calculus lecture part 2

7/27/2019 calculus lecture part 2

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Zeashan ZaidiLecturer BiostatisticsDeptt of Comm Med.ELMC & H

Lucknow


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Algebraic Identitiesand Algebraic Expressions

Identities are the expressionswhich are valid for all real numbers.

Let a and bbe real numbers, then2 2 2

(1) (a + b) = a + 2ab + b2 2 2

(2) (a -b) = a -2ab + b2 2

(3) (a + b)(a -b) = a -b


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Linear equationin one (real) unknownx

is an equation that can be written in the form

where a and b are constants with a 0.

For example following is a linear equation

that gives the solution

ax+ b = 0;

5x+ 4 = 0; x= - 4/5


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An equation in one unknownis an equation

that can be written in the form

Where is any type ofmathematical expression

containing the variable .

x

F(x)

x

F(x) = 0


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x2 16 0- =

xx x

3

21

2 3 51+

+ -=

x x3 25 2 0- + =

Equations - Examples


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Solving linear Equation : ExampleSolve the following equation :

Solution : Using the properties of real numbers

we solve the equation in the following way

So the solution isx= 2

3 8 2 5x x+ = +( )

3 8 2 5

3 8 2 10

3 2 10 8

2

x x

x x

x x

x

+ = +

+ = +

- = -

=

( )


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Example :

Solution :

Solve the following equation :

a x b x c a+ = + b g 3 3,

a x b x cax ab x c

ax x c ab

x a c ab

xc ab

a

+ = ++ = +

- = -

- = -

=-

-

b g

b g

33

3

3

3

So the solution is

xc ab

a=

-

- 3


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Solving Quadratic Equations

A quadratic equation is an equation of the form2

ax + bx + c = 0

where a; b; and c are constants and a 0.

To solve this equation we can use the

or the.

Factorization Method

Quadratic Formula


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Factorization Method :

2x-7x+12 = 0

x=4 x= 3

We factorize the expression and make

two linear equations.

Example : SolveSolution :

2x -7x+12 = 0

2x -(4+3)x+12 = 0

2x -4x-3x+12 = 0

x(x-4) -3(x-4) = 0

(x-4)(x-3) = 0

or are the solutions.


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Note : A quadratic equationhas .

They can be either or .

alwaystwo solutions

equal unequal


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Quadratic Formula :A quadratic equation of the type

2ax + bx + c = 0

have a formula for solutionswhich is given by

x b b aca

= - -2 4

2


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Example :

Solution :

Solve the equationusing

In this equation we have

and

now substituting all these values in the formula

2x-7x+12 = 0

quadratic formula

a = 1, b = -7 c = 12

x =- - - -

7 7 4 1 12

2 1

2b g b g x

x

x

=- - -

=

=

7 49 48

2

7 1

2

4 3

b g

,


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Example : Solve the equationusing quadratic formula

In this equation we have

andnow substituting all these values in the formula

22x-7x-9 = 0

a = 2, b = -7 c = -9

Solution :

x =- - - - -

7 7 4 2 9

2 2

2b g b g b gx

x

x

=- - +

=

= -

7 49 72

16

7 121

16

857

8

b g

,


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Logarithms and exponential equations :many equations have the terms of exponents

and logarithms and their solution finding ismuch difficult.Some simple examples are shown here :


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31

81

1y =

3 181

3 13

3 31

4

1

4

1 1

1

4

4

y y

y

y

y

= =

= = -

= -

-

Example : Solve

We can solve the equation in the following way


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Example : Solve


log log4 7 2x x= -

log log log log

log log log

log log log

log log log

log

log log

4 7 4 2 7

4 7 2 7

4 7 2 7

7 4 2 7

2 7

7 4

2x x x x

x x

x x

x

x

= = -

= -

- = -

- =

=-

- b g

b g


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Example : Solve


log log4 2 3x x= +

log log

log log

log log

4 2

2 2

2 2 3 2

2 3

3

3

2 3

x x

x x

x x

x x

x

=

=

= +

= +

=

+

+

b g


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Real Functions

set ofinputs set of permissible outputs

A real function is a rule which assigend

an exactly one real numbercorrespondingto any real number

belonging to some set

which is called domain.

,

a function is a relation aand awith the property that

.

Alternatively

between

each input is related to exactly one output


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Example

function ,

which relates an

If the input is -3, then the output is 9and we write f(-3) = 9.

2f(x) =x

f(x)

inputx to its square

The output of the function f corresponding

to an inputx is denoted by

Inputx Output f(x)Function


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Example

Let be the set consisting of four apples,say two red ones, a yellow, and a green one,

and let be the set consisting of four colors :red, green, blue, and yellow.

X

Y

Assigning to each apple its coloris a function fromXto Y:


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Color

Red

Green

Blue

Yellow

Apple

A-1

A-2

A-3

A-4


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The input and output of a function

ordered pair

2f(x) =x

are expressed as an ,ordered so that the first element is the input,the second the output. Consider the example

,

we have the ordered pair .(-3, 9) or (2, 4)


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This can be viewed asthe of a point

on the of the function.

ordered pairCartesian coordinates

graph

-4, 16

-3, 9

-2, 4

-1, 10, 0

1, 1

2, 4

3, 9

4, 16

5, 25

0

5

10

15

20

25

30

-5 -4 -3 -2 -1 0 1 2 3 4 5 6


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A variable that represents the

for a function is calledan .

A variable that represents the is called

a because its value

depends on the value ofthe independent variable.

input numbersindependent variable

output numbersdependent variable


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Example :

x2

x + 3 y

Consider the functionf, given by

We may also writeto represent this function.

For each input , the function gives

exactly one output , which is .If , then ;

if , then etc.

2f(x) =x + 3:

2y =x + 3

x = 2 y = 7x = 4 y = 19


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Finding the value of a function :

f(x)

a x

If a function is defined,

and we want to find the value ofthe function at some given value of ,

then this value can be evaluated by

just .puttingx= a in the right side off(x)


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2Example : Letf(x) =x + 3x 7. Find the following:

(1)f(5)(2)f(a + 2)

2(3)f(x)

Solution : We can find the values in the following way

2Given that f(x) =x + 3x 7

(1) 2Thereforef(5) = 5 + 35 - 7= 25 + 15 - 7= 33


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(2)

(3)

2f(a + 2) = (a + 2) + 3(a + 2) 72

= a + 4a + 4 + 3a + 6 72

= a + 7a + 3

2 2 2 2f(x) = (x) + 3(x) 74 2

=x + 3x 7


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Example : Let

Find the following:(1)f(3)

(2)

(3)

f x xb g = 10

f xlogb g

f f3 3b g b g -


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Solution : We can find the values inthe following way

f x

f ff f

f f

f f

f f

xb g

b g b gb g b g

b g b g

b g b gb g b g

=

= - = - =

- =

- =

- =

-

-

-

10

3 10 3 103 3 10 10

3 3 10

3 3 10

3 3 1

3 3

3 3

3 3

0

,

f x

f x

f x

x

x

b g

b gb g

=

=

=

10

10

1

log

log

log

f x

f

f

xb g

b g

b g

=

=

=

10

3 10

3 1000

3

(1)

(2)

(3)


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Example :2

Letf(x) =x + 3x 7.

Find the value of

and hence evaluate

Df x hf x h f x

h,b g

b g b g=

+ -

Df h5,b g

Solution : We can find the value in the


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Solution : We can find the value in thefollowing way

Df x hf x h f x

h

x h x h x x

h

hx h h

h

x h

,b g b g b g

b g b g

=+ -

=+ + + - - + -

=+ +

= + +

2 2

2

3 7 3 7

2 3

2 3

Df h h h5 2 5 3 13,b g = + + = +So,

2


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2Example : Letf(x) =x + 3x,

x 10and g(x) = e +x .

Find the value off[g(x)] andg[f(x)]

Solution :

f g x f e x

e x e x

e x x e e x

e x x e x

x

x x

x x x

x x

b g

c h

= +

= + + +

= + + + +

= + + + +

10

102

10

2 20 10 10

2 20 10 10

3

2 3 3

3 2 3


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and

g f x g x x

e x x

x x

b g

c h

= +

= + +

+

2

3 210

3

3

2

calculus lecture part 2

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