5–24 Sketch the graph of the equation by translating, reflecting, compressing, and stretching the graph of
y  x 2 , y  x , y  1/ x , y  x , or y  3 x appropriately. Then use a graphing utility to confirm that
your sketch is correct.
Solution:
Translate left 1 unit, stretch vertically by a factor of 2, reflect over x-axis,
translate down 3 units.
Solution:
Translate right 3 units, compress vertically by a factor of 1/2, and translate
up 2 units.
Solution:
y = (x + 3)2 = 9; translate left 3 units and down 9 units.
Solution:
y = 1/2 [(x - 1)2 + 2]; translate right 1 unit and up 2 units, compress vertically by
a factor of 1
Solution:
Translate left 1 unit, reflect over x-axis, translate up 3 units.
Solution: Translate right 4 units and up 1 unit.
Solution:
Compress vertically by a factor of 1/2 , translate up 1 unit.
Solution:
Stretch vertically by a factor of
3 and reflect over x-axis.
Solution:
Translate right 3 units.
Solution:
Translate right 1 unit and reect over x-axis.
Solution:
Translate left 1 unit, reflect over x-axis, translate up 2 units.
Solution:
y = 1-1/x; reflect over x-axis, translate up 1 unit.
Solution:
Translate left 2 units and down 2 units.
Solution:
Translate right 3 units, reflect over x-axis, translate up 1 unit.
Solution:
Stretch vertically by a factor of 2, translate right 1/2 unit and up 1 unit.
Solution:
y = |x -2|; translate right 2 units.
Solution:
Stretch vertically by a factor of 2, reflect over x-axis, translate up 1 unit.
Solution:
Translate right 2 units and down 3 units.
Solution:
Translate left 1 unit and up 2 units.
Solution:
Translate right 2 units, reflect over x-axis.
25. (a) Sketch the graph of y = x + |x| by adding the corresponding y-coordinates on the graphs of y = x
and y = |x|.
(b) Express the equation y = x + |x| in piecewise form
with no absolute values, and confirm that the graph you
obtained in part (a) is consistent with this equation.
Solution:
26. Sketch the graph of y = x + (1/x) by adding corresponding y-coordinates on the graphs of y = x and y =
1/x. Use a graphing utility to confirm that your sketch is correct.
Solution:
Questions 27–28: Find formulas for f + g, f − g, fg, and f /g, and state the domains of the functions.
x
1
, g ( x) 
28. f ( x) 
2
1 x
x
Solution:
2
x
1 x  x  1 x
x2  1  x2
2x2  1
f g
 


1  x2 x
x 1  x2
x 1  x2
x 1  x2








Domain of f; (-ꝏ ,+ ꝏ)
Domain of g; x≠ 0 ,or (-ꝏ,0)U(0 ,+ ꝏ)
Domain f+g =Domain of f Ⴖ Domain of g
Domain of f+ g = (-ꝏ ,+ ꝏ) Ⴖ (-ꝏ,0)U(0 ,+ ꝏ)
Domain of f+ g = x≠ 0 or (-ꝏ,0)U(0 ,+ ꝏ)
2
x
1 x  x  1 x
x2 1  x2
1
f g 
 


2
2
2
1 x x
x 1 x
x 1 x
x 1  x2






Domain f - g =Domain of f Ⴖ Domain of g
Domain of f - g = (-ꝏ ,+ ꝏ) Ⴖ (-ꝏ,0)U(0 ,+ ꝏ)
Domain of f - g = x≠ 0 or (-ꝏ,0)U(0 ,+ ꝏ)
1
 x  1 
fg  

2 
2
 1  x  x  1  x
Domain f g =Domain of f Ⴖ Domain of g
Domain of f g = (-ꝏ ,+ ꝏ) Ⴖ (-ꝏ,0)U(0 ,+ ꝏ)
Domain of f g = x≠ 0 or (-ꝏ,0)U(0 ,+ ꝏ)
x
f 1  x2
x2


1
g
1  x2
x
Domain f/ g = (Domain of f Ⴖ Domain of g) \g(x)=0
Domain of f/ g =[ (-ꝏ ,+ ꝏ) Ⴖ (-ꝏ,0)U(0 ,+ ꝏ)] \1/x=0
Domain of f/ g = x≠ 0 or (-ꝏ,0)U(0 ,+ ꝏ)
Solution of Q.27:
Solution of Q.28:
Solution of Q.29:


Question 30. Let g  x   x . Find
(a). g (5s  2)
1
(d ).
g ( x)
(b). g ( x  2)
(c).3g (5 x)
( f ).  g ( x)   g ( x 2 )
2
(e). g ( g ( x))

( g ). g (1/ x )
(h). g  x  1
Solution:
We have given
g  x  x
2

 g  x  x
(a). g (5s  2)  5s  2
(b). g ( x  2) 
(i). g ( x  h)
 g  x  x
x 2
(c).3g (5 x)  3 5 x
 g  x  x
1
1

g ( x)
x
 g  x  x
(d ).
(e). g ( g ( x))  g ( x ) 
( f ).  g ( x)   g ( x 2 ) 
2
( g ). g (

1
)
x
(h). g  x  1
2
 g  x  x
x
 x
2
 x2  x  x  0
1
x

 g  x  x
 g  x  x
 x  1
 g  x  x
 x 1
2
 g  x  x
(i ). g ( x  h)  x  h
Question 31–34: Find formulas for f g and g f , and state the domains of the compositions.
Solution of Q.31:
Solution of Q.32:
f
g
 x  
f ( x)  x  3,
f ( g ( x))  f
As f ( x)  x  3
f
Domain of
Or
g
 x  
f
x2  3



x2  3 
x2  3  3
x2  3  3
x 2  3  3 : x2  3  3  0
x2  3  3 ,
Domain of

g ( x)  x 2  3
x 2  3  9,  x 2  3  3  9  3,  x 2  6,  x  6

  6,  
g :  ,     ,  6 U 
x 2  3  3 , ,  6 U
Then, Domain of f
 
 
6,   ,  6 U
6, 

g
f  x   g ( f ( x))  g
As g ( x 2 ) 

x 3
g
f
 x  

2

x 3

3  x 3  3  x
x
3, 
x2  3  0, hence  ,  
x  : x  0,  0,  
Domain of f: x  3  0, x  0,
Domain of g:
Domain of

Then
Domain of g f :, 3,    0,    3,  
Solution of Q.33:
x
1
, g ( x) 
2
1 x
x
1
 f g  x   f ( g ( x))  f  
 x
1
1
x
1
x
As f   

 2
2
2
 x 1  x 1
 x
1
x 2 
1  
 x
 x 
x
 f g  x   2
x 1
Solution of Q.34:
f ( x) 
Domain of f: x2  1  0,  ,  
Domain of g: x  0, hence  ,0   0,  
Then
Domain of f g :  , 0    0,     ,     , 0    0,  
 x 
f  x   g ( f ( x))  g 
2 
 1 x 
x
1
1  x2 1
)


 x
As g (
x
1  x2
x
x
2
1 x
1
g f  x
x
g
 , 
x  0, hence  ,0   0,  
Domain of f: x2  1  0,
Domain of g:
1

Domain of   x  : x  0,  , 0    0,  
x

Then
Domain of g f :,  ,     , 0    0,     , 0    0,  
Question 35–36: Find a formula for f ◦g◦h.
Solution of Q.35:
f
36: Find
g h  ( x)
if
f ( x) 
1
,
1 x
g ( x)  3 x ,
h( x ) 
1
x3
Solution:
We have  f g h  ( x)  f ( g (h( x)))
h( x ) 
1
x3
g( )  3 
f
 
 g (h( x) )  3  h( x)

1
1 

 f  g ( h( x ) )  

1
1
1
x
 g ( h( x ) )  3

1 1

x3 x
1
x

x 1 x 1
x
x
x 1
Question 37–42: Express f as a composition of two functions; that is, find g and h such that f = g◦h. [Note:
Each exercise has more than one solution.]
Solution of Q.37:
f
g h  ( x)  f ( g (h( x))) 
Solution of Q.38:
(a). f ( x)  x2  1  h  x   1, and
Then g ( x)  x  1, and
g h  g  h  x  1
h  x   x2
1
1
1

, and g h  g  h( x)  
x  3 h( x )
h( x )
1
Then g ( x)  , and h  x   x  3
x
Solution of Q.39:
(b). f ( x) 
Solution of Q.40:
(a). f ( x)  3sin( x2 )  3sin  h( x)  , and
Then g ( x)  3sin x, and
g h  g  h( x)   3sin  h( x) 
h  x   x2
(b). f ( x)  3sin 2 x  4sin x  3  h( x)   4  h( x)  , and
2
Then g ( x)  3x2  4 x, and
Solution of Q.41:
h  x   sin x
g h  g  h( x )   3  h ( x )   4  h ( x ) 
2
Solution of Q.42:
(a). f ( x) 
1
1

and
2
1  x 1  h( x )
Then g ( x) 
1
, and
1 x
(b). f ( x)  5  2 x  5  h( x)
g h  g  h( x )  
h  x   x2
and
Then g ( x)  5  2 x , and
1
1  h( x )
g h  g  h( x)   5  h( x)
h  x   2x
63. In each part, classify the function as even, odd, or neither.
Solution: