Section 1.7

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Section 1.7
Combinations of Functions
Composition of Functions
Domain Revisited
• Recall that the domain of a function is the
set of first coordinates when the function is
expressed as a set of ordered pairs.
• The domain can also be thought of as the
set of input values for a function.
• Some functions have restricted domains.
Functions involving Square Roots
• It is not possible to take the square root of
a negative number.
• The radicand, or expression underneath
the radical, must be 0 or greater.
• To find the domain of a function with a
radical, set the radicand > 0.
• The domain will be the solution to that
inequality (be sure to use interval
notation).
Find the domain. Use interval
notation.
f ( x)  x  2
f ( x)  24  2 x
Functions involving Denominators
• It is not possible to divide by zero. Division
by 0 is undefined.
• To find the domain of a function with a
denominator, set the denominator equal to
0 and solve.
• The domain will be all real numbers except
the solution(s). Be sure to use interval
notation.
Find the domain. Use interval
notation.
2
f ( x) 
x5
2
f ( x)  2
x  x  12
1
f ( x) 
x 3
Most other functions (at least for
now)
• The domain will be all real numbers. Be
sure to use interval notation.
Find the domain. Use interval
notation.
f ( x)  2( x  5)
f ( x)  x  x  12
2
Sum, Difference, Product, Quotient
Let f and g be functions.
( f  g )( x)  f ( x)  g ( x)
( f  g )( x)  f ( x)  g ( x)
( fg )( x)  f ( x)  g ( x)
f
f ( x)
 ( x) 
, g ( x)  0
g ( x)
g
Example
• Find the sum, difference, product, and
quotient for the following pairs of functions.
Give the domain for your answers.
f ( x)  6 x  x  1, g ( x)  x  1
2
f ( x)  x  4 , g ( x)  x  1
Composition of Functions
• The output of one function is the input of
another function.
Composition
 f  g x  f g x
The output of g becomes the input for f.
g  f x  g  f x
The output of f becomes the input for g.
Example
• For each of the following pairs of
functions, find:  f  g 2 
g  f 2
 f  g x 
g  f x 
f ( x)  3x  4, g ( x)  x  2
f ( x)  7 x  1, g ( x)  2 x  9
2
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