Combining Functions
COMBINING FUNCTIONS
Students should be able to create a new function by using addition, subtraction, multiplication and division to combine two different
linear and/or exponential functions. They will also need to be able to do simple compositions of functions. Discussions with students
should include the impact on the domain and range when two functions are combined together. This can be done by looking at the
graphs, the equations or even tables of values. Using a graphing calculator or other technology can be very useful when looking at the
graphs of these combined functions. It will allow students to examine graphs that are more difficult to create by hand and might
otherwise be inaccessible.
Examples:
Let f(x) = 4x – 5 and g(x) = 3x. Find f ( x)  g ( x), f ( x)  g ( x), f ( x) g  x  , and

f ( x)  g ( x)  4 x  5  3x

f ( x)  g ( x)  4 x  5  3x

f ( x) g ( x)  (4x  5) 3x

f ( x) 4 x  5
 x
g ( x)
3
f ( x)
.
g ( x)
Composing Functions
COMPOSING FUNCTIONS
The term "composition of functions" (or "composite function") refers to the combining of functions in such a way that the output from
one function
becomes the input for the next function. In math terms we’d say that the range (or y-values) from one function becomes the domain (or
x-values) of the next function.
The notation used for composition is
(f ◦ g)(x) = f(g(x)) and is read “f composed with g of x” or “f of g of x”.
Notice how the letters stay in the same order in each expression for the composition. f (g(x)) clearly tells you to start with function g
(innermost parentheses are done first).
**ELL note: order here is particularly confusing for ELL students. In English we say the “red car” in Latin based languages (for
example) the literal translation is the “car red”. Hence, explicit discussion with ELL students on this issue is important.
Composition of functions can be thought of as a series of function machines to find your values. The example below shows functions f
and g working together to create the composition (f ◦ g)(x).
NOTE: The starting domain for function g is being limited to the four values 1, 2, 3, 4 and 5 for this example.
g(x) = 2x
f(x) = x + 1
Together they create (f ◦ g)(x).
Domain of g(x)
(-∞, ∞)
Domain of (f ◦ g)(x)
-2
-1
0
1
Domain of f(x)
(0, ∞)
2
1/4
1/2
g(x) = 2x
I’m function g(x).
I’ll take the
input, perform
the function, and
drop them off in
the output
(yellow) area.”
1/4
1/2
1
2
Range of g(x)
(0, ∞)
1
2
4
f(x) = x + 1
4
I’m function
f(x). I’ll pick up
the output from
the 1st function
(yellow),
perform the 2nd
function, and
drop them off in
the output.”
5/4
3/2
2
3
5
Range of (f ◦ g)(x)
(1, ∞)
Just remember that (f ◦ g)(x) is the same as f(g(x)), so start with the innermost parentheses.
The input or domain of g(x) is all real numbers. The output or range of g(x) is all real numbers greater than 0. The
output or range of g(x) is now the input or domain of f(x). Hence, the output or range of f(x) or f(g(x)) is all real
numbers greater than 1.
COMPOSING FUNCTIONS (continued)
Now, suppose that we wish to write our composition as an algebraic expression.
f
g  x   f  g  x  
 
 f 2x
 2x  1
1. Substitute the expression for function g (which in this case is
2x) in for g(x) in the composition. By putting this into
function f as the input, it will clearly show you the order of
the substitutions that will need to be made.
2. Now, substitute this expression (2x) into function f in place
of the x-value and then simplify.
You will find that the concept of composite functions is widely used.
Examples:
1. Given the functions f  x   5x and g  x   x  1 , find
Answers:
f
g  x  and  g f  x  .
f
g  x   f  g  x    f  x  1  5  x  1  5 x  5
g
f  x   g  f  x    g  5 x    5 x   1  5 x  1
Notice that  g f  x  and
f
g  x  do not necessarily yield the same answer. Composition of functions is not commutative.
2. Given the functions p  x   x  2 and h( x)  2 x , find  h p 3 and  h p  x 
Answers:
h
h
p  3  h  p  3  where p(3) gives an answer of 5 and h(5) then gives an answer of 32. The answer is 32.
p  x   h  p  x    h  x  2   2 x  2
Comparing Linear & Exponential Functions
COMPARING LINEAR AND/OR EXPONENTIAL FUNCTIONS
Comparing linear functions:
Graph
Rate of
Change
linear
y = mx + b
slope of m = 0
constant function
y = 0x – 3
therefore, y = -3
slope of
zero
Table
x-int.
x
-3
-2
-1
0
1
2
3
y
-3
-3
-3
-3
-3
-3
-3
x
-3
-2
-1
0
1
2
3
y
-7
-4
-1
2
5
8
11
x
-3
-2
-1
0
1
2
3
y
3.5
3
2.5
2
1.5
1
.5
y-int.
Domain
x values
(0, -3)
none
or
Range
y values
y = -3
all real
numbers
(0, b)
or
y=b
(hence b = -3)
linear
y = mx + b
slope of m > 0
linear increasing
increasing
at a
constant
rate of m
units
y = 3x + 2
next =
now + m
starting at
b
linear
y = mx + b
decreasing
at a
constant
rate of m
units
slope of m < 0
linear decreasing
y = -.5x + 2
next =
now + m
starting at
b
 2 
 ,0
 3 
or
 b 
  ,0
 m 
(0, 2)
or
all real
numbers
all real
numbers
all real
numbers
all real
numbers
(0, b)
(4, 0)
(0, 2)
or
 b 
  ,0
 m 
or
(0, b)
COMPARING LINEAR AND/OR EXPONENTIAL FUNCTIONS
Comparing exponential functions:
Graph
exponential
growth
y = abx
b>1
*growth
factor b
y = 2(3)x
Exponential
decay
y = abx
0<b<1
*decay
factor b
y = 2(.5)x
Rate of
Change
increasing
at an
increasing
rate
next =
now • 3
starting at
2
decreasing
at a
decreasing
rate
next =
now • .5
starting at
2
Table
x
0
1
2
3
4
5
6
y
2
6
18
54
162
486
1458
x
0
1
2
3
4
5
6
y
2
1
.5
.25
.125
.0625
.03125
x-int.
none
x-axis is
an
asympto
te
none
x-axis is
an
asympto
te
y-int.
Domain
x values
Range
y values
all real
numbers
y>0
all real
numbers
y>0
(0, 2)
or
(0, a)
(0, 2)
or
(0, a)
In a linear equation mx represents additive change; (b + m + m + m + m + ……….)
 the slope m is the constant rate of change between any two points on the linear graph/table
y  y 
y

 x1 , y1  and  x2 , y2  = 2 1 , also called
x
 x2  x1 
*In
an exponential equation, bx represents multiplicative change; (a • b • b • b • b • ……….)
 Growth factor b = 100% + growth rate (percent)
Example: Value of savings account is growing by 2.5% per year, the growth fact is 1.025. y = balance (1.025)x
 Decay factor b = 100% – decay rate (percent)
Example: Value of the dollar is declining each year by 3%, the decay factor is 97%. y = 1(.97)x
 The average rate of change of points in an exponential graph/table will not be constant
 The average rate of change between any two points is the slope of the secant line between two points (linear or non-linear).
 y2  y1 
x
 x1 
As students explore the similarities and differences between linear and exponential function, use correct terminology for constant rate of
change (slope of a line) and average rate of change (slope of secant line).
2
Help Links
Composition of Functions:
 http://www.purplemath.com/modules/fcncomp3.htm
 http://www.regentsprep.org/Regents/math/algtrig/ATP7/compositionfunctions.htm
 http://www.mathwarehouse.com/algebra/relation/composition-of-function.php
 Video: http://www.youtube.com/watch?v=S4AEZElTPDo&feature=channel&list=UL