[PDF] Continuity and differentiability of a function

Continuity and differentiability of a function

PAUL MILAN 4 TERMINALE S 1 CONTINUITY OF A FUNCTION 1 6 Continuity and equation Theorem 3 : Intermediate value theorem (IVT) Let f be a continuous function on an

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LATEST EDITIONJUNE1, 2015AT17:37

Continuity and differentiability

of a function

Proofreading of English byLaurence Weinstock

Contents

1 Continuity of a function2

1.1 Finite limit at a finite point. . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Continuity at a point. . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 Continuity of common functions. . . . . . . . . . . . . . . . . . . . 3

1.4 Fixed point iteration theorem. . . . . . . . . . . . . . . . . . . . . . 3

1.5 Continuity and differentiability. . . . . . . . . . . . . . . . . . . . . 4

1.6 Continuity and equation. . . . . . . . . . . . . . . . . . . . . . . . . 5

2 Differentiability6

2.1 Definition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2 Interpretations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.2.1 Graphical interpretation. . . . . . . . . . . . . . . . . . . . . 8

2.2.2 Numerical interpretation. . . . . . . . . . . . . . . . . . . . 8

2.2.3 Kinematic interpretation. . . . . . . . . . . . . . . . . . . . 8

2.3 Monotonicity and sign of the derivative. . . . . . . . . . . . . . . . 9

2.4 Derivative and relative extrema. . . . . . . . . . . . . . . . . . . . . 9

2.5 Derivatives of common functions and derivative rules. . . . . . . . 11

2.5.1 Derivative of common functions. . . . . . . . . . . . . . . . 11

2.5.2 Derivative rules. . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.5.3 Examples. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

PAUL MILAN1TERMINALE S

1. CONTINUITY OF A FUNCTION

1 Continuity of a function

1.1 Finite limit at a finite point

Definition 1 :A functionfhas a

finite limit?ataif any open interval containing?contains all values off(x) by choosingxclose enough toa- i.e. all values ofxin an interval of width 2η centered ata,]a-η;a+η[. The limit is denoted : lim x→af(x) =? a a+ηa-ηC f O?? Note :Sometimes, the functionfdoes not have a limit ata, but does have a right- and left-hand limit such as the integer part function, denotedEin French andintin English (see below). For instance : limx→2-E(x) =1 and limx→2+E(x) =2

1.2 Continuity at a point

Definition 2 :Letfbe a function defined on an interval I. Letabe an element of I. The functionfis said to becontinuousataif, and only if : lim x→af(x) =f(a) The functionfis said to becontinuous on the interval Iif, and only if,fis continuous at each point in I. Note :Less formally, a function is continuous when its graph is a single unbro- ken curve i.e. you could draw it without lifting your pen from the paper. 123

1 2 3 4 5-1

]Cf O

Functionfdiscontinuous at 2

limx→2+f(x) =3?=f(2) 123

1 2 3 4 5-1

Cf O

Functionfcontinuous on[-1,5 ; 5,5]

On the left, the functionfhas a "jump" discontinuity. It is the case for example of the integer part function or more concretely of the function that represents the postal rates (abrupt rate change between letters below 20 g and for them between

20 g and 50 g in France).

PAUL MILAN2TERMINALE S

1. CONTINUITY OF A FUNCTION

There are other types of discontinuities. For example, the discontinuity at 0 of the functionfdefined byf(x) =sin1 xifx?=0 andf(0) =0. ?x?R,?n?Z,n?xTheinteger part functionEis defined by :

E(x) =n(int(x) =nin English)

E(2,4) =2 ;E(5) =5 ;E(-1,3) =-2

For any integer, there is a "jump" on the graph.

So the integer part function is not continuous if

xis an integer. 123
-1 -21 2 3 4-1-2 O

Letfbe the function defined by :???f(x) =sin1

xifx?=0 f(0) =0

The functionfis not continuous at 0 although

no "jump" can be observed here. The function varies increasingly around 0 so that near 0, the function oscillates more and more. We cannot say that asxapproaches 0 the function ap- proaches 0. 1 -11-1O

1.3 Continuity of common functions

Property 1 :The following results can be proven

•All polynomials are continuous onR. •The simple rational functionx?→1xis continuous on]-∞;0[and son]0;+∞[ •The absolute value functionx?→ |x|is continuous onR. •The square root functionx?→⎷xis continuous on[0;+∞[ •The sine and cosine functionsx?→sinxandx?→cosxare continuous onR •Generally speaking, all functions built by algebraic operation(addition, multi- plication) or by composition from the above functions are continuouson their domain, in particular the rational functions.

1.4 Fixed point iteration theorem

Theorem 1 :Fixed point iteration theorem

Let(un)be a sequence defined byu0and a recurrence relationun+1=f(un) converging to?. If the associated functionfis continuous at?, then the limit of the sequence,?, is a solution of the equationf(x) =x. It is said that?is a fixed point off

PAUL MILAN3TERMINALE S

1. CONTINUITY OF A FUNCTION

Proof :

We know that the sequence(un)converges to?so : limn→+∞un=? Moreover, the functionfis continuous at?so : limx→?f(x) =f(?)

By composition, we deduce that : lim

n→+∞f(un) =f(?)?limn→+∞un+1=f(?) but lim

Example :Let(un)be a sequence defined by

?u0=0 u n+1=? 3un+4 In chapter 2, we proved by induction that(un)is positive, increasing and boun- to?. The functionx?→⎷

3x+4 is continuous on[0;4], so?is solution to the

equationf(x) =x.

3x+4=xby squaring

3x+4=x2

x

2-3x-4=0

This equation has two solutions-1 and 4. As the sequence(un)is positive then, according to the fixed point theorem, the sequence(un)converges to 4.

1.5 Continuity and differentiability

Theorem 2 :Differentiability implies continuity

•Iffis differentiable at a pointathen the functionfis continuous ata. •Iffis differentiable on an interval I then the functionfis continuous on I. ?The converse of this theorem is false Note :The converse of this theorem is false. A graph can have no jump discon- tinuity but not have a tangent line at a finite point as in the followingexample :

This function therefore is continuous at

a(no jump), but is not differentiable at a. The graph has no tangent line ata.

The graph is said have a corner orcusp

at the point A A O a? The absolute value functionx?→ |x|is continuous but not differentiable at 0.

PAUL MILAN4TERMINALE S

1. CONTINUITY OF A FUNCTION

1.6 Continuity and equation

Theorem 3 :Intermediate value theorem (IVT)

Letfbe acontinuousfunction on an interval I= [a,b]. For any real numberkbetweenf(a)andf(b), there must be at least one value c?I such thatf(c) =k. Note :The proof of this theorem is not on the syllabus.

This theorem follows from the fact that

theimage of an intervalof a conti- nuous function over an interval ofRis itself an intervalofR

Consider the following graph :kis bet-

weenf(a)andf(b). Note the wording of the theorem "at least one value". This means we could have more. Here for example, we have

3 points wheref(x) =k. The equa-

tionf(x) =ktherefore has 3 solutions : c

1,c2andc3.

abf(a) f(b)k c

1c2c3O

Theorem 4 :IVT with continuous one-to-one functions Letfbe acontinuous and strictly monotonicfunction onI= [a,b]. For anykbetweenf(a)andf(b), the equationf(x) =khas a unique solution in the interval I= [a,b] Proof :The existence is proven by the precedent theorem, and the uniqueness by the monotonicity of the function.

Note :

•This theorem is generalized with an open intervalI=]a,b[.kmust be between limx→af(x)and limx→bf(x) •Whenk=0, we then have to showf(a)×f(b)<0. •Sometimes, in French, the theorem is called the "bijective theorem" because the function is bijective from I tof(I). Example :Letfbe a function defined onRby :f(x) =x3+x-1. Show that the equationf(x) =0 has only one solution onR. Give an approximation to within one unit. Then find , with an algorithm, an approximate value within 10 -6of the solution.

PAUL MILAN5TERMINALE S

2. DIFFERENTIABILITY

123
-1 -20.5 1.0 1.5 Oα

The functionfis acontinuousfunction onRbe-

causefis a polynomial.

The functionfis the sum of increasing functions

x?→x3andx?→x-1, sofisstrictly increasing onR.

We havef(0)=-1 andf(1)=1?f(0)×f(1)<0

hence by the intermediate value theorem, the func- tionfhave an only one solutionα?[0,1]such that f(α) =0.

Algorithm :An algorithm, using thedicho-

tomyprinciple (the interval is divided in two and the operation is repeated) one can find a value ofαrounded to within 10-6.

Consider :

•AandBendpoints of the interval. •Pthe accuracy (natural number). •Nthe number of iterations.

Input :A=0,B=1,P=6 and

f(x) =x3+x-1

We find :A=0,682327,B=0,682328 and

N=20.

20 iterations are required to have an accuracy of

10 -6

Variables:A,B,Creal numbers

P,Nintegersffunction

Inputs and initialization

ReadA,B,P

0→N

Processing

whileB-A>10-Pdo A+B

2→C

iff(A)×f(C)>0(*)then

C→A

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