Download - 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
We can solve the equation in the following way
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
We can solve the equation in the following way
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