Class notes
Limits, Derivatives and introduction to calculus
- Course
- Wtw 114
- Institution
- University Of Pretoria (UP)
Lecture notes with examples with steps
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Unit 1.1.· Ex: xEAor
AUB= x E B (or both) Union
· AB =
Ex:xEAand x EB 3 Intersection
·
AlB Sx:x
=
EA and x 4B3 compliment
· RIG Q =
·
(AUB)(AMB) Sx:x =
EA
or B butnot both (x fARB)3
· (-1,3)r[2,4] [2,3) =
⑳ ⑧
8 0
-
1 234
· (1,3)V[2,43 (-1,4] =
⑳ ⑧
8 0
-
1 234
· Solve (x- ) 0
= using a
sign line.
x +
3
x
0,
=
x7 -3
⑧ ⑧
-
3 o
I
- - t +
x
- -
-
+
x-
- - -
f(x z)3 -
- I + x+ 3
-
t -
I
x(x -
z)3
x+ 3
Solution to x(x z)* =
0is( 3,0]U[2,4).
-
-
x +
3
Solve
· Es where aro is fixed.
A =
-
x 0
x
x=
a -
0
I I x
O
2
9
-
- - x
a - x2
0 is-0,0).
-
Solution to
, Unit 1.2. Absolute values
· (3x 2) -
33x 2 if 3x 2 0 -
=
=
2
-
3x if 3x 25 -
0
· Solve 14x2-51=7 using a number line
4-3 4x 5 -
⑧ ⑧
-> ⑧ 7
x =
- or x 3
=
Solve
ll
·
using number line
-
a
x -
3
-
2x+1
. ⑧
-
2 0 2
x = or x = I
· Solve 12x-1)*3 using a number line.
1
-2x
-
1 2x -
⑧ ⑧ 7
-
3 ⑧ 3
x -
= 1 or x12
The solution set
is (-0,-1JV[2,8)
· Thereom:E (x) = for all real numbers
· Thereom:(x) 0
=
precisely when x 0 =
i.e. (x) 0
=
iff x0 =
· Theream:Let
a,b ER, and n EN.
1. lab) (a)(b)=
2. (B1=; provided bo
3. 1an) 1a) =
4.
Special case of 1: if a 0, then lab) alb =
·
Verify with reasons that lab)=(allb) holds for a, b
ER.
every
Let
a, b ER. Then
lab) (b)2 =
(thereom from class)
- (basic algebra)
= (basic algebra)
=(allb) (thereom from class)
,Simplify:1ax31+13-al
· where rar3.
-
Since -3 ar3 then
-
3 +
35a 3/3 +3 +
hence
0 ra +3 16. Since at10 then 1a+3)=
a+3
Since -35953 then
( 1)( 3) - ( 1)a -
- - - (- 1)(3) hence
-
3r-953 hence 3-353-953+3
hence or 3-a/6. Since 3-a%0 then 13-al=
3-a
.
· Answer:6
· Rewrite 19-32
-59-2
if 9 -
x20
x29 if 9 x=-
0
Simplify9-x1 0
(3 x)(3 x)
+ - 0
=
⑧ ⑧
3
-
3
- t t (3 x)
+
I t - (3 x)
-
-
I - (3 x)(3
+ -
)
We see the solution to aux 0 is [-3,3]
Thus the solution to 9-x-0 is
(-2,3)U(3,4)
Now we can write:
if auto
See 9 x=0
-
9S
x2 xf[ 33]
E x2qif
9 -
if -
x (( -
2,3)V(3,4)
· Thereom:For all a
ER, a <Ia1.
· Proof:Let
a ER be
arbitrary. We consider two cases.
Casel:a 0. Then lal= a (algebraic definition of abs value)
Thus atta) and so a fal
Case 2:aro. Since lalfo, we have arial and so a fal
Since Case 1 and Case 2 all possible conclude
cover cases, we
that 1a) for ER
every
a a
, · Thereom: (Triangle inequality)
For all a, b ER,
|a b) (a) (b)
+
=
+
Proof:Let
·
a, b ER be arbitrary.
|a b12 1(a b)2
+ = +
(thereom from class:(xc (x2)) =
(a
=
+ (Ca+b)0, algebraic definition of absolute value)
b)2
a2 2ab
=>
b2 + (basic algebra)
+
a2 2ab) b2
=> +
(thereom from class x = (x1)
+
-(a2 2(ab) (b4 +
Cao,b20, algebraic definition of absolute
+ value)
-la +21ab) + b12 (thereom from class:(x2) (x2) =
-la +2(allb) (b)2 (thereom from class:(xzy(=(x11y1)
+
(1a) 1bl)
=
2
+
(basic algebral
So la +
b12= (1a) +
1b!)2
Thus (a + b) = (a1 +
1b) (x2[y2 xzy when xx0,y 0) =
· If (x - 1) 3, show that
= 12x-5129 by using the T.I.
12x -
5) 12(x =
-
1) 31
-
(basic algebra)
(2(x 1) ( 3))
= - +
-
(basic algebra)
(2(x 1)1
->
-
13)
+
Itriangle inequality)
2(x 1) 1 3)
=
-
+ -
(thereon from class:(ab) =
alb) when a fol
2(x
=
-
1) +
3 (basic computation)
2.3 3
= +
(by assumption (x-1) = 3)
= G (basic computation)
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