Q1: Determine the domain of the following functions:
1-
f ( x )  3x 4  5 x 3  6 x  2
2-
f ( x) 
3-
f ( x)  x  9
4-
f ( x )  3x 4  5x 3  6 x  2
5-
f ( x )  x 4  x 3  5 x  3x  5
6-
f ( x)  x 2  9
7-
f ( x)  9  x 2
x 1
x4
Q2: Let f ( x )  7 x  2 , g ( x )  x  9 , h( x ) 
1-
f (3) 
2-
g (3) 
3- h(3) 
4- ( f  g )( 3) 
5-
f .h(0) 
Q3: Evaluate the following limits:
x2
x 1
x2
2- lim
x  x  1
x2
3- lim
x  1 x  1
1- lim
x 0
x 2  25
x 5 x  5
x 2  x  12
5- lim
x 4
x 2  16
4- lim
6- lim
x 2  x  12
x2  9
7- lim
x 5  32
x 4  16
x 3
x 2
x 2  x 5  12
x  4 x 5  9 x  7
8- lim
3x  7
, find the following:
x2
Q4: Discuss the continuity of the function:
 3x  4 , x  2
f ( x )   3x  5
,x  2
 x  2
, at x=2
Solutions
Q1: Determine the domain of the following functions:
1-
f ( x )  3x 4  5 x 3  6 x  2
Solution
f (x ) is polynomial function, then its domain is R
2-
f ( x) 
x 1
x4
Solution
Domain f (x ) equal R  {2}
3-
f ( x)  x  9
Solution
Domain f (x ) equal {x : x  9, x  R}
4-
f ( x )  3x 4  5x 3  6 x  2
Solution
Domain f (x ) equal R  {0}
5-
f ( x )  x 4  x 3  5 x  3x  5
Solution
Domain f (x ) equal {x : x 
6-
f ( x)  x 2  9
Solution
Domain f (x ) equal {x : x  3
7-
5
, x  R}
3
f ( x)  9  x 2
Solution
 x  3, x  R}
Domain f (x ) equal {x : 3  x  3, x  R}
Q2: Let f ( x )  7 x  2 , g ( x )  x  9 , h( x ) 
1.
2.
3x  7
, find the following:
x2
f (3)  7(3)  2  19
g (3)  3  9  12
h(3)  3(3)  7  2
3 2
4. ( f  g )( 3)  f (3)  g (3)  19  12
3.
5.
f .h(0)  f (0).h(0)  2 
7 3

2 2
Q3: Evaluate the following limits:
x2 02
2
=
x 1 0 1
x2 2 
 , which is un- determinate value, then divide both numerator and
2. lim
=
x  x  1
 1 
1.
lim
x 0
denominator by x to the greatest power (here x)
x 2
2

1
  1 0  1
The result is: lim x x  lim
x  x
x


1
1 1 0

1
x x

x2
lim
1
x  1 x  1 1  2
3.
=

11
2
4.
5.
6.
x 2  25 52  25 0
=
 which is un- determinate value, then by factorizing the numerator the
x 5 x  5
55
0
( x  5)( x  5)
 lim( x  5)  5  5  10
result will be lim
x 5
x 5
lim
x 2  x  12 4 2  4  12 0
 which is un- determinate value, then by factorizing the
=
x 4
x 2  16
4 2  16
0
( x  4)( x  3) x  3 4  3 7
numerator the result will be lim



x 4 ( x  4)( x  4)
x4 44 8
lim
x 2  x  12 32  3  12 0
 which is un- determinate value, then by factorizing the
=
x 3
x2  9
32  9
0
( x  3)( x  4) x  4 3  4 7
numerator the result will be lim



x 3 ( x  3)( x  3)
x  3 3 3 6
lim
7.
lim
x 2
x 5  32
25  32 0
=
 which is un- determinate va5lue, then by using Caushy' theorem
x 4  16
2 4  16 0
the result will be
8.
lim
x 2
x 5  2 5 5 5 4 5
 ( 2) 
x 4  24 4
2

x 2  x 5  12
, the direct substitution gives us , so you can divide both numerator and
5
x  4 x  9 x  7

5
denominator by x to the greatest power(here x )
lim
x 2 x 5 12
1
12
 5 5
1 5
5
3
x = lim x
x  0 1 0  1
lim x 5 x
x 
9
7
x
x
7 x 
4 4  5 400 4
4 5 9 5  5
x
x
x
x
x
Q4: Discuss the continuity of the function:
 3x  4 , x  2
f ( x )   3x  5
,x  2
 x  2
, at x=2
Solution
To discuss continuity, three conditions will be fulfilled


f ( x  2)  3( 2)  4  2 , defined
lim f ( x ) = lim f ( x ) , lim f ( x ) = lim (3x  4)  2 , but lim f ( x ) 
x 2
lim
x 2
x2
x 2
x 2
x 2 
3x  5 3( 2)  5 11


  , which is mean that the limit doesn't exist, there for
x2
22
0
the function f (x ) discontinuous at x=2.