EMINDER - 2 IMITS AND CONTINUITY€¦ · 2MG Level 1 CALCULUS I Theory Lycée Denis-de-Rougemont...

2MG Level 1 CALCULUS I Theory Lycée Denis-de-Rougemont 2006 – 2007

- 1 - YAy & PRi

TABLE OF CONTENTS

§0 REMINDER __________________________________________ - 2 -

§1 LIMITS AND CONTINUITY _______________________________ - 5 -

1.1 Limits ______________________________________________________________- 5 -

1.2 Limit at infinity of a rational function ___________________________________- 8 -

1.3 Limit of a rational function ____________________________________________- 9 -

1.4 Continuity _________________________________________________________- 11 -

§2 ASYMPTOTES _______________________________________ - 12 -

2.1 Vertical asymptote or “hole”__________________________________________- 12 -

2.2 Horizontal asymptote ________________________________________________- 13 -

2.3 Slant or oblique asymptote____________________________________________- 14 -

2.4 How does the function approach its asymptotes__________________________- 15 -

2.5 Limits of trigonometric functions ______________________________________- 17 -

§3 DIFFERENTIATION ___________________________________ - 19 -

3.1 Derivative__________________________________________________________- 19 -

3.2 Gradient of the tangent to curve at a given point_________________________- 21 -

3.3 The derivative of the powers of x ______________________________________- 23 -

3.4 Differentiation rules _________________________________________________- 23 -

3.5 Examples___________________________________________________________- 26 -

§4 USES OF THE DERIVATIVE _____________________________ - 27 -

5.1 Increasing and decreasing functions ____________________________________- 27 -

5.2 Maxima and minima _________________________________________________- 30 -

§5 STUDY OF A FUNCTION________________________________ - 31 -

§6 OPTIMIZATION ______________________________________ - 34 -

§7 GRAPHING THE DERIVATIVE FUNCTION __________________ - 35 -


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§0 REMINDER

A function from A to B is a relation such that each element of A is associated to exactly

one element or image of B :

4. DOMAIN AND RANGE

The domain of a function ( )f x , written D or fD , is the set containing the values of x for

which the function is defined, that means whose ( )f x exists.

The range of a function is the set containing all the possible outputs y. In fact, it is ( )f D .

Examples

1

( )f xx

= The domain is * \ {0}D = =� � because the division by 0 is forbidden.

The range is *( ) \ {0}f D = =� � , because 1

yx

= can’t be equal to 0.

( ) 4f x x= − [ 4; [D = ∞ because the square root isn’t defined for negative numbers

( ) [0; [f D = ∞ because a square root is always positive or equal to 0.

5. PARITY OF A FUNCTION

Even and odd functions are functions which satisfy particular relations of symmetry.

A function whose graph is symmetrical about the y-axis is called an even function. A

function f is even if ( ) ( )f x f x− = for all real x of the domain fD .

A function whose graph is symmetrical about the origin is called an odd function.

A function f is odd if ( ) ( )f x f x− = − for all real x of the domain fD . Many functions are

neither odd nor even !

Examples

( ) cos( )f x x= is an even function because ( ) cos( ) cos( ) ( )f x x x f x− = − = =

Symmetrical about the y-axis

domain range

:

( )

f A B

x y f x

→

→ =


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3( ) 2f x x x= + is an odd function because :

( )3 3 3( ) ( ) 2 ( ) 2 2 ( )f x x x x x x x f x− = − + ⋅ − = − − = − + = −

Symmetrical about the origin

( ) 4 7f x x= + is neither odd nor even (we say else) because :

4 7 ( ) not even( ) 4 ( ) 7 4 7

4 7 ( ) not oddx f x

f x x xx f x

≠ + =− = ⋅ − + = − +

≠ − − = −

6. ZEROS (ROOTS), EXCLUDED VALUES AND SIGN OF A FUNCTION

Before plotting the function, we look for its possible zeros and excluded values. Then

we establish its signs.

The zeros of a function are the inputs x in D that satisfy ( ) 0f x = . These are the

intersection points with the x-axis.

The excluded values of a function are the inputs x in D for which the function is not

defined.

The sign of a function can be « + », « 0 » or « – ». As the sign can only change after a

zero or an excluded value, we look for them and we solve ( ) 0f x = to find the zeros.

After that the signs are arranged in a table of signs.

Example

4 ( )( )

3 1 ( )

x g xf x

x h x

−= =

− so

( ) 4 0 4 ZERO

1( ) 3 1 0 EXCLUDED

3

g x x x

h x x x

= − = ⇒ =

= − = ⇒ = �

(4) 01

\3

f

D

=

= �


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Table of signs

7. PERIODIC FUNCTIONS AND PERIOD OF A FUNCTION

Among the functions that have been met so far, some have a graph that is repeated :

the entire graph can be formed from copies of one particular portion, repeated at

regular intervals. One speaks about periodic functions. Some examples are cosine,

sine and tangent and their linear combinations.

The period of a function is a number T such that ( ) ( )f x T f x+ = for all x of fD .

Graphically, the period is the smallest horizontal and constant shift between two

identical outputs y. The sum of periodic functions remains periodic. Its period is the

LCM (Lowest Common Multiple) (= PPMC) of the periods involved in the sum.

Example

Here is the graph of a periodic function. Its period is represented by the arrows.

Concrete example

The period of ( ) cos( )f x x= is 2T π= because cos( 2 ) cos( )x xπ+ = . The one of

( ) cos(2 )f x x= is T π= (the graph oscillates twice faster than ( ) cos( )f x x= )

8. LONG DIVISION

When a polynomial p(x) is divided by a non-constant divisor, d(x), the quotient q(x)

and the remainder r(x) are defined by the equality :

( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

p x r xp x d x q x r x q x

d x d x= + ⇔ = +

where the degree of the remainder is less than the degree of d(x).

1

3 4

g + + + 0 –

h – 0 + + +

f – Imp. + 0 – 4

1

3x=


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§1 LIMITS AND CONTINUITY

1.1 Limits

In mathematics, the concept of a "limit" is used to describe the behaviour of a function, as

its argument x gets "close" to a point a, or infinity. Limits are used in calculus and other

branches of mathematical analysis to define derivative and continuity.

As the notion of "limit" is not easy to understand, we consider an example to try to feel

what a limit is.

Example

We consider the function 23 12

( )2

xf x

x

−=

−. The domain of f(x) is { }\ 2fD = � because f(x)

is not defined for x = 2 (division by 0). But we can calculate f(x) for some values of x

close to 2.

We state that the values of f(x) are close to 12 for x close to 2 thus we can say the values

of f(x) approach 12 as x approaches 2.

With mathematical words, we say that f(x) tends to 12 as x tends to 2 and we write

( ) 12 as 2.f x x→ →

Using the word limit we say that the limit of f(x) as x approaches 2 is equal to 12 and

write 2

lim ( ) 12x

f x→

= .

Definition Finite limit

We write lim ( )x a

f x L→

= and say “the limit of f(x) as x approaches a equals L” if we can

make the value of f(x) arbitrarily close to L (as close to L as we like) taking x sufficiently

close to a, but not equal to a.

x 1,9 1,99 1,999 … 2 … 2,001 2,01 2, 1

f(x) 11,7 11,97 11,997 … … 12,003 12,03 12,3


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Remark

The sentence “but not equal to a” in the definition above means that we never consider x =

a. In fact, f(x) is may be not even defined when x = a. Consequently, the limit may not be

equal to the value f(a).

Example

( )2

2lim 3 1x

x→

− = and

2

3

9lim 6

3x

x

x→

−=

−

Definition Left-hand and right-hand limit

We write lim ( ) or lim ( )x a x a

f x L f x L−

<→ →

= = and say “left-hand limit of f(x) as x approaches a” is

equal to L or “the limit of f(x) as x approaches a from the left” is equal to a.

In the same way we define “the right-hand limit of f(x)” and write lim ( ) or lim ( )x a x a

f x L f x L+

>→ →

= = .

Examples

Given1

( )f xx

= .

0lim ( )x

f x+→

= +∞ and 0

lim ( )x

f x−→

= −∞ , thus 0

lim ( )x

f x→

doesn’t exist.

2 2lim 2 0 but lim 2 does not exist because the function square root

is not defined for negative values of

x xx x

x

+ −→ →− = −

Given 1

( )f xx

=

0lim ( )x

f x+→

= +∞ and 0

lim ( )x

f x−→

= +∞ , thus 0

lim ( )x

f x→

= +∞ .

Theorem

lim ( ) if and only if lim ( ) and lim ( )x a x a x a

f x L f x L f x L+ −→ → →

= = =

Definition Infinite limit

The expression lim ( )x a

f x→

= ∞ means that the values of f(x) can be made arbitrarily large by

taking x sufficiently close to a but not equal to a.


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Example

21lim

( 1)x

x

x→= ∞

− (divide by a very small number is the same as to multiply by a very large one)

Definition Limit at infinity

If the values of f(x) become closer and closer to L as x become larger/smaller and

larger/smaller, we write lim ( )x

f x L→∞

= or lim ( )x

f x L→−∞

= .

Examples

3lim 0

2x x→∞=

−

1lim 1

2x

x

x→−∞

−= −

− lim sin( ) does not exist !!!

xx

→∞

Piecewise-defined function

Here is the function « sign » ( ) sgn( )f x x= .

It’s a piecewise-defined function : 1 if 0

( ) sgn( ) 0 if 01 if 0

xf x x x

x

>= = =

− <

Its graphic representation is formed of two semi-lines and one point.

A filled circle represents a point while an empty circle represents a hole. This function is

not continuous.

0lim sgn( ) 1x

x−→

= − and 0

lim sgn( ) 1x

x+→

= .

Thus 0

lim sgn( )x

x→

doesn’t exist.

But (0) sgn(0) 0f = = exists !

To calculate explicitly a limit

Try to guess the limit replacing x by numbers close to a.

Simplify the expression of f(x) (if we have "0"

0)

Using a known limit.


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Example

2

3

(2,5) 2.25; (2,9) 4,41; (2,99) 4,9401; (2,999) 4,994...

( ) 4 lim ( ) ?

(3,5) 8,25; (3,1) 5,61; (3,01) 5,0601; (3,001) 5,006...x

f f f f

f x x f x

f f f f→

= = = =

= − =

= = = =

↗

↘

3lim ( ) 5x

f x→

⇒ =

Limits laws

1) [ ]lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x→ → →

+ = +

2) [ ]lim ( ) lim ( ) ,x a x a

f x f xλ λ λ→ →

⋅ = ⋅ ∈�

3) [ ]lim ( ) ( ) lim ( ) lim ( )x a x a x a

f x g x f x g x→ → →

⋅ = ⋅

4) lim ( )( )

lim if lim ( ) 0( ) lim ( )

x a

x a x a

x a

f xf xg x

g x g x→

→ →

→

= ≠

1.2 Limit at infinity of a rational function

Given a rational function : 1 2

1 2 1 0

1 2

1 2 1 0

...( )( )

( ) ...

m m

n n

n n

n n

p x p x p x p x pp xf x

q x q x q x q x q x q

−

−

−

−

+ + + + += =

+ + + + +, where m and

n are the degrees of the polynomials.

We want to calculate ( )

lim( )x

p x

q x→∞. The wheeze to do this is to divide all the terms by the

highest power of x. Let’s consider an example :

Example 2

2 2 2 2 2

22

22 2

4 3 9 3 94

4 3 9( )

52 52 52

x xx x x x x x xf x

xx

xx x

− + − +− +

= = =−− +

− ++

By property 4 (§1.1), the limit of a quotient is the quotient of the limits, for all that the limit

of the denominator is not equal to zero :

2 2

3 9 3 9lim 4 4 lim lim 4 0 0 4x x xx x x x→∞ →∞ →∞

− + = − + = − + =

and

2 2

5 5lim 2 2 lim 2 0 2x xx x→∞ →∞

− + = − + = − + = −

Finally : 4

lim ( ) 22x

f x→∞

= = −−


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Actually, the limit at infinity of a rational function is equal to the limit of the

quotient of the terms of highest degree of the numerator and the denominator.

Examples

2 2

2 2

4 3 9 4 4lim lim lim 2

22 5 2x x x

x x x

x x→∞ →∞ →∞

− += = = −

−− + −

5 4 5

4 3 4

7 12 5 7 7lim lim lim

8 2 19 8 8x x x

x x x x x

x x x→−∞ →−∞ →−∞

+ − += = = −∞

− +

3 2 3

5 4 5 2

5 7 8 5 5lim lim lim 0

2 17 6 2 2x x x

x x x x

x x x x→∞ →∞ →∞

− + += = =

− +

Conclusion

Given the quotient : 1 2

1 2 1 0

1 2

1 2 1 0

...( )( )

( ) ...

m m

m m

n n

n n



−

−

−

−

+ + + + += =

+ + + + +, where

m and n are the degrees of the polynomials.

Then

0 if

lim ( ) lim if

or if

m

m m

nx xnn

m np x p

f x m nqq x

m n

→∞ →∞

<

= = =

+∞ − ∞ >

1.3 Limit of a rational function

A rational function, 1 2

1 2 1 0

1 2

1 2 1 0

...( )( )

( ) ...

m m

n n

n n

n n



−

−

−

−

+ + + + += =

+ + + + +, is the quotient of two

polynomials. Generally, it is quite easy to find the limit of this kind of functions. We just have

to replace x by numbers close to a.

Example

2

3

2 3lim 3

2x

x

x→

−=

+

But in the following case, this method is not useful :

IF THE LIMIT IS OF THE FORM "0"

0

This means that the numerator and the denominator of f(x) are equal to 0 when x = a.

In this case it is not possible to find directly the limit replacing x by numbers close to a.


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Actually the limits of 2 2

4

4 5 7, and

x x x

x x x as x tends to 0 are equal to 4, 0 and ∞

respectively. They are all different but all in the form "0"

0 To calculate this kind of

limits, we have to simplify the expression of f(x). As we have "0"

0 when x = a (i.e.

x = a is a zero of the numerator and of the denominator) we know that the numerator

and the denominator can be factorized by (x – a) and f(x) can be written as

( ) '( )( )

( ) '( )

x a p xf x

x a q x

− ⋅=

− ⋅ where p’(x) and q’(x) are the quotient of the division of p(x) and

q(x) by (x – a). It is now possible to simplify f(x) by (x – a).

Example

2 4

4 "0 "( ) lim ( ) ? (4) indeterminate

02 9 4 x

xf x f x f

x x →

− = = =

− +

We try to guess this limit :

4

(3,9) 0,1471... ; (3,99) 0,14326... ; (3,99) 0,1428979...

(4,1) 0,1388... , (4,01) 0,14245... ; (4,001) 0,1428163...

lim ( ) 0,1428x

f f f

f f f

f x→

= = =

= = =

⇒ =

This is not really satisfying. To find a better result, we have to simplify the expression

of f(x) using the factor form for 2x2 – 9x + 4. As it is a polynomial of degree two, we

use the Viete’s formula:

2

1,2

4

2

b b acx

a

− ± −= and we find :

1

1,2

2

49 7

41/ 2

x

x

x

=±

=

=

↗

↘

Thus 2 12 9 4 2( 4)( )

2x x x x− + = − − and we can write:

4

2

4 4 1( )

12 9 4 2 12( 4)( )2

xx xf x

x x xx x

≠− −= = =

− + −− −

therefore 4

1lim ( ) 0,142857

7xf x

→= = .

Remark

For the case 1) if the degree of p(x) or q(x) is 2 we use the formula to factorize them

but if the degree is more than 2 we have to make the division explicitly.


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1.4 Continuity

In mathematics, a continuous function is one in which arbitrarily small changes in the

input x produce arbitrarily small changes in the output f(x). If small changes in the input x

can produce a broken jump in the changes of the output f(x), the function is said to be

discontinuous (or to have a discontinuity).

With mathematical words :

f is continuous at x = a if lim ( ) ( )x a

f x f a→

=

The excluded values of a function are always discontinuities.

A function f is continuous on an interval if it is continuous at each number a of the

interval. Roughly speaking, this means that the graph of f has no “holes” on this interval.

Consequently, you can draw the graph of f without removing your pencil from the paper.

Examples

a) 2( ) 2f x x= is continuous on �

It is possible to follow the graph from the left

to the right without “jumping” with our pencil.

b) 1

( )g xx

= is discontinuous at x = 0.

If we follow the graph from the left to the

right we have to “jump” with our pencil.

But this function is continuous on *� .

-4 -2 2 4

10

20

30

40

50

-4 -2 2 4

-4

-2

2

4


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§2 ASYMPTOTES

An asymptote is a line whose distance to a given curve tends to zero. An asymptote may or

may not intersect its associated curve. We have three kinds of asymptotes : vertical,

horizontal and slant or oblique asymptotes.

2.1. Vertical asymptote and “hole”

1. VERTICAL ASYMPTOTE

A vertical asymptote can only exist if the function f has discontinuities, that is to

say excluded values. So we have to find them first by looking for the domain of f.

If x = a is an excluded value of f and satisfies at least one of the following conditions :

1.) lim ( )x a

f x+→

= ±∞ 2) lim ( ) wherex a

f x a−→

= ±∞ ≠ ±∞

Then the vertical line x = a is said to be a vertical asymptote (V.A.) of the function f.

Examples

1 15

4

7 5 "33 / 4 " 5( ) \ lim ( ) . . :

4 5 4 0 4x

xf x D f x V A x

x +

→

+ = = = = +∞ ⇒ =

− �

{ }2 22

2

23

2 2 " 4 "( ) \ 2;3 lim ( )

( 2)( 3) 05 6" 6"

lim ( ) . . : 2 and 30

x

x

x xf x D f x

x xx x

f x V A x x

+

+

→

→

= = = = = +∞− −− +

= = +∞ ⇒ = =

�

2. “HOLE”

If a is an excluded value such that lim ( ) ,with x a

f x a→

±∞ ≠ ∞≠≠≠≠ , then there is a “hole” in

the graph of the function at x a= .


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Example

{ }

2

2 1 22

2

22

2

2 2

2 4( ) ? 2 0 VIETE 2 and 1

22 4

\ 1;2 and ( )( 2)( 1)

"0 "lim ( ) INDETERMINATE FACTORIZATION

02 4 2( 2) 2

2 4 2( 2) ( )( 2)( 1)2

x

x

xf x D x x x x

x xx

D f xx x

f x

x xx x f x

x xx x

→

≠

−= = − − = ⇒ = = −

− −−

⇒ = − =− +

= ⇒ ⇒

− −− = − ⇒ = = =

− +− −

�

First excluded value : = 21x

22 2

21 1 1 1

1 1

1

2 2 2lim ( ) lim HOLE: 2 ;

1 3 3

2 4 6 2lim ( ) lim lim lim

( 2)( 1) 3( 1) 12 2

lim lim Vertical asymptote: 11 1

x x

x x x x

x x

x

f xx

xf x

x x x x

xx x

→ →

→− →− →− →−

→− →−> <

+

= = ≠ ±∞ ⇒

+

− −= = =

− + − + +

⇒ = +∞ = −∞ ⇒ = −+ +

Second excluded value : = -12x

2.2. Horizontal asymptote

We met this kind of asymptotes when we study the behaviour of the function for very large

positive/negative values of x ( )andx x→ +∞ → −∞ .

The horizontal line y = c is said to be a horizontal asymptote (H.A.) if :

lim ( ) where x

f x c c→±∞

= ≠ ∞

As x gets larger/smaller and larger/smaller, the graph of f gets closer and closer to the

horizontal line, but without touching it. Pay attention to the fact that it doesn’t mean that there is

no intersection. This may happened only for relatively small values of x.

Examples

2 3 2 2

lim lim 0,4 . . : 0,45 7 5 5x x

x xH A y

x x→∞ →∞

+ −= = = − ⇒ = −

− + −

0

2 2

3 11 3 3lim lim lim 0 . . : 0

77 9 4 7

x

x x x

x xH A y

xx x x

≠

→∞ →∞ →∞

−= = = ⇒ =

−− + − −

because deg(den) > deg(num)

Remark

If deg(den) = deg(num) then there is a H.A.

If deg(den) > deg(num) then there is a H.A. with equation y = 0


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2.3. Slant or oblique asymptote

If the function ( )

( )( )

g xf x

h x= is such that deg( ) deg( ) 1g h− = , then there is a slant asymptote.

To determine its equation, we have to use the long division of ( ) by ( )g x h x . The result

is : ( )( )

( )( ) ( )

g x rf x ax b

h x h x= = + + , where r is the remainder.

The line y ax b= + is called slant or oblique asymptote.

Example

2

2

2

2 3( ) deg( ) deg( ) 1

5

DIVISION :

2 3 5

(2 10 ) 2 11

11 3

(11 55)

52

x xf x num den

x

x x x

x x x

x

x

r

+ −= − = ⇒

−

+ − −

− − +

−

− −

=

SLANT ASYMPTOTE

Therefore, the slant asymptote is : 2 11y x= +

Graphically

Horizontal asymptote

Vertical asymptote

Slant asymptote


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2.4. How does the function approach its asymptotes

The question here is to determine if the function approaches its asymptotes by superior or

inferior values.

If we make the following long division ( )

( )( )

p xf x

q x= we obtain :

1. FOR THE HORIZONTAL ASYMPTOTE : ( )

( )( ) ( )

p x rf x c

q x q x= = +

Example

2. FOR THE SLANT ASYMPTOTE : 2

( )( )

( ) ( )

p x rf x ax b

q x q x= = + +

Example 2

2

2 3 52( ) 2 11

5 5

x xf x x

x x

+ −= = + +

− −

To determine how the function approaches its asymptote, we have to consider the sign of

the limits lim( )x

r

q x→ +∞ and lim

( )x

r

q x→ −∞.

1. FOR THE HORIZONTAL ASYMPTOTE :

1

1

5,8then 0 then ( ) 0,4

5 75,8

then 0 then ( ) 0,45 7

x f xx

x f xx

→ +∞ → → −− +

→ −∞ → → −− +

< <

> >

2. FOR THE SLANT ASYMPTOTE :

2

2

52then 0 then ( ) 2 11

552

then 0 then ( ) 2 115

x f x xx

x f x xx

→ +∞ → → +−

→ −∞ → → +−

> >

< <

1

2 3 5,8( ) 0,4

5 7 5 7

2 3 5 7

(2 2,8) 0,4

5,8

xf x

x x

x x

x

r

+= = − +

− + − +

+ − +

− − −

=


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Reference example

Study the asymptotic behaviour of the function 2

2

3 5 2( )

4 5 1

x xf x

x x

− +=

− +

First of all we have to determine the domain of f. For that we have to find the excluded

values which are the zeros of the denominator so we have to solve :

1

2

1,2

2

15 3

4 5 1 08

1

4

x

x x x

x

=±

− + = ⇒ = =

=

↗

↘

Now we have to determine whether the values 1

1 and4

x x= = are vertical asymptotes

or not :

x = 1

1

"0 "lim ( )

0xf x

→= Indeterminate ! Simplifying f we obtain :

1

1 1

23( )( 1)

3 2 13lim ( ) lim ( )1 4 1 3

4( )( 1)4

x

x x

x xx

f xx

x x

≠

→ →

− −−

= = = ≠ ± ∞−

− −

As this limit is different from ± ∞ it is not a vertical asymptote but a “hole” with

coordinates 1

1;3

.

x = ��

1

4

lim ( )x

f x→

= ± ∞ so x = 1

4 is a vertical asymptote

Finally, as the degree of the denominator (2) is equal to the degree of the numerator (2),

we know that there is no slant asymptotes, but a horizontal one :

3

lim ( )4x

f x→±∞

= so the line 3

4y = is a horizontal asymptote

We finally have to determine how the function f approaches its horizontal asymptote :

2 2

2

2

3 5 2 4 5 1

1,25 1,25(3 3,75 0,75) 0,75 ( ) 0,75

4 5 1

1,25 1,25

x x x x

xx x f x

x x

x

− + − +

− +− − + ⇒ = +

− +

− +

2

2

1,25 1,25

4 5 1then 0 then ( ) 0,75

then 0 the1,25

n ( ) 0,751,25

4 5 1

x f x

x

x

x

x

f

xx

x x

− +

− +→ +∞ → →

→ −− +

− +∞ → →

< <

> >


- 17 - YAy & PRi

x

1

sin( )x

tan( )x

xP

x

1

sin( )x tan( )x

xP

x

1

sin( )x tan( )x

xP

x

1

sin( )x tan( )x

xP

2.5. Limits of trigonometric functions

Here are some limits, easy to calculate :

22 2

lim tan( ) lim tan( ) lim tan( ) doesn't existxx x

x x xππ π

− +

→→ →

= +∞ = −∞ =

But what can we say about 0

sin( )limx

x

x→ ?

Given the trigonometric circle and a positive angle x (radians).

Then we can write the following inequalities :

1 sin( ) sin( )

2 2

x x⋅= ≤ 21

2 2

x xπ

π⋅ ⋅ = ≤

1 tan( ) tan( )

2 2

x x⋅=

y = 3

4

x = 1

4

Hole at x = 1


- 18 - YAy & PRi

x

xP

sin( )x

tan( )x

1

Thus : sin( ) tan( )x x x≤ ≤

Let’s divide by sin( )x (>0) : 1 cos( )sin( )

xx

x≤ ≤

Let’s take the inverse : sin( ) 1

1cos( )

x

x x≥ ≥ change the direction of the inequality

We take the limit 0x → : 0 0 0 0

sin 1 sinlim 1 lim lim 1 lim 1

cosx x x x

x x

x x x+ + + +→ → → →≥ ≥ ⇔ ≥ ≥ so

0

sinlim 1x

x

x+→= .

For a negative angle :

sin( ) tan( )

2 2 2

x x x≥ ≥ (because tan(x) and sin(x) are negative and sin( ) tan( )x x≥ )

Thus : sin( ) tan( )x x x≥ ≥

Let’s divide by sin( )x (<0) : 1 cos( )sin( )

xx

x≤ ≤

Let’s take the inverse : sin( ) 1

1cos( )

x

x x≥ ≥

We take the limit 0x → : 0 0 0

0

sin( ) 1lim 1 lim lim

cos( )sin( )

1 lim 1

x x x

x

x

x xx

x

− − −

−

→ → →

→

≥ ≥

≥ ≥

so 0

sin( )lim 1x

x

x−→=

Conclusion

0

sin( )lim 1x

x

x→=

From there we obtain that :

0 0 0 0 0

sin( ) sin( )tan( ) sin( ) 1cos( )

lim lim lim lim lim 1 1 1cos( ) cos( )x x x x x

x xx xx x

x x x x x→ → → → →= = = ⋅ = ⋅ =

Conclusion

0

tan( )lim 1x

x

x→=


- 19 - YAy & PRi

Important result

0 sin( ) and tan( )x x x x x≈ ⇒ ≈ ≈

§3 DIFFERENTIATION

3.1 Derivative

The main purpose of the chapter Calculus is to be able to sketch the graph of a given

function f without calculating too many values. For that we have to find information about

the function f. At the present time, for a given function f, we know how to find its domain, to

study its sign and its asymptotic behaviour, but this is not enough to sketch precisely its

graph. Therefore, we need some more information about it.

Imagine that it is late at night and you still have your homework to do for the next day. You

have to sketch the graph of a given function, so you decide to phone one of your friends to

get some help. What are the elements that you need to plot it ?

Domain

Vertical asymptotes

Holes

Other asymptotes

x and y–intercepts

Sign

Maxima and minima

Increasing and decreasing parts

Now we have to find a way to determine the maxima, the minima and the increasing and

decreasing parts. For that we need a new tool called derivative. To define it, we consider

a function f, two points A and B and the chord joining them :

-6 -4 -2 2 4 6

-15

-10

-5

5

10

15


- 20 - YAy & PRi

The gradient of this chord is, by definition :

( ) ( )y y f a h f a

x h h

∆ ∆ + −= =

∆

If we change the value of h to make it close to 0, the two points A and B are very close to

each other. Consequently, the gradient of the chord joining A and B is close to the gradient

of the tangent at A. In the limit, as h tends to 0, we obtain the gradient of the tangent at A.

We call this quantity the derivative of f at x = a and we note it '( )f a .

In general, the derivative of f(x) is :

0

( ) ( )'( ) lim

h

f x h f xf x

h→

+ −=

Geometrical interpretation

The derivative of f at x = a is the value of the gradient of the tangent to f at x = a

Often we call it the derivative in 4 steps :

1ST STEP determine ( ) ( )f x x f x h+ ∆ = +

2ND STEP determine ( ) ( )y f x h f x∆ = + −

3RD STEP calculate

( ) ( )y f x h f x

x h

∆ + −=

∆

4TH STEP calculate

0 0

( ) ( )lim lim '( )x h

y f x h f xf x

x h∆ → →

∆ + −= =

∆

Examples

1. Find the derivative of ( ) 4 5f x x= − :

1ST STEP ( ) 4( ) 5 4 4 5f x h x h x h+ = + + = + −

2ND STEP ( )( ) ( ) 4 4 5 4 5 4 4 5 4 5 4y f x h f x x h x x h x h∆ = + − = + − − − = + − − + =

3RD STEP

( ) ( ) 44

y f x h f x h

x h h

∆ + −= = =

∆

4TH STEP

0 0

( ) ( )lim lim 4 4 '( ) '( ) 4h h

f x h f xf x f x

h→ →

+ −= = = ⇒ =

a

f(a)

x∆ = h

a + h

f(a + h)

f(x)

y∆ = f(a + h) – f(a)

f(a)

B(a + h, f(a + h))

A(a, f(a))


- 21 - YAy & PRi

2. Find the derivative of 2( ) 3f x x= :

1ST STEP 2 2 2( ) 3( ) 3 6 3f x h x h x hx h+ = + = + +

2ND STEP 2 2 2 2( ) ( ) 3 6 3 3 6 3y f x h f x x hx h x hx h∆ = + − = + + − = +

3RD STEP

2( ) ( ) 6 36 3

y f x h f x hx hx h

x h h

∆ + − += = = +

∆

4TH STEP

0 0

( ) ( )lim lim 6 3 6 '( ) '( ) 6h h

f x h f xx h x f x f x x

h→ →

+ −= + = = ⇒ =

Remark

Sometimes we write the derivative like that : d

'( )d

ff x

x= . This is the derivative of f

according to x.

3.2 Gradient of the tangent to curve at a given point

Given the function ( )y f x= . The gradient to this curve at x a= is given by '( )f a . The

coordinates of the contact point are ( ; ( ))a f a . Thanks to these information, it is possible to

find the equation of the tangent to f(x) at x a= in the form y mx b= + . The gradient m is

'( )m f a= (by definition of the derivative) and we have to replace x and y by the

coordinates of the contact point ( ; ( ))a f a to find h.

Example

Find the equation of the tangent to the curve 2y x= at 3x = −

Before all we need the coordinates of the contact point and the derivative :

1) Coordinates of the contact point : 2( 3; ( 3)) ( 3;( 3) ) ( 3;9)f− − = − − = −

2) Derivative :

1ST STEP 2 2 2( ) ( ) 2f x h x h x hx h+ = + = + +

2ND STEP 2 2 2( ) ( ) 2 2y f x h f x x hx h x hx h∆ = + − = + + − = +

3RD STEP

2( ) ( ) 22

y f x h f x hx hx h

x h h

∆ + − += = = +

∆

4TH STEP

0 0

( ) ( )lim lim(2 ) 2 '( ) '( ) 2h h

f x h f xx h x f x f x x

h→ →

+ −= + = = ⇒ =

3) Then we evaluate the derivative at 3x = − to find the value of the gradient :

'( 3) 6f − = − the gradient is –6 6y x b= − +


- 22 - YAy & PRi

4) As the tangent passes through the point ( 3;9)− we replace its coordinates in

6y x b= − + to find b :

9 6 ( 3) 9b b= − ⋅ − + ⇒ = −

Finally, the equation of the tangent to 2y x= at 3x = − is : 6 9y x= − −

3x = −

( 3) 9y f= − =

'( 3) 6f − = −

1

( 3;9)−

h = 9−

6 9y x= − −

2y x=


- 23 - YAy & PRi

3.3 The derivative of the powers of x

We know that the derivative of x2 is 2x, the one of x3 is 3x2 so we can deduce the

derivative of xn :

1( ) then '( ) ,n nf x x f x nx n−= = ∈�

Thanks to this formula, it is now possible to derivate functions like 3, , ...x x . For that we

use the fact that 11

3 32 , , ...x x x x= = . Thus we have the following results :

1 1 1

12 2 2

1 1 1( ) then '( )

2 2 2f x x x f x x x

x

− −

= = = = =

1 1 21

3 3 3 3

3 2

1 1 1( ) then '( )

3 3 3f x x x f x x x

x

− −

= = = = =

Finally we have the general formula :

1 1 11

1

1 1 1( ) then '( )

n

n n n n

nnf x x x f x x x

n n n x

−−

−= = = = =

3.4 Differentiation rules

The method we use to determine the derivative is sometimes quite hard and long. Luckily

we have some rules to differentiate functions :

1. THE SUM RULE

( ) ( ) ( ) then '( ) '( ) '( )f x u x v x f x u x v x= + = +

Proof

0 0 0

( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )'( ) lim lim lim '( ) '( )

x x x

f x u x v x

y u x x v x x u x v x u x x u x v x x v x

x x xu x x u x v x x v x

x xy u x x u x v x x v x

f x u x v xx x x∆ → ∆ → ∆ →

= +

∆ + ∆ + + ∆ − − + ∆ − + + ∆ −= = =

∆ ∆ ∆+ ∆ − + ∆ −

+∆ ∆

∆ + ∆ − + ∆ −= = + = +

∆ ∆ ∆


- 24 - YAy & PRi

2. THE PRODUCT RULE

( ) ( ) ( ) then '( ) '( ) ( ) ( ) '( )f x u x v x f x u x v x u x v x= ⋅ = ⋅ + ⋅

Proof

0

( ) ( ) ( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

( ) ( )( ) (

f x u x v xy u x x v x x u x v x

x x

u x x v x x u x v x x u x v x x u x v x

xu x x v x x u x v x x u x v x x u x v x

x xu x x u x

v x x u xx

=

= ⋅

∆ + ∆ ⋅ + ∆ − ⋅= =

∆ ∆

+ ∆ ⋅ + ∆ − ⋅ + ∆ + ⋅ + ∆ − ⋅=

∆+ ∆ ⋅ + ∆ − ⋅ + ∆ ⋅ + ∆ − ⋅

+ =∆ ∆

+ ∆ −+ ∆ ⋅ +

∆

��

( ) ( ))

( ) ( )lim ( ) ( ) '( )

( ) ( )lim ( ) ( ) '( )

'( ) lim '( ) ( ) ( ) '( )

v x x v x

xu x x u x

v x x v x u xx

v x x v xu x u x v x

xy

f x u x v x u x v xx

+ ∆ −⋅

∆+ ∆ −

+ ∆ ⋅ = ⋅∆

+ ∆ −⋅ = ⋅

∆∆

= = −∆

3. THE MULTIPLYING CONSTANT RULE

( ) ( ) then '( ) '( )f x k u x f x k u x= ⋅ = ⋅

Proof

We use the product rule : �0

'( ) ' ( ) '( ) '( )f x k u x k u x k u x=

= ⋅ + ⋅ = ⋅

4. THE QUOTIENT RULE

2

( ) '( ) ( ) ( ) '( )( ) then '( )

( ) ( ( ))

u x u x v x u x v xf x f x

v x v x

⋅ − ⋅= =

Proof

We write f(x) as a product, 1

( ) ( )( )

f x u xv x

= ⋅ , and we use the product rule.


- 25 - YAy & PRi

5. THE CHAIN RULE OR THE COMPOSITE FUNCTIONS RULE

Reminder

The composite function g f∗ is defined by ( ) ( ( ))g f x g f x∗ = .

First f then g !! Illustration : ( ) ( ) ( ( ))f gx f x u g u g f x y→ = → = =

Question What is the derivative of ( ) 'u v∗ ?

Answer

( ) ( ( )) then '( ) '( ( )) '( )f x u v x f x u v x v x= = ⋅

Proof Too long…

Illustration

( ) ( ) ( ( ))

'( )'( )

f gx f x u g u g f x y

g uf x

→ = → = =

⋅

��

Examples

1. Derivate 2( ) 5 11f x x= +

2

12

5 11

10u

x x u u y

x

→ + = → =

⋅

��

2 2

1 1010

2 5 11 2 5 11

xx

x x= ⋅ =

+ + ��

2

5'( )

5 11

xf x

x=

+

2. Derivate 2( ) cos ( )f x x=

2cos( )

sin( ) 2

x x u u y

x u

→ = → =

− ⋅

��

�� '( ) 2sin( )cos( )f x x x= −

'( ) '( ( )) ( ) '( )f x g f x g f x= ⋅ = ∗

sin( ) 2cos( ) 2sin( )cos( )x x x x= − ⋅ = −


- 26 - YAy & PRi

3. Derivate ( )2

3( ) 2 8 11f x x x−

= + −

2

3 2

3

2 8 11

8 26

x x x u u y

ux

−

−

→ + − = → =

+ −⋅

��

�� 2

3 3

12 16'( )

(2 8 11)

xf x

x x

− −=

+ −

3.5 Examples

1. The multiplying constant rule

3( ) 25f x x= then 2 2'( ) 25 3 75f x x x= ⋅ =

2. The sum rule

4( ) 3 cos( )f x x x= + then 3'( ) 12 sin( )f x x x= −

3. The multiplying rule

( ) sin( )f x x x= ⋅ then '( ) 1 sin( ) cos( ) sin( ) cos( )f x x x x x x x= ⋅ + ⋅ = +

4. The quotient rule

4

5( )

3

xf x

x

+= then

4 3 4 4 3

4 2 8 5

1 3 ( 5) 12 3 12 60 3 20'( )

(3 ) 9 3

x x x x x x xf x

x x x

⋅ − + ⋅ − − − −= = =

5. The chain rule

a) ( ) cos(2 1)f x x= + then ( ) cos( )( ) 2 1

'( ) sin(2 1) 2 2sin(2 1)u x xv x x

f x x x== +

− + ⋅ = − +=

b) 5( ) ( 3 4)f x x= − + then 5

4 4

( ) ( )( ) 3 4

'( ) 5( 3 4) ( 3) 15( 3 4)u xv x x

f x x x==− +

− + ⋅ − = − − +=

c) ( )( ) 4 1

1 2( ) 4 1 '( ) 4

2 4 1 4 1u xv x x

f x x then f xx x=

= −

= − = ⋅ =− −

22 3 3

3 3

12 16(6 8) ( 2(2 8 11) )

(2 8 11)

xx x x

x x

− − −= + ⋅ − + − =

+ −


- 27 - YAy & PRi

Summary of some important derivatives

( )f x c x x2 x3 xn 1

x x sin( )x cos( )x

'( )f x 0 1 2x 3x2 nxn-1 2

1

x

− 1

2 x cos( )x sin( )x−

Remark

Now that we have rules to find derivatives, §3.2 becomes :

Find the equation of the tangent to 2y x= at 3x = −

Before all, we have to find the coordinates of the contact point :

1. Coordinates of the contact point : 2( 3; ( 3)) ( 3;( 3) ) ( 3;9)f− − = − − = −

2. Derivative : '( ) 2f x x=

3. Then we evaluate the derivative at 3x = − to find the gradient of the tangent :

'( 3) 6f − = − the gradient is –6 6y x b= − +

4. The tangent passes through ( 3;9)− consequently, we replace its coordinates

in the equation 6y x b= − + to find b :

9 6 ( 3) 9b b= − ⋅ − + ⇒ = −

Finally, the equation of the tangent to 2y x= at 3x = − is : 6 9y x= − −

§4 USES OF THE DERIVATIVE

4.1. Increasing and decreasing functions

Before all we have to define the meaning of increasing and decreasing functions. Without any

math symbols we can visualize an increasing function as a function that is “going up” and a

decreasing function as a function that is “going down”.


- 28 - YAy & PRi

Increasing function Decreasing function

But all the functions are not as simple as lines. Actually, what can we say about parabolas ?

If we look at the graph of such a function, we observe

that one of its parts is decreasing ( ) and

another one is increasing ( ). Consequently, a

function can be both and we say that a function f is

increasing or decreasing on an interval I. On this

example, the function f is :

decreasing over ] ]; 1− ∞

increasing over [ [1 ; ∞

We have to find a mathematic criterion to be able to decide when a function is increasing and

when it is decreasing. We know how to solve this problem for a line : if the gradient is positive,

the line is increasing if it is negative it is decreasing. We use the same way for the other

functions by considering the gradient of the tangent. This gradient is given by the derivative : if it

is positive, the function is increasing and if it is negative it is decreasing.

It is now possible to give a mathematical definition of increasing and decreasing functions.

Definition Increasing and decreasing functions

If '( ) 0f x > in an interval ] [;p q except at isolated points where '( ) 0f x = , then ( )f x is

increasing over the interval [ ];p q .

If '( ) 0f x < in an interval ] [;p q except at isolated points where '( ) 0f x = , then ( )f x is

decreasing over the interval [ ];p q .

As the property of growth and decrease is given by the sign of the derivative, we have to find

the points where the derivative changes its sign.

1

f

Negative gradient

Positive gradient


- 29 - YAy & PRi

If we look at the graph of the parabola above, we can easily deduce that the important point is

the vertex. What can we say about the other functions ?

For this function there are two important points : A and

B. We observe that before A, the function is increasing

so, by definition, '( ) 0f x > . Between A and B, f is

decreasing so '( ) 0f x < and after B f is increasing again

so '( ) 0f x > . So the derivative changes its sign at these

two points, consequently we have to find a condition to

be able to determine them.

Looking at them, we observe that the points A and B are characterised by a horizontal tangent,

so the gradient is 0 and '( ) 0f x = . The points where '( ) 0f x = are called stationary points.

As for the sign of a function we organize this study in a table called variation table.

Summary

The sign of the derivative determines if a function is increasing or decreasing. This sign may

change at the points were the derivative is equal to zero (stationary points) or if there are

excluded values.

Example

Determine where the function 22 3

( )2

x xf x

x

−=

− is increasing/decreasing

First we have to find the excluded values of f if any : 2 0 2x x− = ⇒ =

Then we calculate the derivative of f :

2

2 2 2

2 2( ) 2 3( ) 2

2 2

2 2 2

(4 3) ( 2) 1 (2 3 ) 4 8 3 6 2 3'( )

( 2) ( 2)

2 8 6 2( 4 3) 2( 1)( 3)

( 2) ( 2) ( 2)

u x x xv x x

x x x x x x x x xf x

x x

x x x x x x

x x x

= −= −

− ⋅ − − ⋅ − − − + − += ==

− −

− + − + − −= =

− − −

The next step is to find the stationary points. For that we solve the equation : '( ) 0f x =

1 22

2( 1)( 3)0 2( 1)( 3) 1 and 3

( 2)

x xx x x x

x

− −= ⇔ − − ⇒ = =

−

-8 -6 -4 -2 2 4

2

4

6

8

10

A

B

f


- 30 - YAy & PRi

Finally we build the table of variations :

x 1 2 3

f’(x) + 0 – Imp – 0 +

f(x) Imp

4.2. Maxima and minima

Since 4.1 it is obvious that the maxima and the minima are characterised by the condition

'( ) 0f x = . Therefore to determine them we have to find the stationary points of the function

and then we use the fact that a maximum is defined by the following condition :

Behaviour of f f increases turning point f decreases

Sign of f’ '( ) 0f x > '( ) 0f x = '( ) 0f x <

Visualization

a minimum by :

Behaviour of f f decreases turning point f increases

Sign of f’ '( ) 0f x < '( ) 0f x = '( ) 0f x >

Visualization

And a level point by :

Behaviour of f f increases turning point f increases

Sign of f’ '( ) 0f x > '( ) 0f x = '( ) 0f x >

Visualization

Or by :

Behaviour of f f decreases turning point f decreases

Sign of f’ '( ) 0f x < '( ) 0f x = '( ) 0f x <

Visualization

We observe these conditions directly in the variation table. Let’s consider the one in 4.1 :

x 1 2 3

f’(x) + 0 – Imp – 0 +

f(x) Imp

Max Min


- 31 - YAy & PRi

Remark

Maxima and minima are called turning points.

§5 STUDY OF A FUNCTION

The main purpose of such a study is to plot the graph of a given function f. We have listed the

main points that we have to study to be able to sketch the graph of a function f. This is the

process to study a function :

1. Domain, intercepts, sign

2. Asymptotic behaviour

a) Vertical asymptotes and holes

b) Horizontal or slant asymptotes

3. Derivative

a) Stationary points

b) Variation table

4. Graph

Example

Study the function 2

( )2 3

xf x

x=

−

1. DOMAIN, INTERCEPTS, SIGN

Domain : 3 3

2 3 0 \2 2

x x D

− = ⇒ = ⇒ =

�

Intercepts : 2

2: Condition : 0 0 0 02 3

x

xf O y x x

x∩ = ⇒ = ⇔ = ⇒ =

−

20

: Condition : 0 020 3

yf O x∩ = ⇒ =−

So the x-inetrcept is ( )0 ; 0xI and the y-intercept is also ( )0 ; 0yI

Sign :

x 0 3/2

( )f x – 0 – imp +


- 32 - YAy & PRi

2. ASYMPTOTIC BEHAVIOUR

a) Vertical asymptotes and holes

As 3

2

lim ( )x

f x+

→

= +∞ or 3

2

lim ( )x

f x−

→

= −∞ then 3

2x = is a vertical asymptote (V.A.)

No holes

b) Horizontal or slant asymptotes

As deg( ) deg( ) 1num den− = there is a slant asymptote. To find its equation we

make the division :

2

2

2 3

3 1 3( )

2 2 4

3

2

3 9

2 4

9

4

x x

x x x

x

x

−

− − +

− −

So

( )

1 3 9 / 4( )

2 4 2 3s x

f x xx

= + +−�

The slant asymptote is the line : 1 3

2 4y x= +

Now we have to determine if f is over or under its asymptote :

9 / 4 1 3If , then 0 so ( )

2 3 2 4

9 / 4 1 3If , then 0 so ( )

2 3 2 4

x f x xx

x f x xx

> >

< <

→ +∞ → → +−

→ −∞ → → +−

Intersection function-asymptote : two possible methods

1) Solve the equation : ( )f x asymptote=

2

2 21 3 9 90

2 3 2 4 4 4

xx x x imp

x= + ⇒ = − ⇒ =

−

2) The function intersects its asymptote when ( ) 0s x = :

9 / 4 90 0

2 3 4imp

x= ⇒ =

−

So no intersection with the asymptote.


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3. DERIVATIVE

a) Stationary points 2 2

2 2 2

2 (2 3) 2 2 6 2 ( 3)'( )

(2 3) (2 3) (2 3)

x x x x x x xf x

x x x

⋅ − − ⋅ − −= = =

− − −

The stationary points satisfy : '( ) 0f x =

1 22

2 ( 3)0 2 ( 3) 0 0 3

(2 3)

x xx x x x

x

−= ⇒ − = ⇒ = =

−

We have to find the y coordinates of these two points. For that we replace the

values of x in f(x) :

2

1 1

2

2 2

00 (0;0)

2 0 3

3 93 (3;3)

2 3 3 3

y S

y S

= = ⇒⋅ −

= = = ⇒⋅ −

b) Variation table

Max Min

4. GRAPH

x 0 3/2 3

'( )f x + 0 – imp – 0 +

( )f x imp

−10 −5 5 10

−10

−5

5

10


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§6 OPTIMIZATION

The purpose of this part of the chapter analysis is to maximize or minimize a given size.

Let’s consider an example.

Example

A new drinks company wants to produce a cylindrical can of 1 litre which has an area as

small as possible. How do you choose the radius r and the height h ?

The volume is given by : 2 1r hπ =

The area by : 22 2rh rπ π+

The function to minimize is : 22 2A rh rπ π= +

As A is a function of more than one variable (r and h) we have

to extract h from the volume and substitute it into A. We obtain

2

2

11r h h

rπ

π= ⇒ =

We substitute : 2 2

2

1 2( ) 2 2 2A r r r r

rrπ π π

π= + = + . With x = r, the function we have to

minimize is 22( ) 2f x x

xπ= + . Consequently we have to find the turning points of ( )f x . For

that we differentiate ( )f x : 2

2'( ) 4f x x

xπ

−= +

The stationary points satisfy :

3 3 32

2 1 1'( ) 0 4 0 2 4 0

2 2f x x x x x

xπ π

π π

−= ⇒ + = ⇒ − + = ⇒ = ⇒ =

Is it a maximum or a minimum ? The variations table gives :

MIN

With this radius we obtain : 20,54dm 1,08dm area 5,53dmr h= = =

The optimal cylinder has a diameter equal to the height.

x 31

0,542π

≅

'( )f x – 0 +

( )f x

h

r


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Optimization process

1) Determine the unknowns of the problem and the equation by which they are

linked.

2) Indicate the interval of the unknowns’ possible values (may be restricted)

3) Determine the function to be optimized : if it has more than one variable, rewrite it, by

substitution, as a single variable function ( )f x , where x is one of the problem’s

unknowns.

4) Determine the turning points of the function by using the derivative.

5) Check that it is the right kind of turning point.

6) Find the solution to the given problem.

§7 GRAPHING THE DERIVATIVE FUNCTION

Given the graph of a function ( )y f x= , how can we plot the graph of '( )y f x= on the

same coordinates system ?

Idea

Find the stationary points : as their derivative equals zero, they determine the

intersections between the x-axis and the graph of '( )y f x= .

Then Choose any abscissa px . Draw the tangent to the curve at the point ( ; )p pP x y

that has the selected abscissa.

Determine the slope of this tangent. For that choose 1x∆ = so that the length of

y∆ is the approximation of '( )pf x .

Finally, place the point '( ; '( ))p pP x f x : the curve '( )y f x= passes through 'P .

and/or

Choose a slope k.

Using the coordinate system, draw a line that has the selected slope.

Translate that line until it is tangent to ( )y f x= .

The tangency point ( ; )p pP x y satisfies '( )pf x k= .

Place the point '( ; )pP x k : the curve '( )y f x= passes through 'P .


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Remark

As the derivative of a polynomial of degree n is a polynomial of degree n – 1, when…

… 2( )f x ax bx c= + + is a quadratic polynomial (represented by a parabola), then

'( ) 2f x ax b= + is a linear function and is represented by a straight line.

… ( )f x ax b= + is a first degree polynomial (represented by a straight line), then

'( )f x a= is a constant and is represented by a horizontal line.

EMINDER - 2 IMITS AND CONTINUITY€¦ · 2MG Level 1 CALCULUS I Theory Lycée Denis-de-Rougemont...

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Transcript of EMINDER - 2 IMITS AND CONTINUITY€¦ · 2MG Level 1 CALCULUS I Theory Lycée Denis-de-Rougemont...