Chapter 1 - Worksheet 3 - NEW FUNCTIONS FROM OLD
1.
NAME Key
141
Answers will vary
Graph that is invertible and at least 2 characteristics. – Either always increasing or always decreasing.
There are no turning points
It is 1 to 1.
For every input there is only one output, every output has only 1 input.
2. For what values of A and K will S (t )  At 3  K be a one-to-one function?
A can be any real number except zero, since A will compress or stretch the function vertically
and K can be any real number, since K will shift the function vertically.
3. The life expectancy, L , of a child can be expressed as a function of the year of birth.
L  f ( y) 
A. find f (10) 
y  66.94
0.01y  1
y  0 corresponds to 1950.
10  66.94
 69.95 years
0.01(10)  1
f (10) means finding the life expectancy of a person born in 1960 (1950+10). A person born in 1960 will
have a life expectancy of 69.95 years.
f 1 ; Start with f ( y ) and solve for y.
y  66.94
L
 (0.01y  1) L  y  66.94  0.01yL  L  y  66.94  0.01yL  y  66.94  L 
0.01y  1
66.94  L
y  f 1 ( L) 
0.01L  1
66.94  (76)
So f 1 (76) 
 37.75 years after 1950.
0.01(76)  1
f 1 (76) tells you the year of birth of a person with a life expectancy of 76 years. A person with a life
B. Find
expectancy of 76 years would have been born in Oct of 1987 or 1987.75
4. Let f ( x) 
(37.75 years past 1950).
x
1
and g ( x ) 
. Find f ( g ( x)) and g ( f ( x)) . Simplify completely.
2
1 x
x 1
1
( x  1) 2
x 1
x 1
 1 
x

1
f ( g ( x))  f 


 2

2
2
2
( x  1)
( x  1)  1 x  2 x  2
 x 1
 1 
1 

 x 1
 x 
g ( f ( x))  f 

2 
1 x 
1
x
1
1 x2

1 x2
1 x2

1 x2 x2  x 1
5. Use the numerical representation of f ( x) below to match the numerical information in
column A with the symbolic representation in column B.
x
f ( x)
-4
5
-2
1
0
6
2
2
4
7
Note: One function in column B has no representation in Column A.
There needs to be some work shown. Show that the new points are correct or explain the transformation
and how it relates to the table.
Column A
Column B
1. __C___
x
-4
7
g ( x)
a.
-2
3
0
8
2
4
4
9
1 
b. f  x 
2 
2. __H___
x
h( x )
f ( x  2)
-2
5
3. __A___
x
-2
5
m( x )
4. _B____
x
-8
n( x )
5
5. __F___
x
-4
-5
g ( x)
6. __D___
x
4
5
k ( x)
7. __E___
x
7
u ( x)
11
-1
1
0
6
1
2
2
7
c. f ( x)  2
0
1
2
6
4
2
6
7
d. f ( x)
-4
1
0
6
4
2
8
7
e. f ( x  3)  4
-2
-1
0
-6
2
-2
4
-7
f.  f ( x)
2
1
0
6
-2
2
-4
7
g. f ( x  2)
5
6
3
10
1
5
-1
9
h.
f (2 x)