = f, \qquad f^+ + f^- = \frac{|f| + f}{2} + \frac{|f| - f}{2} = \frac{2|f|}{2} = |f|. $$

Skills to practice

Computing compositions

III Parity$\virgule$ oddness$\virgule$ periodicity

Symmetry shrinks the work. If you know a function is even, the right half of its graph determines the left half. If you know it is $T$-periodic, one period determines all of $\mathbb{R}$. Two flavours of symmetry: about a line (parity), about a translation (periodicity).

Definition — Even and odd functions

Let $E \subset \mathbb{R}$ symmetric about $0$ (i.e., $\forall x \in E, -x \in E$), and $f : E \to \mathbb{R}$.

$f$ is even if $\forall x \in E, f(-x) = f(x)$. The graph of $f$ is symmetric about the $y$-axis.
$f$ is odd if $\forall x \in E, f(-x) = -f(x)$. The graph of $f$ is symmetric about the origin $O$. (If $0 \in E$, then $f(0) = 0$.)

Example

$x \mapsto x^{2k}$ is even ($k \in \mathbb{N}$) and $x \mapsto x^{2k+1}$ is odd. More generally, every monomial of even degree is even ; every monomial of odd degree is odd.

Example

The function $|\cdot| : x \mapsto |x|$ is even on $\mathbb{R}$ : $|-x| = |x|$. The cosine $\cos$ is even on $\mathbb{R}$ : $\cos(-x) = \cos(x)$. The sine $\sin$, the tangent $\tan$, and the identity $x \mapsto x$ are odd : $\sin(-x) = -\sin x$, $\tan(-x) = -\tan x$, and $-x = -(x)$.

Definition — Periodic function

Let $E \subset \mathbb{R}$ and $T > 0$. We say $E$ is $T$-periodic if $\forall x \in E, x + T \in E$ and $x - T \in E$. A function $f : E \to \mathbb{R}$ defined on a $T$-periodic set is $T$-periodic if $$ \textcolor{colordef}{\forall x \in E, \quad f(x + T) = f(x).} $$ The real $T$ is then called a period of $f$. We say $f$ is periodic if there exists at least one $T > 0$ such that $f$ is $T$-periodic.

Example

$\sin$ and $\cos$ are $2\pi$-periodic on $\mathbb{R}$. $\tan$ is $\pi$-periodic on its domain. The function $x \mapsto \mathrm{e}^x$ is not periodic (it is strictly monotone). If $T$ is a period of $f$, then so is $kT$ for every $k \in \mathbb{N}^*$ ; one therefore speaks of a period, not the period.

Proposition — Operations on periodic functions

Let $T > 0$, $E$ a $T$-periodic subset of $\mathbb{R}$, and $f, g : E \to \mathbb{R}$ both $T$-periodic. Then:

$\textcolor{colorprop}{f + g}$ and $\textcolor{colorprop}{fg}$ are $T$-periodic on $E$ ;
if $g$ does not vanish on $E$, $\textcolor{colorprop}{f/g}$ is $T$-periodic on $E$ ;
for $\omega > 0$, the function $\textcolor{colorprop}{x \mapsto f(\omega x)}$ is $T/\omega$-periodic on the set $\{x \mid \omega x \in E\}$.

(For $\omega \ne 0$ in general, the period is $T/|\omega|$ ; we state the result for $\omega > 0$ to keep the formula uncluttered.)

Proof

Sum. For $x \in E$ : $(f+g)(x+T) = f(x+T) + g(x+T) = f(x) + g(x) = (f+g)(x)$.
Product. For $x \in E$ : $(fg)(x+T) = f(x+T) \cdot g(x+T) = f(x) \cdot g(x) = (fg)(x)$.
Quotient. If $g$ does not vanish on $E$, the same calculation gives $(f/g)(x+T) = f(x)/g(x) = (f/g)(x)$.
Dilation. Set $h(x) = f(\omega x)$ defined on $E' = \{x \mid \omega x \in E\}$. For $x \in E'$, $$ h\!\left(x + \frac{T}{\omega}\right) = f\!\left(\omega \cdot \left(x + \frac{T}{\omega}\right)\right) = f(\omega x + T) = f(\omega x) = h(x), $$ using the $T$-periodicity of $f$.

Method — Reducing the study domain via parity / periodicity

Strategy. Before computing limits or building variation tables, exploit symmetry to halve the work :

If $f$ is even on $E$ (symmetric about $0$), study $f$ only on $E \cap \mathbb{R}_+$ ; the left half follows by symmetry about $Oy$.
If $f$ is odd, study $f$ only on $E \cap \mathbb{R}_+$ ; the left half follows by symmetry about $O$.
If $f$ is $T$-periodic, study $f$ on any interval of length $T$ contained in $E$, for instance $[a, a+T[$ (the half-open interval avoids redundancy of the two endpoints) ; the rest follows by translation.
If $f$ is both even and $T$-periodic, study $f$ only on $[0, T/2]$.

Concrete. For $f(x) = \cos(2x) + \sin^2 x$ on $\mathbb{R}$ : $f$ is even ($\cos$ and $\sin^2$ are) and $\pi$-periodic ($\cos(2x)$ has period $\pi$, $\sin^2 x$ also has period $\pi$). Study domain : $[0, \pi/2]$.

Skills to practice

Proving parity and oddness
Finding a period

IV Affine transformations of the graph

Translating, scaling, or reflecting a function shifts its graph in predictable ways. Mastering this dictionary turns sketching from guesswork into reading-off. Watch the sign trap: $f(x+a)$ shifts the graph left by $a$, not right.

Theorem — Affine transformations of a graph

Let $f : E \to \mathbb{R}$, $a \in \mathbb{R}$, and $\lambda > 0$. The graphs below are deduced from $\Gamma_f$ as follows:

Reflections.
- $\textcolor{colorprop}{x \mapsto -f(x)}$ : reflection of $\Gamma_f$ about the $x$-axis ($Ox$).
- $\textcolor{colorprop}{x \mapsto f(-x)}$ : reflection of $\Gamma_f$ about the $y$-axis ($Oy$).
Translations.
- $\textcolor{colorprop}{x \mapsto f(x) + a}$ : translation of $\Gamma_f$ by vector $a \, \vec{\jmath}$ (vertical shift up by $a$).
- $\textcolor{colorprop}{x \mapsto f(x + a)}$ : translation of $\Gamma_f$ by vector $-a \, \vec{\imath}$ (horizontal shift LEFT by $a$).
Scalings.
- $\textcolor{colorprop}{x \mapsto \lambda f(x)}$ : vertical dilation by factor $\lambda$ ($\lambda > 1$ stretches, $\lambda < 1$ contracts).
- $\textcolor{colorprop}{x \mapsto f(\lambda x)}$ : horizontal contraction by factor $1/\lambda$ ($\lambda > 1$ contracts, $\lambda < 1$ stretches).

For $\lambda < 0$, combine the case $|\lambda|$ with the corresponding reflection.

Example

From $y = \sin x$ to $y = 1 - \dfrac{\sin(2x)}{2}$ in three steps : (1) horizontal contraction by factor $2$ giving $\sin(2x)$ ; (2) vertical contraction by factor $1/2$ giving $\sin(2x)/2$ ; (3) reflection about $Ox$ then translation by $\vec{\jmath}$ giving $1 - \sin(2x)/2$.

Example

For a fixed $a \in \mathbb{R}$ and a function $f : E \to \mathbb{R}$, identify the geometric link between $\Gamma_f$ and the graph of $x \mapsto f(a - x)$.

Answer

Write $f(a - x) = f(-(x - a))$, then read it as $f(-y)$ at $y = x - a$. Step 1: $y \mapsto f(-y)$ has graph symmetric to $\Gamma_f$ about $Oy$ (reflection). Step 2: $x \mapsto f(-(x - a))$ shifts the previous graph by $a$ to the right (translation by $+a \, \vec{\imath}$). Net effect: the graph of $x \mapsto f(a - x)$ is the reflection of $\Gamma_f$ about the vertical line $x = a/2$. Indeed, the points $x$ and $a - x$ are symmetric about $a/2$ since their midpoint is $(x + a - x)/2 = a/2$.

Example

Three elementary transformations: vertical translation, vertical dilation, reflection about $Ox$. Read each one as a single operation on the graph; the chain of transformations of the previous Example combines several of these in sequence.

Method — Sketching $g(x) \equal a + \lambda f(\mu x + b)$ from $\Gamma_f$

Algorithm. Write the inner argument as $\mu x + b = \mu(x + b/\mu)$. Apply the transformations in this order; reordering changes the result.

Horizontal scaling first. Starting from $\Gamma_f$, apply the transformation $x \mapsto \mu x$ (suppose $\mu > 0$ ; for $\mu < 0$ combine $|\mu|$ with a reflection about $Oy$). The point $(X, Y) \in \Gamma_f$ becomes $(X/\mu, Y)$ : horizontal contraction by $1/\mu$ if $\mu > 1$, dilation by $1/\mu$ if $0 < \mu < 1$. The new graph is that of $u : x \mapsto f(\mu x)$.
Horizontal translation. Apply $x \mapsto x + b/\mu$ : translate the previous graph by $-b/\mu \, \vec{\imath}$ (LEFT if $b/\mu > 0$). Indeed $u(x + b/\mu) = f(\mu(x + b/\mu)) = f(\mu x + b)$.
Vertical scaling. Multiply by $\lambda$ : vertical scaling by $\lambda$, with a reflection about $Ox$ if $\lambda < 0$.
Vertical translation. Add $a$ : translate by $a \, \vec{\jmath}$.

Trap. The horizontal translation step shifts left when $b/\mu > 0$, even though $b/\mu > 0$ visually feels like « right ».

Skills to practice

Sketching transformed graphs

V Monotonicity$\virgule$ bounded functions$\virgule$ extrema

Monotonicity tracks how the function moves with its input. Boundedness tracks how far the values can go. Maxima and minima identify specific values that are attained. Together they yield the rough shape of the graph and lay the groundwork for variation tables (next section). The vocabulary here is direct ; the proofs use only the order on $\mathbb{R}$ and basic logic.

Definition — Monotone$\virgule$ strictly monotone function

Let $E \subset \mathbb{R}$ and $f : E \to \mathbb{R}$.

$f$ is increasing on $E$ if $\forall x, y \in E, \ x \le y \implies f(x) \le f(y)$.
$f$ is strictly increasing on $E$ if $\forall x, y \in E, \ x < y \implies f(x) < f(y)$.
$f$ is decreasing on $E$ if $\forall x, y \in E, \ x \le y \implies f(x) \ge f(y)$.
$f$ is strictly decreasing on $E$ if $\forall x, y \in E, \ x < y \implies f(x) > f(y)$.
$f$ is (strictly) monotone on $E$ if it is (strictly) increasing or (strictly) decreasing on $E$.

Example

$f : x \mapsto x^2$ is strictly increasing on $\mathbb{R}_+$ and strictly decreasing on $\mathbb{R}_-$, but not monotone on $\mathbb{R}$. To prove the latter, we contradict both « increasing » and « decreasing ». $f$ is not increasing on $\mathbb{R}$ : $-1 < 0$ but $f(-1) = 1 > 0 = f(0)$. $f$ is not decreasing on $\mathbb{R}$ : $0 < 1$ but $f(0) = 0 < 1 = f(1)$. Hence $f$ is monotone in neither direction. The piecewise behaviour --- monotone on each side of $0$ but not globally --- is typical of even functions.

Example

The function $f : x \mapsto 1/x$ is strictly decreasing on $]0, +\infty[$ and on $]-\infty, 0[$ separately, but not monotone on the union $\mathbb{R}^* = ]-\infty, 0[ \,\cup\, ]0, +\infty[$. $f$ is not decreasing on $\mathbb{R}^*$ : $-1 < 1$ but $f(-1) = -1 < f(1) = 1$. $f$ is not increasing on $\mathbb{R}^*$ either : $1 < 2$ but $f(1) = 1 > 1/2 = f(2)$. Monotonicity on a union of intervals is not the same as monotonicity on each piece.

Proposition — Strict monotonicity $\Rightarrow$ injectivity

If $f : E \to \mathbb{R}$ is strictly monotone on $E$, then $f$ is injective on $E$ : $\textcolor{colorprop}{\forall x, y \in E, \ x \ne y \implies f(x) \ne f(y)}$.

Proof

Let $x, y \in E$ with $x \ne y$. Without loss of generality, $x < y$. By strict monotonicity, either $f(x) < f(y)$ (strictly increasing case) or $f(x) > f(y)$ (strictly decreasing case). In both cases $f(x) \ne f(y)$.

Proposition — Operations on monotone functions

Let $E \subset \mathbb{R}$ and $f, g : E \to \mathbb{R}$.

Sum. If $f$ and $g$ are increasing on $E$, then $\textcolor{colorprop}{f+g}$ is increasing on $E$. Same with « decreasing ».
Product (positive case). If $f$ and $g$ are non-negative and both increasing on $E$, then $\textcolor{colorprop}{fg}$ is increasing on $E$. Without the non-negativity hypothesis the result fails: $f(x) = g(x) = x$ are both increasing on $\mathbb{R}$, but $fg(x) = x^2$ is not.
Composition (rule of signs). For $f : E \to F$ and $g : F \to \mathbb{R}$ both monotone : $\textcolor{colorprop}{g \circ f}$ is increasing if $f$ and $g$ have the same sense of variation, decreasing if they have opposite senses of variation.

Proof

Sum. Let $x, y \in E$ with $x \le y$. By increase, $f(x) \le f(y)$ and $g(x) \le g(y)$ ; adding, $(f+g)(x) \le (f+g)(y)$.
Product (positive case). Let $x, y \in E$ with $x \le y$. By increase, $0 \le f(x) \le f(y)$ and $0 \le g(x) \le g(y)$. Multiplying these two inequalities of non-negative reals : $f(x) g(x) \le f(y) g(y)$.
Composition. Suppose both increasing. Let $x \le y$ in $E$. By increase of $f$, $f(x) \le f(y)$. Then by increase of $g$, $g(f(x)) \le g(f(y))$. The other three sign combinations follow analogously.

Definition — Bounded function$\virgule$ maximum$\virgule$ minimum

Let $f : E \to \mathbb{R}$.

$f$ is bounded above (majorée) if $\exists M \in \mathbb{R}, \forall x \in E, f(x) \le M$. Such $M$ is an upper bound of $f$.
$f$ is bounded below (minorée) if $\exists m \in \mathbb{R}, \forall x \in E, f(x) \ge m$.
$f$ is bounded if it is both bounded above and bounded below ; equivalently, $\exists K \ge 0, \forall x \in E, |f(x)| \le K$.
$f$ admits a maximum at $a \in E$ if $\forall x \in E, f(x) \le f(a)$. The value $f(a)$ is then $\max_E f$.
$f$ admits a minimum at $a \in E$ if $\forall x \in E, f(x) \ge f(a)$. The value $f(a)$ is then $\min_E f$.
Setting $f(E) = \{f(x) \mid x \in E\}$ : $\textcolor{colordef}{\sup_E f = \sup f(E)}$, $\textcolor{colordef}{\inf_E f = \inf f(E)}$ when these exist (i.e., when $f$ is bounded above / below).

Example

$\cos$ on $\mathbb{R}$ is bounded with $\max_\mathbb{R} \cos = 1$ (attained at $x = 0$ and at every $2k\pi$) and $\min_\mathbb{R} \cos = -1$ (attained at $x = \pi$ and at every $\pi + 2k\pi$). Both bounds are attained --- a typical situation for a bounded continuous function on a periodic domain.

Example

The function $f : x \mapsto \dfrac{1}{1+x^2}$ on $\mathbb{R}$ is bounded with $\sup_\mathbb{R} f = 1$ attained at $x = 0$ (so $\max_\mathbb{R} f = 1$) and $\inf_\mathbb{R} f = 0$ not attained (no $x \in \mathbb{R}$ satisfies $f(x) = 0$). The function $g : x \mapsto 1/x$ on $]0, +\infty[$ is bounded below by $0$ (in fact $\inf g = 0$, not attained) but not bounded above (it diverges as $x \to 0^+$).

Skills to practice

Proving monotonicity from the definition
Establishing boundedness

VI Derivative rules

The derivative is the limit of the slope of secants. Lycée established the rules ; we restate them with explicit hypotheses, add the inverse-function rule, and observe that derivability implies continuity. Heavy theorems (Rolle, mean-value theorem) and the formal proofs of all these rules are deferred to the chapter Differentiability.

Definition — Derivability$\virgule$ derived function$\virgule$ tangent

Let $I$ be an interval of $\mathbb{R}$ and $f : I \to \mathbb{R}$ a function.

For $a \in I$, $f$ is differentiable at $a$ (or derivable at $a$) if the limit $$ \textcolor{colordef}{f'(a) = \lim_{x \to a, \, x \ne a} \frac{f(x) - f(a)}{x - a}} $$ exists and is finite. The real $f'(a)$ is the derivative number of $f$ at $a$. Endpoint convention. If $a$ is an endpoint of $I$, the limit is taken relatively to $I$ (one-sided derivative : right-derivative if $a$ is the lower endpoint, left-derivative if $a$ is the upper endpoint).
$f$ is differentiable on $I$ if $f$ is differentiable at every $a \in I$. The map $f' : I \to \mathbb{R}$, $a \mapsto f'(a)$, is the derived function of $f$.
When $f$ is differentiable at $a$, the tangent to $\Gamma_f$ at $(a, f(a))$ is the line of equation $\textcolor{colordef}{y = f(a) + f'(a)(x - a)}$.

Example

The slope of the secant joining $(a, f(a))$ and $(x, f(x))$ is the difference quotient $\dfrac{f(x) - f(a)}{x - a}$. As $x \to a$, the secant tends to the tangent, of slope $f'(a)$.

Proposition — Differentiability $\Rightarrow$ continuity

Let $I$ be an interval and $f : I \to \mathbb{R}$. If $f$ is differentiable at $a \in I$, then $f$ is continuous at $a$.

Proof

For $x \in I$ with $x \ne a$, $$ f(x) - f(a) = \frac{f(x) - f(a)}{x - a} \cdot (x - a). $$ As $x \to a$, the first factor tends to $f'(a)$ (a finite real, by hypothesis) and the second factor tends to $0$. Hence $f(x) - f(a) \to 0$, i.e., $f(x) \to f(a)$ : $f$ is continuous at $a$.

Theorem — Operations on differentiable functions

Let $I$ be an interval, $f, g : I \to \mathbb{R}$ differentiable on $I$, and $\lambda, \mu \in \mathbb{R}$.

Linearity. $\lambda f + \mu g$ is differentiable on $I$ and $\textcolor{colorprop}{(\lambda f + \mu g)' = \lambda f' + \mu g'}$.
Product. $fg$ is differentiable on $I$ and $\textcolor{colorprop}{(fg)' = f'g + fg'}$.
Quotient. If $g$ does not vanish on $I$, $f/g$ is differentiable on $I$ and $\textcolor{colorprop}{\left(\dfrac{f}{g}\right)' = \dfrac{f' g - f g'}{g^2}}$.
Composition. Let $J$ be an interval, $f : I \to J$ differentiable on $I$, and $h : J \to \mathbb{R}$ differentiable on $J$. Then $h \circ f$ is differentiable on $I$ and $\textcolor{colorprop}{(h \circ f)' = (h' \circ f) \cdot f'}$.

(Statements ; rigorous proofs in the chapter Differentiability.)

Example

Sketch the derivation of the product rule $(fg)' = f'g + fg'$ from the difference quotient.

Answer

For $a \in I$ and $x \ne a$ in $I$, write the difference quotient of $fg$ using the algebraic identity $(fg)(x) - (fg)(a) = f(x) g(x) - f(a) g(a) = f(x)\bigl(g(x) - g(a)\bigr) + g(a)\bigl(f(x) - f(a)\bigr)$. Dividing by $x - a$, $$ \frac{(fg)(x) - (fg)(a)}{x - a} = f(x) \cdot \frac{g(x) - g(a)}{x - a} + g(a) \cdot \frac{f(x) - f(a)}{x - a}. $$ As $x \to a$ : $f(x) \to f(a)$ (by the Proposition « Differentiability $\Rightarrow$ continuity » above, since $f$ is differentiable at $a$, hence continuous at $a$) ; $\dfrac{g(x) - g(a)}{x - a} \to g'(a)$ ; $\dfrac{f(x) - f(a)}{x - a} \to f'(a)$. The right-hand side tends to $f(a) g'(a) + g(a) f'(a)$. Therefore $(fg)'(a) = f'(a) g(a) + f(a) g'(a)$.

Proposition — Derivative of the inverse function

Let $I$ be an interval, $f : I \to f(I)$ continuous, strictly monotone, and differentiable at $a \in I$ with $f'(a) \ne 0$. Then $f^{-1}$ is differentiable at $b = f(a)$ and $$ \textcolor{colorprop}{(f^{-1})'(b) = \frac{1}{f'(a)} = \frac{1}{f'(f^{-1}(b))}}. $$ (Admitted ; the continuity of $f^{-1}$ on the interval $f(I)$ relies on the strict-monotone TVI, proved in Limits and continuity, and the full proof is in Differentiability.)

Example

Compute $f'$ for $f : x \mapsto x \sin x$ on $\mathbb{R}$.

Answer

$f$ is the product of $u(x) = x$ (with $u'(x) = 1$) and $v(x) = \sin x$ (with $v'(x) = \cos x$). By the product rule, $$ f'(x) = u'(x) v(x) + u(x) v'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x. $$

Example

Compute $f'$ for $f : x \mapsto \sqrt{x^2 + 1}$ on $\mathbb{R}$.

Answer

Set $u(x) = x^2 + 1 > 0$ on $\mathbb{R}$ and $h(t) = \sqrt{t}$ on $\mathbb{R}_+^*$. Then $f = h \circ u$ with $u'(x) = 2x$ and $h'(t) = 1/(2\sqrt{t})$. By the chain rule, $$ f'(x) = h'(u(x)) \cdot u'(x) = \frac{1}{2 \sqrt{x^2 + 1}} \cdot 2x = \frac{x}{\sqrt{x^2 + 1}}. $$

Example

Compute $f'$ for $f : x \mapsto \dfrac{x+1}{x^2+3}$ on $\mathbb{R}$, then factorize.

Answer

$f$ is a quotient of two differentiable functions, with $g(x) = x^2 + 3 \ge 3 > 0$ on $\mathbb{R}$. The numerator $u(x) = x+1$ and $u'(x) = 1$ ; the denominator $g'(x) = 2x$. By the quotient rule, $$ \begin{aligned} f'(x) &= \frac{u'(x) g(x) - u(x) g'(x)}{g(x)^2} && \text{(quotient rule)} \\ &= \frac{1 \cdot (x^2 + 3) - (x+1) \cdot 2x}{(x^2 + 3)^2} && \text{(substitution)} \\ &= \frac{x^2 + 3 - 2x^2 - 2x}{(x^2 + 3)^2} && \text{(expansion)} \\ &= \frac{-x^2 - 2x + 3}{(x^2 + 3)^2} && \text{(simplification)} \\ &= \frac{-(x+3)(x-1)}{(x^2 + 3)^2} && \text{(factorization of $-x^2 - 2x + 3 = -(x-1)(x+3)$).} \end{aligned} $$

Method — Differentiating a quotient: factorize $f'$ to read its sign

Doctrine. The quotient rule yields $f' = (u'g - ug')/g^2$. The denominator $g^2$ is always non-negative, so the sign of $f'$ is the sign of the numerator $u'g - ug'$. Always factorize this numerator before reading signs : a factored form yields the sign at a glance, an expanded form usually does not. Recipe.

Apply the quotient rule.
Expand the numerator only as far as needed to combine like terms.
Factorize the numerator (look for a common factor, a discriminant, or a remarkable identity).
Read off the sign of $f'$ from the sign of the factored numerator.

Skills to practice

Computing derivatives
Using the inverse derivative formula

VII Sign of derivative and variation tables

The link between the sign of $f'$ and the monotonicity of $f$ is the practical engine of single-variable analysis. We state it here (admitting the proof, which uses the mean-value theorem) and use it to build variation tables. The crucial restriction is that the result holds on an interval ; on a union of intervals, one must apply the theorem on each piece separately.

Theorem — Sign of derivative $\Leftrightarrow$ monotonicity on an interval

Let $I$ be an interval and $f : I \to \mathbb{R}$ differentiable on $I$.

$f$ is \textcolor{colorprop}{constant on $I$} $\iff$ $f' = 0$ on $I$.
$f$ is \textcolor{colorprop}{increasing on $I$} $\iff$ $f' \ge 0$ on $I$.
$f$ is \textcolor{colorprop}{decreasing on $I$} $\iff$ $f' \le 0$ on $I$.

Strict-monotonicity corollary. If $f' \ge 0$ on $I$ and $f'$ is not identically zero on any non-trivial subinterval (in particular if $f' > 0$ except at finitely many points of $I$), then $f$ is \textcolor{colorprop}{strictly increasing on $I$}.
(Admitted ; the proof rests on the mean-value theorem, treated in Differentiability.)

Warning -- the domain must be an interval

The theorem above applies only on an interval. Counterexample : the function $f : x \mapsto 1/x$ on $\mathbb{R}^* = ]-\infty, 0[ \,\cup\, ]0, +\infty[$ has derivative $f'(x) = -1/x^2 < 0$ everywhere, yet $f$ is not decreasing on $\mathbb{R}^*$ : $f(-1) = -1 < f(1) = 1$ contradicts the definition of decreasing. The theorem applies on each interval $]-\infty, 0[$ and $]0, +\infty[$ separately, where $f$ is indeed strictly decreasing. Rule of thumb : when the domain is not an interval, split it into intervals before applying the sign-of-derivative theorem.

Example

Build the variation table of $f : x \mapsto \dfrac{x+1}{x^2 + 3}$ on $\mathbb{R}$.

Answer

The domain $\mathbb{R}$ is an interval. The derivative was computed in the previous Method's Example : $f'(x) = \dfrac{-x^2 - 2x + 3}{(x^2 + 3)^2} = \dfrac{-(x+3)(x-1)}{(x^2 + 3)^2}$. Since $(x^2 + 3)^2 > 0$, the sign of $f'$ is the sign of $-(x+3)(x-1)$. Roots at $x = -3$ and $x = 1$ ; signs: $$ \begin{array}{c|ccccccc} x & -\infty & & -3 & & 1 & & +\infty \\ \hline -(x+3)(x-1) & & - & 0 & + & 0 & - & \\ \hline f'(x) & & - & 0 & + & 0 & - & \\ \hline f(x) & 0 & \searrow & -\frac{1}{6} & \nearrow & \frac{1}{2} & \searrow & 0 \end{array} $$ We compute $f(-3) = (-3+1)/((-3)^2 + 3) = -2/12 = -1/6$ and $f(1) = 2/4 = 1/2$. The limits at $\pm\infty$ are $0$ since the degree of the denominator exceeds that of the numerator. Hence $f$ is strictly decreasing on $]-\infty, -3]$, strictly increasing on $[-3, 1]$, strictly decreasing on $[1, +\infty[$, with global minimum $-1/6$ at $x = -3$ and global maximum $1/2$ at $x = 1$.

Example

Deduce that $\dfrac{x+1}{x^2 + 3} \in \left[-\dfrac{1}{6}, \dfrac{1}{2}\right]$ for every $x \in \mathbb{R}$.

Answer

From the variation table, $f$ attains $\min_\mathbb{R} f = -1/6$ at $x = -3$ and $\max_\mathbb{R} f = 1/2$ at $x = 1$. Hence for every $x \in \mathbb{R}$, $-1/6 \le f(x) \le 1/2$, i.e., $f(x) \in [-1/6, 1/2]$.

Method — Building a variation table

Algorithm.

Determine the domain $D_f$. If $D_f$ is not an interval, split it into intervals before continuing.
Reduce the study domain via parity / periodicity if applicable.
Compute $f'$ on the interior of each interval.
Factorize $f'$ so the sign reads at a glance (cf. previous Method).
Build the sign table of $f'$ on each interval.
Deduce the monotonicity of $f$ on each interval (theorem above).
Compute the limits of $f$ at the endpoints of each interval.
Compose the synthesis table : line for $x$, line for the sign of $f'$, line for the variations of $f$ (arrows).

Skills to practice

Building a variation table
Proving inequalities by function study

VIII Higher-order derivatives$\virgule$ class $C^k$

Once differentiable, a function may admit a derivative that is itself differentiable, and so on. The iterative vocabulary --- $f^{(k)}$ for the $k$-th derivative, classes $C^k$ and $C^\infty$ --- structures the regularity hierarchy. We work on an interval $I$ throughout. The iterative definition rests on the previous-derivative being itself differentiable ; the class $C^k$ adds the continuity of $f^{(k)}$.

Definition — Successive derivatives

Let $I$ be an interval and $f : I \to \mathbb{R}$. We define the successive derivatives of $f$ by induction : $$ \textcolor{colordef}{f^{(0)} = f, \qquad f^{(k+1)} = (f^{(k)})' \text{ if } f^{(k)} \text{ is differentiable on } I.} $$ $f$ is $k$-times differentiable on $I$ if $f^{(k)}$ is well-defined on $I$. We write $f^{(1)} = f'$, $f^{(2)} = f''$, $f^{(3)} = f'''$ for the first three orders.

Definition — Class $C^k$ and class $C^\infty$

Let $I$ be an interval and $k \in \mathbb{N}$.

$f$ is of class $C^k$ on $I$, written $\textcolor{colordef}{f \in C^k(I, \mathbb{R})}$, if $f$ is $k$-times differentiable on $I$ and $f^{(k)}$ is continuous on $I$. (Here $C^0(I, \mathbb{R})$ denotes the set of functions continuous on $I$ ; the full theory of continuity is developed in Limits and continuity.)
$f$ is of class $C^\infty$ on $I$, written $\textcolor{colordef}{f \in C^\infty(I, \mathbb{R})}$, if $f \in C^k(I, \mathbb{R})$ for every $k \in \mathbb{N}$.

We have the inclusion chain $$ \textcolor{colordef}{C^\infty(I, \mathbb{R}) \subset \dots \subset C^2(I, \mathbb{R}) \subset C^1(I, \mathbb{R}) \subset C^0(I, \mathbb{R}) \subset \mathcal{F}(I, \mathbb{R}).} $$

Example

For $n \in \mathbb{N}$, the function $f : x \mapsto x^n$ is of class $C^\infty$ on $\mathbb{R}$, with successive derivatives : $$ f^{(k)}(x) = \begin{cases} \dfrac{n!}{(n-k)!} x^{n-k} & \text{if } k \le n, \\ 0 & \text{if } k > n. \end{cases} $$ In particular, after $n+1$ successive derivations, every polynomial of degree $\le n$ becomes the zero function.

Example

Compute the successive derivatives of $f : x \mapsto \dfrac{1}{x-1}$ on $I = ]1, +\infty[$.

Answer

Write $f(x) = (x-1)^{-1}$. Then $f'(x) = -(x-1)^{-2}$, $f''(x) = 2(x-1)^{-3}$, and by induction : $$ f^{(k)}(x) = (-1)^k k! \cdot (x-1)^{-(k+1)} = \frac{(-1)^k \, k!}{(x-1)^{k+1}}, \qquad k \in \mathbb{N}. $$ $f$ is of class $C^\infty$ on $I$ since each $f^{(k)}$ is continuous on $I$ (the denominator does not vanish there).

Proposition — Operations preserve class $C^k$

Let $I$ be an interval, $k \in \mathbb{N} \cup \{\infty\}$, and $f, g \in C^k(I, \mathbb{R})$. Then :

$\textcolor{colorprop}{\lambda f + \mu g \in C^k(I, \mathbb{R})}$ for all $\lambda, \mu \in \mathbb{R}$ (linearity) ;
$\textcolor{colorprop}{fg \in C^k(I, \mathbb{R})}$ (product) ;
if $g$ does not vanish on $I$, $\textcolor{colorprop}{f/g \in C^k(I, \mathbb{R})}$ (quotient) ;
if $J$ is an interval, $f : I \to J$ in $C^k(I, \mathbb{R})$, $h \in C^k(J, \mathbb{R})$, then $\textcolor{colorprop}{h \circ f \in C^k(I, \mathbb{R})}$ (composition).

(Admitted ; proofs in Differentiability.)

Example

Show that the inclusion $C^1(\mathbb{R}, \mathbb{R}) \subset C^0(\mathbb{R}, \mathbb{R})$ is strict by exhibiting a function in $C^0$ but not in $C^1$.

Answer

The function $f : x \mapsto |x|$ is continuous on $\mathbb{R}$, hence $f \in C^0(\mathbb{R}, \mathbb{R})$. It is differentiable on $\mathbb{R}^*$ with $f'(x) = 1$ if $x > 0$ and $f'(x) = -1$ if $x < 0$. At $x = 0$, the difference quotient $\dfrac{|h| - 0}{h - 0} = \dfrac{|h|}{h}$ equals $+1$ for $h > 0$ and $-1$ for $h < 0$, so the limit does not exist : $f$ is not differentiable at $0$. Hence $f \notin C^1(\mathbb{R}, \mathbb{R})$. The same construction generalizes : the family $f_k : x \mapsto x^k |x|$ is of class $C^k$ on $\mathbb{R}$ but not $C^{k+1}$, witnessing the strict inclusion $C^{k+1} \subsetneq C^k$ for every $k \ge 0$. For $k = 1$ : $g(x) = x|x|$ equals $x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$, hence $g'(x) = 2x$ for $x > 0$ and $g'(x) = -2x$ for $x < 0$ ; the difference quotient at $0$ gives $g'(0) = 0$, so $g'(x) = 2|x|$ on $\mathbb{R}$, which is continuous (so $g \in C^1$) but not differentiable at $0$ (so $g \notin C^2$). The pattern continues for higher $k$.

Inverse function and class $C^k$ -- forward reference

If $f \in C^k(I, \mathbb{R})$ is strictly monotone with $f' \ne 0$ on $I$, then $f^{-1} \in C^k(f(I), \mathbb{R})$ (admitted ; full proof in Differentiability). This result will be used systematically in the next chapter to establish the regularity of $\arccos$, $\arcsin$, $\arctan$, and $\ln$ as inverses of suitable restrictions of $\cos$ (to $[0, \pi]$), $\sin$ (to $[-\pi/2, \pi/2]$), $\tan$ (to $]-\pi/2, \pi/2[$), and $\exp$ (already bijective $\mathbb{R} \to \mathbb{R}_+^*$). Beware of the two roles of the endpoints : the bijection (hence the continuity of the inverse) needs the closed interval, but the hypothesis $f' \ne 0$ holds only on its open interior --- $\cos' = -\sin$ and $\sin' = \cos$ vanish at $0$, $\pi$, $\pm\pi/2$. So $\arccos$ and $\arcsin$ are continuous on $[-1, 1]$ but differentiable (indeed $C^\infty$) only on $]-1, 1[$, with vertical tangents at $\pm 1$.

Skills to practice

Computing successive derivatives