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Basics of Functions Continued…
• Domain• Evaluating from…
– An equation– A graph – A table
• Composition– Add/Subtract/Multiply/Divide– Difference Ratio
Domain
2
4(1)
2 3
xf x
x x
2(2) 9g x x
(3) 3 2h x x
2 35. For the functions 2 3 4 1
find the following:
f x x g x x
– 4 – 2 2
5
0
9
4
( )y x f
( )y xg
a.
6. Given the graph, if possible, evaluate (f + g)(1).f(1) = (think when x = 1 on
the f(x) graph, what is the y-value?)
g(1) = (think when x = 1 on the g(x) graph, what is the y-value?)
(f + g)(1) = add the two parts above together
x
y
– 4 – 2 2
5
0
9
4
( )y x f
( )y xg
a.
6. Given the graph, evaluate (f + g)(1).
f(1) = 3 g(1) = 1
(f + g)(1) = 4
x
y
– 4 – 2 2
5
0
9
4
( )y x f
( )y xg
7. Given the graph, if possible, evaluate (f / g)(0).f(0) = (think when x = 0 on
the f(x) graph, what is the y-value?)
g(0) = (think when x = 1 on the g(x) graph, what is the y-value?)
(f / g)(0) = divide
x
y
– 4 – 2 2
5
0
9
4
( )y x f
( )y xg
7. Given the graph, evaluate (f / g)(0).
f(0) = 1
g(0) = 0
(f /g)(0) = 1/0, thus (f/g)(0) is undefined because the denominator
is 0.
x
y
x (x) g(x)
– 2 – 3 undefined
0 1 0
1 3 1
-1 -1 undefined
4 9 2
8. Given the table, if possible, use the given representations of f and g to evaluate (f·g)(1).f(1), when x = 1 in the table, what is the value of f(x)?
g(1), when x = 1 in the table, what is the value of g(x)?
(f · g)(1), evaluate the product
x (x) g(x)
– 2 – 3 undefined
0 1 0
1 3 1
-1 -1 undefined
4 9 2
8. Given the table, if possible, use the given representations of f and g to evaluate (f · g)(1).f(1) = 3
g(1) = 1
(f · g)(1) = 3(1) = 3
x (x) g(x)
– 2 – 3 undefined
0 1 0
1 3 1
-1 -1 undefined
4 9 2
9. Given the table, if possible, use the given representations of f and g to evaluate (f – g)(-2).f(-2), when x =-2 in the table, what is the value of f(x)?
g(-2), when x = -2 in the table, what is the value of g(x)?
(f – g)(-2), evaluate the difference
x (x) g(x)
– 2 – 3 undefined
0 1 0
1 3 1
-1 -1 undefined
4 9 2
9. Given the table, if possible, use the given representations of f and g to evaluate (f – g)(-2).f(-2) = -3
g(-2) = undefined
(f – g)(-2) = undefined
Note: looking back at the graph on slide 7, g(x) is not defined at x = -2. In simplistic terms, that means there is no graph for the value of x.