Continuity of functions

Definition. Let $f:D\to \mathbb{R}$, with ${{x}_{0}}$ an accumulation point of D. f is continuous at ${{x}_{0}}$ if $\displaystyle \underset{x\to {{x}_{0}}}{\mathop{\lim }}\,f(x)=f({{x}_{0}})$; ${{x}_{0}}$ is called a point of continuity.

Definition Let $\alpha \in D$. $\alpha$ is a point of discontinuity of the first kind if the one-sided limits at $\alpha$ exist and are finite, but the function is not continuous at $\alpha$.

Definition Let $\alpha \in D$. $\alpha$ is a point of discontinuity of the second kind if it is not of the first kind.

Theorem. If $I$ is an interval and f is continuous on $I$, then $J=f(I)$ is an interval (a function continuous on an interval has the Darboux property on that interval).

Differentiable functions

Definition of the derivative at a point. Let $f:E\to \mathbb{R}$, ${{x}_{0}}\in E$, with ${{x}_{0}}$ an accumulation point of $E$, $\displaystyle {f}'({{x}_{0}})=\underset{x\to {{x}_{0}}}{\mathop{\lim }}\,\frac{f(x)-f({{x}_{0}})}{x-{{x}_{0}}}=\underset{\begin{smallmatrix} h\to 0 \\ {{x}_{0}}+h\in E \end{smallmatrix}}{\mathop{\lim }}\,\frac{f({{x}_{0}}+h)-f({{x}_{0}})}{h};$

$\displaystyle {{{f}'}_{s}}({{x}_{0}})=\underset{\begin{smallmatrix} x\to {{x}_{0}} \\ x<{{x}_{0}} \end{smallmatrix}}{\mathop{\lim }}\,\frac{f(x)-f({{x}_{0}})}{x-{{x}_{0}}}, {{{f}'}_{d}}({{x}_{0}})=\underset{\begin{smallmatrix} x\to {{x}_{0}} \\ x>{{x}_{0}} \end{smallmatrix}}{\mathop{\lim }}\,\frac{f(x)-f({{x}_{0}})}{x-{{x}_{0}}};$

${f}'({{x}_{0}})={{{f}'}_{s}}({{x}_{0}})={{{f}'}_{d}}({{x}_{0}}).$

Geometric interpretation:

1. if ${f}'({{x}_{0}})\in \mathbb{R}$, $y-f({{x}_{0}})={f}'({{x}_{0}})(x-{{x}_{0}})$ is the equation of the tangent to the graph of the function $f$ at the point $A({{x}_{0}},f({{x}_{0}}));$

2. if $f$ is continuous at ${{x}_{0}}$, ${{{f}'}_{d}}({{x}_{0}})=+\infty , {{{f}'}_{s}}({{x}_{0}})=-\infty$ or conversely, ${{x}_{0}}$ is a cusp of the graph;

3. if $f$ is continuous at ${{x}_{0}}$ and the one-sided derivatives at ${{x}_{0}}$ exist, at least one being finite, but $f$ is not differentiable at ${{x}_{0}}$, then ${{x}_{0}}$ is a corner point of the graph.

Rules of differentiation

$f,g:E\to \mathbb{R},$ $f,g$ differentiable at $x\in E$, then:

1. ${{\left( f+g \right)}^{\prime }}(x)={f}'(x)+{g}'(x);$

2. ${{\left( cf \right)}^{\prime }}(x)=c{f}'(x), c\in \mathbb{R};$

3. ${{\left( f\cdot g \right)}^{\prime }}(x)={f}'(x)\cdot g(x)+f(x)\cdot {g}'(x);$

4. if $g(x)\ne 0$, ${{\left( \frac{f}{g} \right)}^{\prime }}(x)=\frac{{f}'(x)\cdot g(x)-f(x)\cdot {g}'(x)}{{{g}^{2}}(x)};$

5. if $f:I\to J, g:J\to \mathbb{R},$ $f$ is differentiable at ${{x}_{0}}\in I$ and $g$ is differentiable at ${{y}_{0}}=f({{x}_{0}})$, then ${{\left( g\circ f \right)}^{\prime }}({{x}_{0}})={g}'({{y}_{0}})\cdot {f}'({{x}_{0}});$

6. if $f:I\to J,$ is continuous, bijective and differentiable at ${{x}_{0}}$ with ${f}'({{x}_{0}})\ne 0$, then ${{f}^{-1}}:J\to I,$ is differentiable at ${{y}_{0}}$, ${{y}_{0}}=f({{x}_{0}})$ and ${{\left( {{f}^{-1}} \right)}^{\prime }}({{y}_{0}})=\frac{1}{{f}'({{x}_{0}})}.$

Derivatives of the elementary functions

Elementary function

Composite function

Domain of definition

${c}'=0\text{ }(c=\text{constant }\!\!\breve{\mathrm{a}}\!\! )$

$R$

${x}'=1$

$R$

$({{x}^{n}}{)}'=n{{x}^{n-1}}\text{, }(n\in {{N}^{*}})$

${{\left( {{u}^{n}} \right)}^{\prime }}=n\cdot {{u}^{n-1}}\cdot {u}'$

$R$

${{\left( \frac{1}{x} \right)}^{\prime }}=-\frac{1}{{{x}^{2}}}$

${{\left( \frac{1}{u} \right)}^{\prime }}=-\frac{{{u}'}}{{{u}^{2}}}$

$R*$

${{\left( \sqrt[n]{x} \right)}^{\prime }}=\frac{1}{n\cdot \sqrt[n]{{{x}^{n-1}}}}$

${{\left( \sqrt[n]{u} \right)}^{\prime }}=\frac{{{u}'}}{n\cdot \sqrt[n]{{{u}^{n-1}}}}, n\in {{\mathbb{N}}^{*}}$

$(0,\infty )$

${{\left( \sqrt{x} \right)}^{\prime }}=\frac{1}{2\sqrt{x}}$

${{\left( \sqrt{u} \right)}^{\prime }}=\frac{{{u}'}}{2\sqrt{u}}$

$(0,\infty )$

${{\left( {{e}^{x}} \right)}^{\prime }}={{e}^{x}}$

${{\left( {{e}^{u}} \right)}^{\prime }}={{e}^{u}}\cdot {{u}^{'}}$

$R$

${{\left( {{a}^{x}} \right)}^{\prime }}={{a}^{x}}\ln a, 0<a\ne 1$

${{a}^{u}}={{a}^{u}}\cdot {u}'\cdot \ln a$

$R$

${{\left( \ln x \right)}^{\prime }}=\frac{1}{x}$

${{\left( \ln \left( u \right) \right)}^{\prime }}=\frac{u'}{u}$

$(0,\infty )$

${{\left( {{\log }_{a}}x \right)}^{\prime }}=\frac{1}{x\ln a}, 0<a\ne 1$

${{\left( {{\log }_{a}}u \right)}^{\prime }}=\frac{{{u}'}}{u\ln a}$

$(0,\infty )$

${{\left( \sin x \right)}^{\prime }}=\cos x$

$\sin (u)=\cos (u)\cdot {u}'$

$R$

${{\left( \cos x \right)}^{\prime }}=-\sin x$

$(\cos (u){)}'=-\sin (u)\cdot {u}'$

$R$

${{\left( tgx \right)}^{\prime }}=\frac{1}{{{\cos }^{2}}x}$

${{\left( tg\left( u \right) \right)}^{\prime }}=\frac{{{u}'}}{{{\cos }^{2}}u}$

$\cos x\ne 0$

${{\left( ctgx \right)}^{\prime }}=-\frac{1}{{{\sin }^{2}}x}$

${{\left( ctg\left( u \right) \right)}^{\prime }}=-\frac{{{u}'}}{{{\sin }^{2}}u}$

$\sin x\ne 0$

${{\left( \arcsin \left( x \right) \right)}^{\prime }}=\frac{1}{\sqrt{1-{{x}^{2}}}}$

$(\arcsin (u){)}'=\frac{{{u}'}}{\sqrt{1-{{u}^{2}}}}$

$(-1,1)$

${{\left( \arccos x \right)}^{\prime }}=-\frac{1}{\sqrt{1-{{x}^{2}}}}$

${{\left( \arccos u \right)}^{\prime }}=-\frac{{{u}'}}{\sqrt{1-{{u}^{2}}}}$

$(-1,1)$

${{\left( arctgx \right)}^{\prime }}=\frac{1}{1+{{x}^{2}}}$

${{\left( arctgu \right)}^{\prime }}=\frac{{{u}'}}{1+{{u}^{2}}}$

$R$

${{\left( arcctg\left( x \right) \right)}^{\prime }}=-\frac{1}{1+{{x}^{2}}}$

${{\left( arcctg\left( u \right) \right)}^{\prime }}=-\frac{{{u}'}}{1+{{u}^{2}}}$

$R$

Higher-order derivatives

The function $f:D\to \mathbb{R}$ is (n+1) times differentiable at x₀$\in D\cap D'$ if:

1. there exists V$\in v({{x}_{0}})$ such that V$\subset$ ${{D}_{{{f}^{(n)}}}}$_;

2. the derivative function ${{f}^{(n)}}:{{D}_{{{f}^{(n)}}}}\to \mathbb{R}$ is differentiable at x₀;

In this case the limit is written ${{f}^{(n+1)}}({{x}_{0}})\text{ }(sau\frac{{{d}^{n+1}}f({{x}_{0}})}{d{{x}^{n+1}}} sau {{D}^{n+1}}f({{x}_{0}}))$ and is called the derivative of order (n+1) of the function f at the point x₀.

Leibniz's rule

Given the functions $f,g:D\to \mathbb{R}$ that are n times differentiable, $\displaystyle {{(fg)}^{(n)}}=\sum\limits_{k=0}^{n}{{{C}_{n}}^{k}{{f}^{(n-k)}}{{g}^{(k)}}.}$

General properties of differentiable functions

Let $f:D\to \mathbb{R}$ be a function; x₀$\in D$ is called a point of absolute (global) minimum of the function f if: f(x₀)≤ f(x), $\forall x\in D$. x₀$\in D$ is called a point of absolute (or global) maximum of the function f if: f(x)≤ f(x₀), x₀$\in D$_.

Let $f:I\to \mathbb{R}$ be a function and ${{x}_{0}}\in I$.

x₀ is called a point of absolute (global) minimum of the function f if there is a neighbourhood V of x₀ such that f(x)≤ f(x₀), $\forall x\in V\cap I$; f(x₀) is called a relative maximum of the function.

x₀ is called a point of relative (or local) minimum of the function f if there is a neighbourhood V of x₀ such that f(x)≤ f(x₀), $\forall x\in V\cap I$; f(x₀) is called a relative minimum of the function.

${{x}_{0}}\in I$ is a point of relative (or local) extremum of f if it is a point of relative (or local) minimum or maximum.

Fermat's theorem

Let $I\subset \mathbb{R}$ be an interval and $f:I\to \mathbb{R}$ a function differentiable at a point of local extremum x₀ in the interior of the interval I (${{x}_{0}}\in I$ and not an endpoint of the interval); then $f'({{x}_{0}})=0$.

Geometric interpretation. If the graph of the function has a tangent at a point of extremum which does not coincide with the endpoints of the graph, then the tangent at that point is parallel to the Ox axis.

Rolle's theorem. Let $f:[a,b]\to \mathbb{R}$ be a function with the following properties:

1. f is continuous on [a,b]

2. f is differentiable on (a,b)

3. f(a)=f(b)

Then $\exists c\in (a,b)$ such that $f'(c)=0$.

Geometric interpretation Let A(a, f(a)), B(b,f(b)); since f(a)=f(b) it follows that AB $\parallel$Ox.

If the graph of the function f has a tangent at every point (with the possible exception of the endpoints A and B) and the line joining the endpoints A, B is parallel to Ox, then there is at least one point of the graph at which the tangent is parallel to the Ox axis.

Lagrange's theorem (the mean value theorem). Let $f:[a,b]\to \mathbb{R}$ be a function with the properties:

1. f is continuous on [a,b]

2. f is differentiable on (a,b)

Then $\exists c\in (a,b)$ such that $\frac{f(a)-f(b)}{b-a}=f'(c)$.

Geometric interpretation. If the graph of f has a tangent at every point (with the possible exception of the endpoints), there is at least one point on the graph at which the tangent is parallel to the chord [AB] joining the endpoints.

Consequences of Lagrange's theorem

1. If $f:I\to \mathbb{R}$ is differentiable and $f'(x)=0$, $\forall x\in I$, then f is constant on I.

2. If $f,g:I\to \mathbb{R}$ are differentiable on the interval I and $f'=g'$ then

g-f=constant.

3. Monotonicity of functions. A function $f:I\to \mathbb{R}$ differentiable on the interval I with $f'>0$ is strictly increasing, and if $f'<0$ then f is strictly increasing.

Cauchy's theorem. If the functions $f,g:[a,b]\to \mathbb{R}$ satisfy the conditions:

1. f and g are continuous on the closed interval [a,b];

2. f and g are differentiable on the open interval (a,b);

3. $g'(x)\ne 0,\forall x\in (a,b)$;

Then g(a)≠g(b) and $\exists c\in (a,b)$ such that $\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}$.

Darboux's theorem. If f is a function differentiable on the interval I, then $f'$ has the Darboux property on I.

L'Hôpital's theorem. Let a,b $\in \overline{\mathbb{R}}$, a<b and $I\subset \mathbb{R}$ an interval with (a,b)$\subset I\subset [a,b]$. If ${{x}_{0}}[a,b]$ and $f,g:I\backslash \{{{x}_{0}}\}\to \mathbb{R}$ are functions with the properties:

1) $\displaystyle \underset{x\to {{x}_{0}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{x}_{0}}}{\mathop{\lim }}\,g(x)=0$ (respectively $\displaystyle \underset{x\to {{x}_{0}}}{\mathop{\lim }}\,\left| g(x) \right|=+\infty$);

2) f and g are differentiable and $g'(x)\ne 0,\forall x\in I\backslash \{{{x}_{0}}\}$;

3) there exists $\displaystyle \underset{x\to {{x}_{0}}}{\mathop{\lim }}\,\frac{f'(x)}{g'(x)}\in \overline{\mathbb{R}}$, finite or infinite;

Then $g(x)\ne 0,\text{ (}\forall )x\in I\backslash \{{{x}_{0}}\}$ and there exists $\displaystyle \underset{x\to {{x}_{0}}}{\mathop{\lim }}\,\frac{f(x)}{g(x)}=\underset{x\to {{x}_{0}}}{\mathop{\lim }}\,\frac{f'(x)}{g'(x)}$.

Real functions. Introductory notions

Let E and F be two sets. We say a function has been defined on E with values in F if to each element x∈E there corresponds one and only one element y∈F. A function is the whole formed by the sets E and F together with the correspondence from the elements of E to the elements of F. The set E is called the domain of definition of the function, and the set F is called the set in which the function takes its values (the codomain).

A function may be written f:E→F. A generic element x of the domain E is called an argument or variable of the function f. The element of F corresponding to an element x∈E through the function f is written f(x) and is called the image of x under f, or the value of the function f at x.

Sketching the graph of a function

To sketch the graph of a twice-differentiable function f, proceed as follows:

1. Determine the maximal domain of definition:

1. for rational expressions, the denominator of the fraction must be non-zero;

2. the quantity under a radical of even index must be greater than or equal to zero;

3. the base of an exponential function must be strictly positive;

4. the arcsine and arccosine functions must be defined on [-1,1];

5. the number to which the logarithm is applied must be strictly positive, and the base of the logarithm must be strictly positive and different from 1.

2. Make explicit the functions: modulus, maximum, minimum, signum, integer part and fractional part (if the function contains them).

3. Determine whether the function is even or odd:

1. if the function is even $f(-x)=f(x)$, then the graph of the function is symmetric about the axis of ordinates (the Oy axis);

1. if the function is odd $f(-x)=-f(x)$, then the graph of the function is symmetric about the origin; so it is enough to sketch the graph on the positive Ox semi-axis and then reflect it;

3. the graph of a function $f$ is symmetric about the line if $f(2a-x)=f(x)$ I is symmetric about the point $(a,0)$ if $f(2a-x)=-f(x)$;

4. if the function is periodic with period p, it is enough to draw the graph of the function for x∈E belonging to an interval of length p, since the graph repeats.

4. Determine the intersections with the coordinate axes:

1. $y=0\Rightarrow f(x)=0$, and if the solutions of the equation $f(x)=0$ exist, then the graph meets the $Ox$ axis at the points of the form $({{x}_{1}},0), ({{x}_{2}},0),...$

2. $x=0\Rightarrow y=f(0)\Rightarrow$ the point at which the graph meets the axis of ordinates, of the form $(0,f(0))$.

5. Determine the limits at the ends of the intervals and the continuity of the function

6. Determine the asymptotes:

1. vertical. Vertical asymptotes are defined for unbounded functions, even if they are defined on bounded sets. They should be looked for at the points of discontinuity of the function, that is at the points where the function f is not defined.

1. oblique. If $\displaystyle \underset{x\to +\infty }{\mathop{\lim} }\,f(x)$ $\displaystyle \left( \underset{x\to -\infty }{\mathop{\lim} }\,f(x) \right)$ exists and equals $+\infty$ $\left( -\infty \right)$, then compute, if it exists, the limit $\displaystyle m=\underset{x\to \infty }{\mathop{\lim} }\,\frac{f(x)}{x},$ $\displaystyle \left( {m}'=\underset{x\to -\infty }{\mathop{\lim} }\,\frac{f(x)}{x} \right).$ If m (respectively ${m}'$) is finite, compute, when it exists, $\displaystyle n=\underset{x\to \infty }{\mathop{\lim} }\,\left[ f(x)-mx \right]$ $\displaystyle \left( {n}'=\underset{x\to -\infty }{\mathop{\lim} }\,\left[ f(x)-{m}'x \right] \right).$ If n (respectively ${n}'$) is finite, then the line with equation $y=mx+n$ (respectively $y={m}'x+{n}'$) is an oblique asymptote to the branch towards $+\infty$ $\left( \text{respectiv }-\infty \right)$ of the graph of the function.

Notes:

1. if m exists and is finite, but n does not exist or is infinite, the graph of the function has no oblique asymptote at ±∞;

2. if m does not exist or is infinite, the graph of the function has no oblique asymptote at ±∞.

3. horizontal. If $\displaystyle \underset{x\to \pm \infty }{\mathop{\lim} }\,f(x)=a$ exists and is finite, then the line y=a is an asymptote at ±∞, parallel to the Ox axis.

Notes:

1. If the graph has a horizontal asymptote, then it cannot also have an oblique asymptote at ±∞, and conversely;

2. For periodic functions, a graph may have infinitely many vertical asymptotes;

3. There may be horizontal asymptotes towards ±∞ and oblique ones towards ±∞;

4. For the inverse circular functions, the graph may have infinitely many horizontal asymptotes;

5. If the line y=a is a horizontal asymptote of the graph of the function f, then the distance between the graph and the asymptote, measured vertically, decreases continually as the point on the graph moves away.

7. The first derivative

1. Determine the set of points at which f is differentiable,

2. Compute ${f}'$. At the points where the function is defined but not differentiable, determine (if they exist) the left derivative ${{{f}'}_{s}}$ and the right derivative ${{{f}'}_{d}}$, in order to establish the one-sided tangents.

3. Determine the real roots of the equation ${f}'(x)=0$ and draw up the sign table of the derivative ${f}'$.

4. Determine the intervals of monotonicity and the points of local extremum of the function:

   • If ${f}'(x)>0$, for $x\in (a,b)$, then f is strictly increasing on $(a,b)$.

   • If ${f}'(x)>0$ for $x\in (c,d)$, then f is strictly decreasing on $(c,d)$.

   • If ${f}'(x)=0$ for $x\in (e,g)$, then f is constant on $(e,g)$.

5. If f is continuous at ${{x}_{0}}\in E$ and there is an interval $(a,b)$ containing ${{x}_{0}}$ such that f is increasing (decreasing) on $(a,{{x}_{0}})\cap E$ and decreasing (increasing) on $({{x}_{0}},b)\cap E$, then ${{x}_{0}}$ is a point of local maximum (minimum) of f relative to E.

8. The second derivative

   1. Compute the second derivative, ${f}''$.

   2. Solve the equation ${f}''(x)=0.$ The roots of this equation will be possible points of inflection.

   3. Determine the intervals on which the second derivative keeps a constant sign.

      • If the second derivative ${f}''$ is strictly positive (+) on an interval $I\subset E$, then the function is convex on $I$.

      • If the second derivative ${f}''$ is strictly negative (-) on an interval $I\subset E$, then the function is concave on $I$.

      • If the second derivative ${f}''$ does not vanish on an interval $I\subset E$, then ${f}''$ keeps the same sign on $I$.

9. Draw up a table of values of the function f – a table in which, for the sake of order, the results obtained above are recorded:

• In the first row go the notable values of $x$ obtained earlier;

• In the second row go the corresponding values of the first derivative, ${f}'(x)$, and the sign of ${f}'$;

• In the third row go the corresponding values of the function $f(x)$, together with the arrows $\nearrow \text{ sau }\searrow$ marking the increase or decrease of the function;

• In the fourth row go the corresponding values of the second derivative, ${f}''(x)$, and the sign of ${f}''$.

$x$	$-\infty$	$0$	$\infty$
${f}'(x)$
$f(x)$
${f}''(x)$

10. Sketch the graph of the function

The notable points $(x,f(x))$, which follow from the table, are then plotted in the plane $xOy$. The asymptotes are drawn if they exist.

Taking into account what the first and second derivatives indicate, the points are joined by a curve.

Standard Taylor series

The expansion of the function $f(x)$ as a Taylor series at the point $x={{x}_{0}}$ has the following general form: $\displaystyle f(x)=f({{x}_{0}})+\sum\limits_{n\ge 1}{\frac{{{f}^{(n)}}({{x}_{0}})}{n!}\cdot {{(x-{{x}_{0}})}^{n}}}$.

1. $f(x)={{e}^{x}}=1+\frac{x}{1!}+\frac{{{x}^{2}}}{2!}+...+\frac{{{x}^{n}}}{n!}+...$

2. $f(x)=\cos (x)=1-\frac{{{x}^{2}}}{2!}+\frac{{{x}^{4}}}{4!}-\frac{{{x}^{6}}}{6!}+...$

3. $f(x)=\sin (x)=\frac{x}{1!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-...$