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Chapter A3
Functions
Remember that :
(1) Natural numbers N = { 0 , 1 , 2 ,3 ,……. }
(2) integers
Z = { ……..-3 , -2 , -1 , 0 , 1 , 2 , 3 , …..}
(3) Real numbers(R) : all numbers exists including fractions .
Notes:
1) ∞ & -∞ are not real numbers .
2) ∞ is greater than any real number .
3) -∞ is less than any real number .
The domain
The domain of any function is all the real numbers R except the set of
zeroes of the function in the denominator .
Example (1)
5x
What is the domain of :
3
X-2
Answer
The domain is R – { 2 }
The range (image set)
The range is the set of numbers between the lowest and the highest
number in y – axis or the f(x) axis .
Intervals and ranges :
Let a and b be real numbers with a < b .
* the closed interval [ a , b ] means :
a ≤ x≤b
this means that the closed interval includes a , b .
( x is between a and b including them )
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* the open interval ( a, b ) means :
a<x<b
this means that the open interval doesn’t include (a,b)
ex.
F(x) = 2x+1 ( -1 ≤ x ≤ 1 )
This means that [ -1 , 1 ] which means that x is from -1 to 1 .
f (-1) = 2 (-1)
f (0) = 2 (0) + 1 = 1
f (1) = 2 (1) + 1 = 3
Interval notation:
inequality
Closed interval:
interval
a≤ x≤b
a≤ x
x≤ a
[a,b]
[a,∞)
(∞,a]
Open interval :
a<x<b
a<x
x<a
(a,b)
(a,∞)
( -∞ , a )
Half open or half closed :
a≤ x<b
a<x≤b
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[a,b)
(a,b]
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Example (2)
Find the domain for each of the following:
(a) f ( x ) =
x–1
Answer
Square root is error for –ve number , so to avoid this we must start
putting x =1 , then x = 2 then x = ……….. n .
Then the domain ( image set ) is [ 1 , ∞ ].
(b) f ( x )
=
1
x-2
+
1
x+3
Answer
The expression
1
x–2
gives real numbers for all (x) in R except
for x = 2
.
Also ,
1
gives all real numbers except x = -3.
x+3
So, the domain is :
(-∞ , -3 ) , ( 3, 2 ) , ( 2 , ∞ ) or you can write it as R – { -3 , -2 } .
(c) y =
1
x
Answer
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Eng.sherif yehia
0101560322
Note: the modulus ( absolute value ) of a real number is the
regardless of it’s sign .
Even and odd functions
(Not included in MST 121 course)
The function is said to be even if : f (-x) = f (x).
And the function is said to be odd if : f (-x) = - f (x).
Example:
(1)
If f (x) = x² - 1
So for f (2) = 3 & f (-2) = 3
So ָ f (x) = f (-x) , then this function is an even function .
(2)
If f (x) = 1 / x
So for f (1) = 1 & f (-1) = -1
So ָ f (x) = - f (x) , then this function is an odd function .
How to sketch an even or odd function
Remarks:
1- The graph of the even function is symmetric about y – axis.
2- The graph of the odd function is symmetric about the origin.
Example (3)
Sketch the function f (x) = x³ ( -1 ≤ x ≤ 1 )
x³
-1
0
F(x)
-1
0
Answer
-1
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1
1
1
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Graphs of quadratic functions
The equation of any quadratic equation is :
Y = a ( x – b )² + c
Where:
 The graphs are open upward when a > 0 ( positive ).
 And open downward when a < 0 ( negative ).
 These graphs are symmetric about y – axis if a > 1 then the
parabola gets narrower , and if a < 1, then the parabola
gets wider .
 The vertex is ( b , c )
Example (4)
Sketch the following quadratic function :
f (x) = -2 ( x – 4 )² + 2
Answer:
a = -2 , b = 4 , c = 2
so for a < 0 ( -ve )
the parabola looks downward and th
ָvertex is ( 4 , 2 ) , draw the axis of symmetry around x = 4.
a = -2 therefore;
a = 2 which is > 1.
So the parabola gets narrower, and then plot to points:
X
F (x)
2
-6
3
0
2
4
2
5
0
6
-6
(4,2)
4
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The functions
Eng.sherif yehia
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Example (5):
Sketch the following quadratic function:
y = x² + 1
Answer:
ָ a > 0 ( positive ) , then the parabola looks upward
ָ
for ( x = 0 ) , ( y = 1 )
So the vertex of the parabola is ( 0 , 1) and the axis of symmetry
is around x = 0 (origin) .
Plot 2 points before and after the axis:
x
Y= x² + 1
-2
5
-1
2
0
1
1
2
2
5
Then draw the parabola
3
Y= x² + 1
2
1
-2
unit A3
-1
0
(0, 1)
1
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0101560322
After expanding the equation , it becomes :
Y = ax² + bx + c
Here the vertex is : x = The x-coordinate of the vertex is – b then
gets y-coordinate by substitution.
2a
x= –b
2a
How can we re-arrange the equation above to the general equation
of parabola?
Equation of circle ‫هيا نتذكر طريقة الدائرة‬
Example (6)
Re-arrange -2x² + 4x – 1
Answer
Take -2 ( x² - 2x + ½ )
-2[( x – 1 ) – 1 + ½ ]
-2[ ( x – 1 ) – ½ ]
Example (7)
Sketch the following quadratic function :
f (x) = 4x² - 56x + 192
Answer
f(x) = 4[ x² - 14x + 48 ]
4 [ ( x – 7 )² - 7² + 48 ]
4 [ ( x – 7 )² - 1 ]
Therefore f(x) = 4( x – 7 )² - 4
The vertex is ( 7 , -4 ) , then the parabola is :
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Increasing & decreasing
Functions
I want to know here whether the function is increasing or decreasing
when I substitute in x .
Example (8)
(a) h(x) = sin x
( -½ π ≤ x ≤ ½ π )
Answer
When substitute x by -½ π which means -90º and ½ π which is 90º .
So, the interval is between sin -90º and sin 90º .
Therefore, the interval of the function h(x) is [ -1 , 1 ].
Then the function is increasing function…. This is called one – one.
One to one.
1
-1
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x
(b) g(x) = ( ½ )
The function is decreasing function, this also is called one – one.
(c) f(x) = sin x
The function in neither increasing nor decreasing , it’s called
many – one .
-2π
unit A3
-π
0
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π
2π
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Inverse functions
Inverse function happens if the function is only one – one function .
It’s used to convert any function f(x) to it’s opposite way or f ‫־‬¹ .
Example (9)
Show that the function f(x) = 2x + 1 , where x in [ -1 , 1 ] has an
inverse function f ‫־‬¹ , find the rule of f ‫־‬¹ and sketch the graph .
Answer
first , we sketch the graph of f , this is the graph of y = 2x + 1,
restricted to the close interval [ -1 , 1 ].
f(-1) = 2(-1) + 1 = -1 , f(1) = 2(1) +1 = 3
y
3
image set
From the graph :
* the function f(x)is increasing and so one – one.
* the image set is [ -1 , 3 ].
So, f has inverse function f ‫־‬¹ with domain [-1,3]
And image set [-1,1].
-1
We can find the rule of f ‫־‬¹ by solving the equation :
-1
y = f(x) = 2x + 1
y = f(x)
= 2x + 1
x
1
domain
to obtain x in terms of y :
y = 2x + 1 , so x = ½ ( y – 1 )
so , the inverse function expressed in terms of x is :
f ‫־‬¹ (x) = ½ ( x – 1 ) ( x in [ -1 , 3 ] )
finally , the graph of y = f ‫־‬¹ (x) is obtained by reflecting the graph
of y = f(x) in the 45º line :
y
f (-1) = ½ (-1 – 1 ) = -1
f(3)=½(3–1)=1
y = f ‫־‬¹(x) = ½ ( x – 1 )
1
-1
1
2
3
-1
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The functions
x
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Example (10)
Find the inverse function f ‫־‬¹ and sketch the graph of y = f ‫־‬¹ (x) of
f(x) = 4x² - 56x + 192 ( x in [ 0 , 6 ] ).
Answer
first , sketch the graph of f , this is the graph of y = 4x² - 56x + 192,
restricted to the close interval [ 0 , 6 ].
f(0) = 4(0) – 56(0) + 192 = 192
f(6) = 4 (6)² – 56(6) + 192 = 0
y
192
y = f (x)
0
x
6
From the graph :
* the function f(x)is decreasing and so one – one.
* the image set is [ 0 , 192 ].
So, f has inverse function f ‫־‬¹ with domain [0,192], and image set [0,6].
We can find the rule of f ‫־‬¹ by solving the equation : y = f(x) =4x² - 56x + 192
. To obtain x in terms of y : 4x² - 56x + 192 = 0
So ,
x =
56 ± √ (-56)² - 16 (192 – y )
8
= 1
8
=
56 ± √ 64 + 16 y
7± √1+¼ y
Now , we need to choose the solution of this equation which lies in
[0,6] , the domain of f .
Therefore we choose : x = 7 - √ 1 + ¼ y
so , the inverse function expressed in terms of x is :
f ‫־‬¹ (x) = 7 - √ 1 + ¼ x
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( x in [0,192] )
The functions
Eng.sherif yehia
0101560322
finally , the graph of y = f ‫־‬¹ (x) is obtained by reflecting the graph
of y = f(x) in the 45º line :
y
f (0) = 7 - √ 1 + ¼ (0) = 6
6
f ( 192 ) = 7 - √ 1 + ¼ (192) = 0
y=f
0
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‫־‬¹(x)
192
x
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Laws of logarithms
* Log x = y
a
y
Means a = x
* Log x
means log x
10
* Log a = 1 .
a
* Log 1 = 0 .
n
* Log x = n log x
a
a
* Log a.b = log a + log b
*Log a /b = log a –log b
very important note
Lin functions has the same effect as the logarithmic function .
Example (1)
What is the value of log ½
2
Answer
unit A3
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Examples
Find the value of (x) of the following :
(1) 2
x+5
=8
Answer
We can solve by using two ways :
The first way:
x+5
3
2
= 2  x + 5 =3
Therefore x = 3 – 5 = -2
x = -2
The second way:
Log both sides
x+5
log 2 = log 8
( x + 5 ) log 2 = log 8
( x + 5 ) = log8 / log 2
(x+5)=3
x = 3 – 5 = -2
x
(2) 2 = 100
Answer
Log both sides :
x
log 2 = log 100
x log 2 = log 100  x = log100 / log2 = 6.64 (to nearest 2 decimal )
3
(3) 2 log x = 6
Answer
( 2 x 3 )log x = 6
6 log x = 6
1
Therefore log x = 1 , Therefore 10 = x
X = 10
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(4) Verify the equation :
Ln
x+1
e
x² + x³
= x + 1 – 2ln x – ln (x+1)
Answer
(5) log 6 + log 8 – log 2 – log 24
Answer
(6) Show that :
Log
2
4
x . 4
2 x²
3x
= 4 log x + 6x – x²
Answers
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