Download 1 Sets of real numbers

1
Sets of real numbers
Outline
• Sets of numbers, operations, functions
• Sets of natural, integer, rational and real numbers
• Operations with real numbers and their properties
• Representations of real numbers
• Elements of topology
We denote by
N = {0 1 2   }
Z = {
©    −2 −1 0 1 2   ª}
Q= 
 :   ∈ Z,  6= 0
R
(the
(the
(the
(the
set
set
set
set
of
of
of
of
natural numbers)
integer numbers)
rational numbers)
real numbers)
In the following chapters, we will also be concerned with the sets R2 , R3 , or more generally R ( ≥ 1), defined
as follows
R = { = (1  2       ) : 1  2      ∈ R} 
Given a point  = (1  2       ) ∈ R , 1  2       are called the coordinates of . In the cases  = 1 2
and 3 we can represent  by a point on the real line, in the real plane, respectively in the real space, as indicated
in the figure below. In the cases  ≥ 4 we cannot represent graphically the point  (it is difficult for example to
make a sketch in a 4-dimensional space!), but we can still think  a being a point in R .
x
x2
0
x3
R2
R
x
x
0
x1
0
R3
x1
x2
Figure 1: Graphic representation of a point (vector)  in R, R2 and R3 .
Alternately, the point  ∈ R can also be viewed as a vector in R , from the origin to the corresponding point
 (see Figure 1 above).
Recalling the addition of vectors, we can define the addition of points in R as follows: if  = (1       ) and
 = (1       ), we define  +  by
 +  = (1 + 1       +  ) 
(1)
There are two multiplication operations which can be performed with a point/vector in R :
i) multiplication by a scalar: if  = (1       ) ∈ R and  ∈ R, we define the scalar product  ∈ R by
 = (1       )
(2)
ii) dot product: if  = (1       ),  = (1       ) ∈ R , we define the dot product  ·  ∈ R by
 ·  = 1 1 +     
1
(3)
Using Pythagora’s theorem, it is not difficult to see that the length of the vector  = (1       ) ∈ R (denoted
by ||||) is given by
q
√
|||| = 21 +    + 2 =  · 
(4)
The length kk of a vector in R is called in mathematics the norm of . It has the following properties:
Proposition 1.1 For any   ∈ R we have the following:
i) kk ≥ 0
ii) kk = 0 implies  = 0 = (0     0).
iii) kk − kk ≤ k − k ≤ kk + kk
iv) k−k = kk
v) kk =  kk for any   0.
Proof. Exercise.
y
x−y
x
0
Figure 2: The distance between the points  and  is the length k − k of the vector  − .
From Figure 2 we can see that the distance between the points  = (1       ),  = (1       ) ∈ R is
equal to the length of the vector  − , hence we define
q
 ( ) = || − || = (1 − 1 )2 +    + ( −  )2 
(5)
Remark 1.2 Note that in the case  = 1, the above formula becomes
q
2
 ( ) = || − || = (1 − 1 ) = |1 − 1 | 
in which we recognize the usual formula for the distance between two points on the real line: the distance is just the
absolute value of their difference.
Also, in the cases  = 2 and  = 3 the above formula becomes the familiar formulae from analytic geometry
q
 ( ) = || − || = (1 − 1 )2 + (2 − 2 )2 
respectively
 ( ) = || − || =
q
(1 − 1 )2 + (2 − 2 )2 + (3 − 3 )2 
Thus, in general, we will write k − k for the distance between two points in R (perhaps in the case  = 1 it
is more common to write | − | instead k − k, although both notations are acceptable).
Often, we are interested in describing the points which are “close” to a given point  ∈ R . Recalling that the
distance between  and  is just k − k, this leads to the following definition:
2
R
x
r
B(x, r)
x−r
x
x+r
R2
B(x, R)
x1
0
R3
x
0
r
B(x, R)
Figure 3: The ball  ( ) centered at  of radius   0 in R, R2 and R3 .
Definition 1.3 We define the open ball with center  ∈ R and radius   0 by
 ( ) = { ∈ R : k − k  } 
(6)
Remark 1.4 In the case  = 1, we obtain
 ( ) = { ∈ R : | − |  } = ( −   + ) 
so in this case the ball is just the interval centered at  of radius .
In the case  = 2, we obtain
n
o
2
2
 ( ) = { ∈ R : || − ||  } =  = (1  2 ) ∈ R2 : (1 − 1 ) + (2 − 2 )  2 
so in this case the ball is just the disk centered at  of radius .
In the case  = 3, we obtain
n
o
 ( ) = { ∈ R : || − ||  } =  = (1  2  3 ) ∈ R3 : (1 − 1 )2 + (2 − 2 )2 + (3 − 3 )2  2 
so in this case the ball is the usual 3-dimensional ball centered at  of radius .
In the higher dimensional cases (  3), even though we cannot represent graphically the ball  ( ), we will
still refer to  ( ) as the open ball of center  and radius .
The notion of an open set, extends the notion of an open interval. Recall that in R, an open interval ( ) is
the set of all points between (and not including) the endpoints  and , that is
( ) = { ∈ R :     } 
What makes the interval ( ) open, is the following property:
∀ ∈ ( )
∃  0 
( −   + ) ∈ ( ) 
which can also be written in the following form
∀ ∈ ( )
∃  0   ( ) ∈ ( ) 
In this form, this is exactly the definition of an open set in R :
Definition 1.5 A set  ⊂ R is called open if for any point  ∈  there exists an   0 such that  ( ) ⊂ .
A set  ⊂ R is closed if the complement  = R −  is an open set.
A neighborhood of a point  ∈ R is any set  =  ⊂ R for which there exists   0 such that  ( ) ⊂  .
Note that by definition, the empty set ∅ and R are both open sets and closed sets.
Example 1.6 For any   ∈ R,   , the open interval ( ) is an open set, and the closed interval [ ] is a
closed set. Note that the intervals ( ] and [ ) are neither open nor closed. Conversely, it can be shown that a
set  ⊂ R is open if and only if it is a (countable) union of open intervals, that is  = ∪≥1 (   ).
3
Some properties of the open/closed sets are contained in the following:
Proposition 1.7 Finite intersections of open sets and arbitrary unions of open sets are open. Finite unions and
arbitrary intersections of closed sets are closed sets.
Proof. Exercise.
The following definition is meant to indicate the “relative position” of a point with respect to a set.
Definition 1.8 A point  ∈ R is called an:
i) interior point of  if there exists   0 such that  ( ) ⊂ ;
ii) closure point of  if for any   0  ∩  ( ) 6= ∅;
iii) boundary point of  if for any   0,  ∩  ( ) 6= ∅ and  ( ) * ;
iv) accumulation point / limit point of the set  if for any   0,  ∩ ( ( ) − {}) 6= ∅;
v) isolated point if  ∈  but  is not an accumulation point of .
Remark 1.9 The notion of accumulation / limit point of a set has the following equivalent definition. A point
 ∈ R is a limit point of  if and only if there exists a sequence 1  2     ∈  − {} of points in  − {} such
that lim→∞  = .
In other words, a limit point point  of  is a point which can be obtained as the limit of points in  − {} (both
requirements that  ∈  and  6=  are essential here).
Definition 1.10 Given a set  ⊂ R , we define:
◦
i) the interior of the set  (denoted ) as the set of all interior points of , that is
◦
 = { ∈ R :  interior point of }
ii) the closure of the set  (denoted ̄) as the the set of all closure points of , that is
 = { ∈ R :  closure point of }
iii) the boundary of  (denoted ) by:
◦
 = ̄ − 
◦
Remark 1.11 It can be shown that  is the largest open set contained in , more precisely
◦
=
[

\

⊂−
and ̄ is the smallest closed set containing , that is
̄ =
⊂−
We have the following:
◦
Proposition 1.12 Given an arbitrary set  ⊂ R , the interior  is an open set and the closure ̄ is a closed set,
and we have:
◦
 ⊂  ⊂ ̄
Example 1.13 For any   ∈ R with   , the interior of the set ( ] is ( ) and its closure is [ ].
◦
For the set  = (1 2] ∪ {3}, the set of interior points is  = (1 2), the set of closure points is ̄ = [1 2] ∪ {3},
the boundary points are  = 1, 2 and 3, the set of accumulation points is [1 2], and the only isolated point is  = 3.
4
1.1
Functions
An important notion in mathematics is that of a function. Recall that a function is a rule which assigns to each
number in a certain set (called the domain of the function) a unique number in a certain set (called the range of
the function). If  denotes the “rule” which describes the correspondence between the points in the domain  and
the range  of the function, it is customary to write  :  →  (read as “ defined on  with values in ”) for
the corresponding function. In this notation,  () represents the point in  which corresponds to the point  ∈ .
Note that according to the definition, a function is a triple (  ), that is a rule, a domain, and a range. In
particular, two functions 1 : 1 → 1 and 2 : 2 → 2 are equal if and only they have the same domain, the
same range, and the same rule, that is 1 = 2 , 1 = 2 and 1 () = 2 () for all  ∈ 1 = 2 .
The graph of the function  :  →  is the set  = {(  ()) :  ∈ }. In the case when  ⊂ R (or  ⊂ R2 )
and  ⊂ R, we can represent graphically the graph  , by plotting the points with coordinates (  ()) for all
 ∈ .
Given two functions  :  →  and  :  →  (note that the range of  equals the domain of ), we may define
the composition of  and  as the function  ◦  :  → , ( ◦  ) () =  ( ()),  ∈ .
The composition of functions is associative (meaning that ( ◦ ) ◦  =  ◦ ( ◦  ), whenever the composition of
  and  makes sense), but is not in general commutative (meaning that  ◦  6=  ◦ ).
For a given function  :  →  it may happen that distinct points  are mapped into distinct points  () when this happens, we say that the function is injective. Also, it may happen that any point in the range of the
function is the image under  of a point in its domain - when this happens, we say that  is a surjective function.
The formal definitions are the following.
Definition 1.14 We say that the function  :  →  is
i) injective if
∀1  2 ∈ , 1 6= 2 =⇒  (1 ) 6=  (2 ) 
or equivalently
∀1  2 ∈ ,  (1 ) =  (2 ) =⇒ 1 = 2 
ii) surjective if
∀ ∈  ∃ ∈  s.t.  () = 
iii) bijective if it is both injective and surjective.
The important consequence of the fact that a function  :  →  is bijective is that to each  ∈  corresponds
exactly one  ∈  such that  () =  (why?). This allows to define the inverse function of  , denoted by
 −1 :  → , by defining  −1 () =  if  () = .
If  is a bijective function, then
 −1 ◦  () =  for all  ∈ 
(7)
and
 ◦  −1 () =  for all  ∈ 
(8)
Next, we present a brief overview of the most common mathematical functions.
1.1.1
Linear functions
A linear function is a function  : R → R,  () = +, where   ∈ R are constants. The graph of a linear function
is a line, and it can therefore be sketched by ploting two arbitrary distinct points (1   (1 )) and (2   (2 )), and
joining them by a line. Note that in the case when the domain of the linear function is just a subset of R (instead
of all real numbers), the graph of the function consists only of a part of a line (instead of a whole line).
Recall that the slope  of a line is defined as the tangent of the angle  between the line and the horizontal
axis, that is
 = tan .
(9)
If (1  1 ) and (2  2 ) are two points of the line, simple geometric consideration show that the slope  can be
found by using the formula
2 − 1
=

(10)
2 − 1
5
If two lines 1 and 2 have slopes 1 , respectively 2 and they are parallel, then 1 = 2 . If they are
perpendicular, their slopes verify 1 · 2 = −1.
Finally, note that when the equation of the line is written in the form  =  + , its slope is exactly the
coefficient of , that is  = .
Example 1.15 Sketch the graph of the function  : R → R,  () = 2 + 3. Redo the exercise in the cases when
 : [0 ∞) → R and  : (−1 3] → R.
Conversely, if the graph of a linear function is known, we can find its equation by using the formula for the
equation of a line determined by two distinct points (1  1 ) and (2  2 )
 − 1
 − 1
=

2 − 1
2 − 1
(11)
or the equation of a line determined by a point (1  1 ) and its slope 
 − 1 =  ( − 1 ) 
(12)
Solving the above equations for , the corresponding linear function is given by  () = , for  in the domain of
the function (note that the domain of the function can be “read” from its graph, but its range cannot be explictly
determined from the graph - why? When the range of a function is not explicitly stated, we usually consider the
range to be R, R2 , aso).
Example 1.16 Find the equation of the line passing through the points (1 5) and (2 7).
1.1.2
Quadratic functions
A quadratic function is a function of the form  : R → R,  () = 2 +  + , where    ∈ R and  6= 0 (if
 = 0, the function is linear). Its graph is a parabola:
• oriented upwards if   0 and downwards if   0,

• symmetric with respect to the line  = − 2
,
´
³
 2 −4
 4
;
• with vertex at the point  − 2
• -intercepts (points where the graph crosses the -axis): if ∆ = 2 − 4 ≥ 0 and 12 are the solutions of the
equation  () = 01 , the parabola crosses the  axis at the points with coordinates (1  0) and (2  0).
Example 1.17 Sketch the function  : R → R,  () = 2 −2+3. Is this function injective, surjective or bijective?
Can you modify the domain/range of  such that  is bijective? What is in this case its inverse function?
1.1.3
Trigonometric functions
¡
¢
For angles  ∈ 0 2 , the trigonometric functions of the angle  are defined as follows. Consider a right triangle
∆ with angles  = ,  = 2 and  = 2 − , as in Figure 4 below.
The trigonometric functions are defined as follows
sin  =
tan  =

 
=




 
=


 

 
=




 
cot  =
=


 
cos  =
The definition is then extended to any angle  ∈ [0 2) by
⎧
if  ∈ [ 2  )
⎨ sin ( − ) 
− sin ( − )  if  ∈ [ 3
sin  =
2 ) 
⎩
− sin (2 − )  if  ∈ [ 3

2)
2
√
−± 2 −4
solutions of the equation 2 +  +  = 0 are given by 12 =
if ∆ = 2 − 4 ≥ 0. The same formula also works
2
in the case when ∆  0, but the solutions 12 are in this case complex numbers.
1 The
6
C
π
2
A
α
−α
B
Figure 4: A right triangle with angle ∠ = .
and similar formulae for cos , tan  and cot  (recall the trigonometric circle!). Finally, the definition of sin  and
cos  is extended to any  ∈ R by 2 periodicity, i.e. sin (2 + ) = sin . The tangent and cotangent functions
are extended by periodicity  (note tan 2 and cot 0 are not defined).
Some trigonometric formulae:
sin2  + cos2  = 1
sin ( ± ) = sin  cos  ± cos  sin 
cos ( ± ) = cos cos  ∓ sin  sin 
±
∓
sin  ± sin  = 2 sin
cos
2
2
+
−
cos  + cos  = 2 cos
cos
2
2
+
−
cos  − cos  = −2 sin
sin
2
2
sin 2 = 2 sin  cos 
cos 2 = cos2  − sin2 
1 − cos (2)
sin2  =
2
1 + cos (2)
2
cos  =
2

£
£
¤
¤
The function  : − 2  2 → [−1 1],  () = sin  is bijective. Its inverse  −1 : [−1 1] £→ − 2¤ 2 is called the
arcsine (inverse sine) function, and is denoted  −1 () = arcsin , in other words for  ∈ − 2  2 and  ∈ [−1 1]
we have
sin  =  ⇐⇒ arcsin  = 
Similarly, the arccosine (cosine inverse) function arccos : [−1 1] → [0 ] is defined as the inverse of the cosine
function, restricted to the interval [0 ].
1.2
Exponential and logarithmic functions
The exponential function (with base   0) is the function  : R → R,  () =  .
When   1 the function is increasing, when  = 1 is constant, and it is decreasing when 0    1 (see the
graphs below).
7
(a > 1)
y
y = ax
x
y=a
y
x
(0 < a < 1)
x
Figure 5: The graph of  =  in the case   1 (increasing) and 0  0  1 (decreasing).
Some properties of the exponential function
 · 



( · )
³  ´

( )
= +
= −
=  · 

=

= ·
for any    0 and   ∈ R.
From the graphs above it can be seen that for any value of   0,  6= 1, the function  : R → (0 ∞),  () = 
is bijective. Its inverse,  −1 : (0 ∞) → R,  −1 () = log  is called the logarithm (with base ) of . Note that
this means
log  =  ⇐⇒  = 
for any   0,  6= 1,   0 and  ∈ R.
Some properties of the logarithms
log ( · ) = log  + log 
µ ¶

log
= log  − log 

log ( ) =  log 
log 
log  =
log 
for any    0,   6= 1,   0 and  ∈ R.
8
1.3
Exercises
1. Evaluate.
h
¡ 6 ¢0 i−1
(a) 7 − 5 13
(b)
(c)
−4
−2
− 34
0
4
−18·3−3
5
3−2 +( 23 )
(
)
( )
( 12 )−2 −5(−2)−2 +( 23 )−2
( 38 )−1 ( 32 )3 +(− 13 )−1
2. Simplify the given expressions.
(a)
(b)
(c)
4−2 −9 −2
−
·
−2 −−2
2−1 +3−1
+
−3 −−3
−1 + −1 · −2 +−1 −1 +−2
³
´−2 ³
´−2 ³
´2
+
−
+
−
+
−
h
i−1 h
i−1
(−2)+0
(d) − (1 − )−1
2
 −+1
3. Without using a calculator, determine which of the given number is larger.
¡ ¢4
¡ 1 ¢3
(a) − 18 or − 32
¡ ¢4
¡ ¢6
(b) − 67 or 36
49
(c) 81150 · 8200 or 3600 · 1675
¡ ¢−65
(d) 365 or 13
4. Evaluate the following.
(a)
(b)

³
33 1
3
22 ·52
+ 23

2
1
· 43 : √13 +01: 200
12−2( 34 +025)·3−2
√1
2
´2
+
+
√ √
√8+ √20
2+ 5
5
4
1
7
10
3 18 :[(4 12
−3 13
24 )· 7 (3 18 −2 12 )·1 17 ]
√
1
1
2 5 + 1 − 38025+01
1
3
·
√1
2
√
3
1
[15:3 35 +(10 12 : 12 )· 14
]·(1 23
125
52 −1 4 )
√
+
1
3
5
5
(2 4 +025·8 7 )− 14
√
√
1
5
p
1
(15−9 ):2
071−
289
(d) √80
− 071+ 14 · 19 2 −113 7 ·99 : 16
− 0 1 (6) :
5
( 3
4
9 ) 71
³ √ ´2
42
53
3 13
:
+
6
−2375
(
)
56
2
· 1227
: 23
(e) 15 45
1+ 1
(c)
7
12
+
1
19
3+ 4
5
+
8−
2
7
5−15
5. Factor completely the given expressions.
(a) 52  + 15 − 5
(b) 83 − 122 2 + 42 
(c) 3  + 2 3 + 2 
(d) −103 − 6 + 2
(e) 92 − (2 + 3)
2
(f) 2 + 6 + 9
(g) 8 − 64  3 + 9 6
(h) −302 + 254 + 92
(i) −62 3 + 1 + 94 6
(j) 3 + 183  3 + 275  6
9
√
3
2
·
√
8
6. Simplify the given expressions.
(a)
24 −83 +82  2
4 −23 
(b)
205 2 +163 2
754 +1202 2 +483
(c)
252 +152 −30
203 +122 −242 
(d)
653 −602  2
392 3 −36 4
(e)
125 −273 2
83 −122  2
7. Solve the given equations.
√ 867
√
3
75 −(− 5 )
841·12
√
= √2523·
(a)

3
(b)
(c)
√
√
√
3 722·
√ 2 = 589

38 3

√
√ 
845
7 5
13 − − 7

√
= 10
7 5
8. Solve the given inequalities.
(a)
(−2)
−3
1
1
2
+ (−2)(+3)
+ (−3)
1
−2 + 1 + −2 + −3  3 − 4
h¡ ¢ ¡ ¢ ³ ´ ¡ ¢ ³ ´ ³ ´i
1
−3
4
(b) − 12 − 34 − −5
− (−2 + 5)   1
+ −2
2
−5
−3
(c)
(3)+(−7)+(−5)+(10)+(−4)+(4)
(−5)+(−7)−(4)+(+10)+(−2)
+

8
 5
7+(−2)−(−3)−4+(+5)−(−1)
(−3)(−2)(−1)
+ 4  − 3
8
¡ 1 ¢ ¡ 3 ¢ £ 5 ¡ 2 ¢ ¡ −3 ¢
¤ 3
5
(e) − 2 − 4 : 6 − 3
(−3)
−
5
8 − 8
(d)
(f)
(−5):(+5)+(+3):(+ 35 )
3
5:
( −2
5 )
(g) 1 1 (36)  +
2
(h)
(i)
5
22
3
+(−2) ]
(−1)100 +(−2)2
[(−3)


200(0005)−001
099
1
5
1
+ 2 (90) − 1 11

2
11

3
+2
+
 1 (99)
9. Consider  = {0 1} and  = {  }. Determine all the functions  :  → , and decide which of these are
injective. Are any of these functions surjective or bijective?
10. Consider   : R → R,  () = | − 1| + 2 and  () = | − 2| + 1. Find  ◦  and  ◦  .
11. Consider    : R → R,  () = 2 + 3,  () = ||   () = 2 .
(a) Find  ◦  and  ◦ . Are they equal?
(b) Find  ◦  and  ◦  . Are they equal?
(c) Find  ◦  and  ◦  . Are they equal?
(d) Find ( ◦ ) ◦  and  ◦ ( ◦  ). Are they equal?
(e) Which of the functions    is injective? Surjective? Bijective?
(f) For the functions that you determined that are bijective, find the corresponding inverse.
12. Consider the functions   : R → R defined by
½
3 + 1  ≤ −1
 () =
−2
1
and
 () =
(a) Sketch the graphs of  and .
(b) Are the functions  and  injective? Surjective? Bijective?
10
½
−3
 ≤ −2

 − 1   −2
(c) Find  ◦  and  ◦  .
13. Consider the functions   : R → R defined by  () = 24 + 4 and  () = 3 +  + 2. Show that  is not
injective and  is injective.
14. Sketch the indicated functions
(a)  :  → R,  = 1 2 3 where 1 = R, 2 = [1 ∞), 3 = (−1 3] and 1 () = 2 () = 3 () = 2 + 1.
Are any two of the three functions equal?
(b)  :  → R,  = 1 2 3 where 1 = R, 2 = (0 ∞), 3 = (−∞ 1] and 1 () = 2 () = 3 () = 3 − 2.
Are any two of the three functions equal?
(c) 12 : {−1 0 1} → R2 , 1 () = 2 and 2 () = ||. Are these functions equal?
15. Find the equation of the line through the points with coordinates (2 1) and (1 2).
16. Find the equation of the line through the point (2 1) and parallel to the line through (5 2) and (0 45).
17. Find the equation of the line through origin and perpendicular to the line with equation  = 2 + 3.
18. Are the lines 2 + 3 = 4 and 3 − 2 = 4 parallel or perpendicular to each other?
19. Find the point of intersection of the lines 2 + 3 = 4 and 3 − 2 = 4
(a) Graphically
(b) Numerically
20. Solve the equation −22 + 3 + 5 = 0.
21. Factor the polynomial 42 − 8 − 5.
22. For which value of  do the solutions of the equation 22 +  + 5 = 0 coincide?
23. Determine the axis of symmetry, the vertex and the -intercepts (if any) of the following quadratic functions,
then sketch their graphs.
(a)  : R → R,  () = −22 +  + 1
(b)  : R → R,  () = 2 +  + 1
(c)  : R → R,  () = −2 + 3 − 12
(d)  : R → R,  () = 12 2 − 13  + 14 
24. Determine the maximum or minimum of the following quadratic functions.
(a)  : R → R,  () = −2 + 2 − 1
(b)  : R → R,  () = 052 +  + 1
(c)  : R → R,  () = 2 + 3 + 2
(d)  : R → R,  () = −22 + 4 + 5
25. Determine the sign of the following quadratic functions.
(a)  : R → R,  () = 2 − 6 + 5
(b)  : R → R,  () = 2 −  +
1
4
(c)  : R → R,  () = −2 − 5 − 6
(d)  : R → R,  () = −042 − 5
26. Sketch the graph of the function  : R → R defined by
⎧
  −2
⎨ − + 2
2 
−2 ≤  ≤ 2
 () =
⎩
+2
2
11
27. Find the values of  and  for which the quadratic function  : R → R,  () = −2 +  +  reaches its
maximum value 5 at  = 3.
28. Find the equation of the parabola through the points (−1 10), (1 3), and (2 55).
29. Solve the following inequalities.
(a) −2 + 3 + 30 ≤  ( − 1)
(b) 32 + 5 + 3  1
(c) −22 + 6 − 3  5
2 + 2 − 2
3
2 − 3 + 4
5 + 7 2 + 21
2
(e)
−

−2
+7
3
(d)
30. Solve the given systems.
½ −1
1
−1 = 5
(a)
+4
2
+4 = 5
½
0 25 + 0 04 = 2
(b)
4 + 25 = 641
½
 + 5−1
= − 23
3
(c)
2+3
3 − 2 = − 21
2
½ 5
4
−1 : −1 = 25 : 24
(d)
2
3
+1 : +1 = 7 : 12
½ +
3 +  = 15
(e)
 − −
5 =6
½
10 − 3 2 = 48
(f)

 − 3 = 0
31. Solve the following systems of equations.
½
 + 2 = 5
(a)
2 + 2 2 −  = 7
½
 − 3 = 0
(b)
22 + 6 − 9 2 =  + 2
32. Solve the following systems of inequalities.
½
72 + 13  2
(a)
2
−5 + 6 − 1  0
⎧
3 − 1 ≤ 0
⎨
32 − 7 + 2 ≤ 0
(b)
⎩
−2 + 2 − 5  0
½
2 ( − 1) ≥ 4 ( + 1)
(c)
2 + 4  0
33. Consider the function  :  → ,  () = 2 + 5 + 6 Determine a choice for the sets  and  such that the
function  is bijective, and then find its inverse function  −1 :  → .
34. Sketch the graphs of the exponential functions   : R → R defined by  () = 2 and  () = 05 .
√
6
35. Decide which of the numbers 1+
√ √
2+ 3
and 
is larger.
36. Solve the given inequalities.
12
2
(a) 5 −−2  1
³ √ ´+3 µ 1 ¶|−4|

(b) 2 2
8

2 −4
(c)
0
32 − 8 + 5
37. Solve the following equations.
(a) 3+1 + 3 = 108
(b) 2+1 − 3 · 2−2 + 32 = 0
(c) 2+2 − 2−1 = 28
(d) 52+1 − 52−1 = 120
(e) 2 + 2+1 + 2+2 = 6 + 6+1
(f) 3+1 + 5+2 = 3+5 − 3 · 5+3
(g) 2 · 22 − 3 · 2 + 1 = 0
(h) 2 + 4 = 272
16
1
(i) 1 + 
= −3
3 −1
3
38. Solve the following equations.
¢
¡
(a) log 32 − 5 = 1
¢
¡
(b) log+1 32 + 2 − 3 = 2
(c) lg ( + 6) − 2 =
1
2
lg (2 − 3) − lg 25
(d) lg2  − 4 lg  + 3 = 0
1
(e) lg4  − 3 lg2  + 8 = 0
4
(f) log3 +4 = 243
(g) log2  + log3  = 1
39. Solve the given equations.
¡
¢
(a) 4 2 − 1 = 4 − 1
2
+2
(b)
+
=2
+2
2
+1
2 − 1
=
(c)
+1
−2
+8
24
(d)
−
=2
−8 −4
1
 +6

− = +
(e)
−6 2
6 6−
40. Solve the inequalities
(a) ( + 1) ( − 2) ≥ 0
(b) −22 + 6 − 3  5
(c) 52 + 3 + 1  0
2 + 2 − 2
3
2 + 3 − 4
5 + 7 2 + 21
2
(e)
−

−2
+7
3
(d)
13
41. For which values of  is the function  () = ( − 2)  +  increasing? Decreasing? Constant?
42. Determine the sign of the given functions  : R → R.
(a)  () = 2 − 6 + 5
1
(b)  () = 2 −  +
4
2
(c)  () = − − 5 − 6
(d)  () = 2 + 3 + 13
(e)  () = 0 42 − 5
43. Determine the sign of the given functions (also indicate the largest domain of definition of the function).
(a)  () =
(b)  () =
−3
+2
43 − 32
3 + ( + 2)3
44. Determine the number of solutions of the equation 2 −2 ( + 1) +8 = 0 in terms of the parameter  ∈ R.
45. For what values of  ∈ R we have ( − 1) 2 +  +  + 1  0 for any  ∈ R?
46. For the given function, find the vertex, axis of symmetry and -intercepts (if any), then graph.
(a)  () = 2 − 6 + 8
(b)  () = −22 + 7 − 5
(c)  () = −2 + 6 − 9
(d)  () = 22 − 3 + 1
(e)  () = −2 + 25
47. Solve the given equations.
√
(a)  + 9 +  + 8 = 0
√
(b) 24 − 10 = 3 − 4
√
(c) 1 − 13 + 32 = 2
p
√
(d) 1 − 4 −  =  − 1
48. Determine the quadratic function  () = 2 +  +  such that its graph passes through the points (1 2),
(−1 6) and (2 3).
49. Determine the quadratic function  () = 2 +  +  with vertex at the point  (4 −4) and -intercept
(0 12).
50. For the given functions  : R → R, determine if they are (or not) increasing, decreasing, injective, surjective,
bijective. If they are bijective, also find the inverse function.
½
 + 1   0
(a)  () =
2 + 1  ≥ 0
½

≤1
(b)  () =
2 − 2 + 2   1
½
−2 + 1  ≤ 0
(c)  () =
−2 + 1   0
14