Transcript (f/g)(x)

Operations on Functions
Section 1-8
Objectives
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I can add, subtract, multiply, and
divide functions
I can find the domains of newly
formed functions
Arithmetic of
Functions
NOTATION
f(x) + g(x) means
Add the f(x) function to
the g(x) function.
Arithmetic of
Functions
NOTATION
f(x) - g(x) means
Subtract the g(x)
function from the f(x)
function.
Arithmetic of
Functions
NOTATION
(fg)(x) means
f(x)

g(x)
Multiply f(x) times g(x)
Arithmetic of
Functions
NOTATION
(f/g)(x) means
f(x)  g(x)
Divide f(x) by g(x)
Let f(x) = 3x + 4 & g(x) = 2x – 3
find:
a) f(x) + g(x)
b)
f(x) – g(x)
c)
f(x)g(x)
d) f(x)/g(x)
5x + 1
x+7
6x2 – x - 12
3x  4
3
;x 
2x  3
2
Operations on Functions:
Addition: (f + g)(x) = f(x) + g(x)
Example: If f(x) = 3x – 6 and g(x) = -3x2 + 3x + 4 then
(f + g)(x) = 3x –6 + -3x2 + 3x + 4 = -3x2 + 6x –2
Subtraction: (f – g)(x) = f(x) – g(x)
Example: Use the same functions as above. Then
(f - g)(x) = 3x – 6 – (-3x2 + 3x + 4 )
(f – g)(x) = 3x2 - 10
Operations on Functions:
Multiplication: (fg)(x) = f(x)g(x)
Example: If f(x) = 3x – 6 and g(x) = -3x2 + 3x + 4 then
(fg)(x) = (3x –6)(-3x2 + 3x + 4) = -9x3 + 9x2 + 12x +
18x2 – 18x – 24 = -9x3 + 27x2 – 6x - 24
Division: (f/g)(x) = f(x)/g(x)
Example: Use the same functions as above. Then
(f/g)(x) = (3x – 6)/(-3x2 + 3x + 4) except for x – values
that make denominator 0
Finding the domains of our new
functions
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*
Finding the domains of these new
functions is a little complicated.
The first step is to find the domain of
each function.
The second step is to find the
intersection of the domains. This
intersection is the domain of the new
function*.
Unless the operation is division in which case you must
also exclude x-values that make the denominator zero
If f and g are functions with domains
A and B:
Their sum f + g is the function given
by
(f + g)(x) = f(x) + g(x)
Domain of f + g
A B
Lets do an example; Let f ( x)  x  2 and g ( x)  x
a) Find (f + g)(x) and b) it’s domain.
(f + g)(x) =
x2  x
Step 1: The domain of f(x) is [-2,).
The domain of g(x) is [0,).
Step 2: [-2,)  [0,) =
[0,).
Since this is not a quotient I don’t have to worry about
division by zero.
Their difference f - g is the
function given by
(f - g)(x) = f(x) - g(x)
Domain of f - g
A B
Lets do an example; Let f ( x)  3x  2 and g ( x)  x 2  4
a) Find (f - g)(x) and b) it’s domain.
(f - g)(x) =  x 2  3 x  2
Step 1: The domain of f(x) is (- ,).
The domain of g(x) is (- ,).
Step 2: (- ,)  (- ,) =
(- ,).
Since this is not a quotient I don’t have to worry about
division by zero.
Their product f  g is the
function given by
(f  g)(x) = f(x)  g(x)
Domain of f  g
A B
Example; Let f ( x)  3x  2 and g ( x)  x  4
a) Find (f · g)(x) and b) it’s domain.
(f · g)(x) = 3 x  10 x  8
2
Step 1: The domain of f(x) is (- ,).
The domain of g(x) is (- ,).
Step 2: (- ,)  (- ,) =
(- ,).
Since this is not a quotient I don’t have to worry about
division by zero.
Their quotient f / g is the
function given by
(f / g)(x) = f(x) / g(x), g(x)  0
Domain of f / g
A  B g ( x)  0
Another example: f ( x) 
x and g ( x)  x  10
a) Find (f /g)(x) and b) it’s domain.
f 
 ( x) 
g
x
x  10
Step 1: Domain of f(x) is [0,).
Domain of g(x) is [10, )
Step 2: [0,)  [10, ) =
[10, )*
*
Division so we can’t plug in 10 so Domain is (10, ).
Example
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Given f(x) = x – 5 and g(x) = x2 -1, find (f+g)(x),
(f - g)(x), (fg)(x) and (f/g)(x)
(f+g)(x) = f(x) + g(x) = (x – 5)+(x2 –1)
=x2 + x - 6
Domain (f+g)(x): (- , )
(f – g)(x) = f(x) – g(x) = (x – 5) – (x2 –1 )
= x – 5 – x2 + 1
= -x2 + x – 4
Domain (f – g)(x) = (- , )
Example
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Given f(x) = x – 5 and g(x) = x2 -1, find (f - g)(x),
(fg)(x) and (f/g)(x)
(fg)(x) = f(x)g(x) = (x – 5)(x2 –1)
= x3 – x – 5x2 + 5
= x3 – 5x2 – x + 5
Domain (fg)(x) : (- , )
(f/g)(x)=f(x)/g(x) = (x – 5)/(x2 – 1)
Domain (f/g)(x): {All real except x 1 or –1}
(- , -1) U (-1,1) U (1, )
Homework
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WS 2-1