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4. In each part, compare the natural domains of f and g.
2
(a). f  x   x  x ; g ( x)  x
x 1
(b). f  x  
x x x
;
x 1
g ( x) 
x
Solution Q.4.
(a).
x 1  0
x  1
(b).
x 1  0
x  1
Therefore, f(x) is defined for x  0 . g(x) is defined
for x  0
Therefore, f(x) is defined for all x except -1. g(x) is all
set of x.
Domain of f ( x)  {x; x  1}
Domain of g ( x)  x  R
Domain of f ( x)  0,  {x  R; x  0}
Domain of g ( x)  0, 
Question 9: Find the natural domain and determine the range of each function. If you have a graphing
utility, use it to confirm that your result is consistent with the graph produced by your graphing utility.
[Note: Set your graphing utility in radian mode when graphing trigonometric functions.]
x
1
(a). f  x  
(b). F  x  
x
x3
(c). g  x  
x2  3
(d). g  x  
x2  2x  5
(e). h  x  
1
1  sin x
(f). H  x  
x2  4
x2
(a). f  x  
1
x3
Solution:
1
x3
x  3  0 or
Domain: f  x  
x3
x3
Domain  ,3U  3,   or x  3
Range: y  f  x 
f 1  y   f 1 f  x 
or
x  f 1  y 
y
1
x 3
x 3 
x
From eq (1)
If y=0
(1)
y  x  3  1
or
1
y
1
3
y
1
f 1  y    3
y
f 1  0  
1
3 3 
0
Range  y  0 or
 ,0U  0, 
(b). F  x  
x
x
Solution:
Domain: F  x  
x
x
x  0 or x  0
Domain  ,0U  0,   or x  0
Range: y  F  x 
F 1  y   F 1 F  x 
or
x  F 1  y 
y
y
x
x
x
 1
x
1,1
Range  y  1 or
(c). g  x  
x2  3
Solution:
Domain:
g  x 
x2  3
 x2  3  0
Or
x2  3  0
x2  3
x 3
Domain of g x  3 or x   3
Range: y  g  x 
g 1  y   g 1 g  x 
or
x  g 1  y 
 3
y  x2  3 
2
3  33  0
So range is zero or greater than zero. Range  0
(d). g  x  
x2  2x  5
Solution:
Domain: g  x   x 2  2 x  5
 x2  2x  5  0
x2  2x  5  0
x2  2x  1  4  0
 x  1
2
40
g  x   x2  2x  5 
 x 1
2
4
Domain for all values of “x”: Domain of g ( x) is all x
2
Range: g  x   x  2 x  5 
 x 1
2
4
for “x” we get y=2, and so on.
y
 x 1
Range  2 Or
2
4 
 0 1
2
4  4 2
Range   2,  
(e). h  x  
1
1  sin x
Solution:
1
1  sin x
or
1  sin x  0
Domain: h  x  


1  sin x  0

sin x  1
,..... then sin x  1
If
x
Or
 1

x    2n     2n  
2
 2

2

, 3
2
, 5
2
Where n  0, 1, 2,.....
sin x  1
Or
1

Domain : x    2n  
2


1

sin x    2n  
2

 n  0, 1, 2,.....
Range : 1  sin x  0
If sin x  1
1  sin x  1 1  0
If sin x  1
(1)
1  sin x  1   1  1  1  2
Range : y 
1
1

1  sin x 2
x2  4
(f). H  x  
x2
Solution:
Domain: H  x  
x 2  22
x2
x 2  0 x  2
x2
H  x 
x 2  22

x2
 x  2  x  2
x20
x  2
x2
 x2
x  2  0
 x  2
Domain of x :  2, 2 U  2,  
 x  2  x  2
x 2  22

 x2
Range: y 
x2
x2
The minimum value we for “x” we put in above eq is “-2” we get y=0.
y  x  2  2  2  0
And above “0” all values +ve values of “x”.
Range  0,  
Q.10.
(a). f  x   3  x
(b). F  x   4  x 2
(c). g  x   3  x
(d). G  x   x  2
(e). h  x   3sin x
(f). H  x   sin x
3


2
(a). f  x   3  x
Solution:
Domain: f  x   3  x
3  x  0 or
x3
Domain
x3
Range: y  f  x 
3 x
f 1  y   f 1 f  x 
or
x  f 1  y 
y  3 x
(1)
if x=3  y  3  3  0
Range  0
(b). F  x   4  x 2
Solution :
Domain: F  x   4  x 2
4  x2  0
4  x2
or
x  2
Domain: x  2 0r 2  x  2
Range: y  f  x 
f 1  y   f 1 f  x 
or
x  f 1  y 
(1)
F  x   4  x2
Range  2
0r
if x=2  y  4  4  0
0 y2
(c). g  x   3  x
Solution :
Domain: g  x   3  x
x 0
x0
Domain: x  0
Range: y  g  x 
or
g 1  y   g 1 g  x 
x  g 1  y 
y  3 x
(1)
if
x 0
 y  3  0  3 and so on
Range  3
3
(d). G  x   x  2
Solution :
3
Domain: G  x   x  2
Domain: all x
Range: y  G  x 
y  x3  2
Range all y
or
G 1  y   G 1 G  x  Or
x  G 1  y 
(1)
(e). h  x   3sin x
Solution:
Domain: h  x   3sin x
Domain: all x
Range: h  x   3sin x or
h 1  y   h 1 h  x 
x  h1  y 
(1)
y  3sin x
We know that sin x is between -1, and 1. So y=3 sin x is between -3, and 3.
Range  3  y  3

(f). H  x   sin x

2
Solution:
x exist for x  0 . H(x) exist for sin x  0 , Therefore
Domain : x  0, x   n  for n  1, 2, .
2

For x: 0  sin x  1 . So, 0  sin x
Range : y  1

2
1
x  n for n  0,1, 2,.......