Download 5.7.2 Operating on Functions Building

Introduction
As is true with linear and exponential functions, we can
perform operations on quadratic functions. Such
operations include addition, subtraction, multiplication,
and division. This lesson will focus on adding,
subtracting, multiplying, and dividing functions to create
new functions. The lesson will also explore the effects of
dividing a quadratic by one of its linear factors.
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5.7.2: Operating on Functions
Key Concepts
Operations with Functions
• Functions can be added, subtracted, multiplied, and
divided.
• For two functions f(x) and g(x), the addition of the
functions is represented as follows:
(f + g)(x) = f (x) + g(x).
• For two functions f(x) and g(x), the subtraction of the
functions is represented as follows:
(f - g)(x) = f (x) - g(x).
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5.7.2: Operating on Functions
Key Concepts, continued
• For two functions f(x) and g(x), the multiplication of
the functions is represented as follows:
(f • g)(x) = f (x) • g(x).
• For two functions f(x) and g(x), the division of the
functions is represented as follows:
æfö
f ( x)
ç g ÷ ( x) = g ( x) .
è ø
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5.7.2: Operating on Functions
Key Concepts, continued
• Adding and subtracting linear expressions from a
quadratic will yield a quadratic.
• Multiplying and dividing a quadratic by anything other
than a constant will not yield a quadratic.
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5.7.2: Operating on Functions
Key Concepts, continued
Restricted Domains
• When considering the division of a quadratic by a
linear factor, it is possible to create a linear expression
with a restricted domain. For example:
æfö
• For f(x) = x2 + 5x + 6 and g(x) = x + 3, ç g ÷ ( x )
è ø
æfö
f ( x)
can be found such that ç ÷ ( x ) =
=
g ( x)
è gø
x 2 + 5x + 6
x+3
x + 2) ( x + 3 ) ( x + 2) ( x + 3 )
(
=
=
= x + 2.
( x + 3)
( x + 3)
5.7.2: Operating on Functions
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Key Concepts, continued
æfö
• In simpler terms, ç ÷ ( x ) = x + 2.
è gø
• Remember that the denominator of a fraction cannot
equal 0.
• Set the denominator equal to 0 and solve for x to find
the restricted value(s) in the domain: x + 3 = 0, so
x ≠ –3.
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5.7.2: Operating on Functions
Key Concepts, continued
• Given the similar function h(x) = x + 2, the domain is
all real numbers, and the range is the same. However,
æfö
since f(x) is divided by g(x), the domain of ç ÷ ( x ) = x + 2
è gø
from the preceding example is all real numbers except
for x = –3 and the range is all real numbers except for
y = –1.
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5.7.2: Operating on Functions
Key Concepts, continued
• This is because when the restricted value of the
domain (–3) is substituted into the simplified form of
æfö
ç g ÷ ( x ) = x + 2 and solved for y, we get:
è ø
• Therefore, since x ≠ –3, then y ≠ –1.
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5.7.2: Operating on Functions
Common Errors/Misconceptions
• forgetting to restrict the domain when dividing functions
• not realizing that functions must be of the same
variable for like terms to be combined
• having difficulty moving from the formal notation to a
workable problem where functions can be used with
operations
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5.7.2: Operating on Functions
Guided Practice
Example 1
Let f(x) = x2 – 3x + 4 and g(x) = x2 + 6x – 3. Build a new
function, h(x), for which h(x) = (f + g)(x).
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5.7.2: Operating on Functions
Guided Practice: Example 1, continued
1. Expand the new function, h(x), into a form
where substitution can be used.
h(x) = (f + g)(x) = f(x) + g(x)
The new function is expanded as h(x) = f(x) + g(x).
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5.7.2: Operating on Functions
Guided Practice: Example 1, continued
2. Add the functions.
f(x) = x2 – 3x + 4 and
g(x) = x2 + 6x – 3
Given functions from
problem statement
h(x) = f(x) + g(x)
Expanded notation
h(x) =
(x2
– 3x + 4) +
h(x) = 2x2 + 3x + 1
(x2
Substitute values for
+ 6x – 3)
f(x) and g(x).
Combine like terms.
The new function is h(x) = 2x2 + 3x + 1.
✔
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5.7.2: Operating on Functions
Guided Practice: Example 1, continued
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5.7.2: Operating on Functions
Guided Practice
Example 3
æfö
For f(x) =
+ 13x – 10 and g(x) = x + 5, find ç ÷ ( x ) .
è gø
æfö
What type of function is the quotient of ç ÷ ( x ) ? Are
è gø
there restrictions on the domain and range of the
æfö
function ç ÷ ( x ) ?
è gø
3x2
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5.7.2: Operating on Functions
Guided Practice: Example 3, continued
1. Since the functions are being divided,
write the functions f(x) and g(x) as a
fraction.
æfö
f ( x ) 3x 2 + 13x - 10
ç g ÷ ( x) = g ( x) =
x+5
è ø
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5.7.2: Operating on Functions
Guided Practice: Example 3, continued
2. Factor the quadratic function, f(x).
3x 2 + 13x - 10
x+5
3x - 2) ( x + 5 )
(
=
x+5
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5.7.2: Operating on Functions
Guided Practice: Example 3, continued
3. Simplify the equation and define the type
of equation of the simplified form.
Divide away the monomial (x + 5) from the top and
bottom of the fraction:
( 3x - 2) ( x + 5) = ( 3x - 2) ( x + 5) = 3x - 2
x+5
x+5
æfö
The function ç ÷ ( x ) is a linear equation that graph
è gø
the line y = 3x – 2.
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5.7.2: Operating on Functions
Guided Practice: Example 3, continued
4. Look at the original fraction to see if there
are restricted values on the domain.
In this case, x ≠ –5 because (–5) + 5 = 0 and division
by 0 is undefined. Next, substitute x = –5 into the final
equation to determine the restricted value(s) of y.
3x – 2 = 3(–5) – 2 = –17
Since x ≠ –5, then y ≠ –17.
✔
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5.7.2: Operating on Functions
Guided Practice: Example 3, continued
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5.7.2: Operating on Functions