Download f of g

6.3
Power Functions
and
Function Operations
Add.
(3x2 + 9x + 3) + (5x2 – 2x + 7)
8x2 + 7x + 10
Let’s say the first half above is f(x).
So, f(x) = 3x2 + 9x + 3
Let’s say the second half above is g(x).
So, g(x) = 5x2 – 2x + 7
What is f(x) + g(x) ?
2
3x + 9x + 3
+
5x2 – 2x + 7
=
8x2 + 7x + 10 !!
You try.
Let f(x) = 8x2 + 11x + 1
Let g(x) = x2 - 2x + 4
Find f(x) + g(x)
8x2 + 11x + 1
+
x2 - 2x + 4
=
9x2 + 9x + 5
Domain: All Real Numbers
It works the same for the operations too!
Try subtraction…
Let f(x) = 8x2 + 11x + 1
Find f(x)
2
(8x + 11x + 1)
-
-
Let g(x) = x2 - 2x + 4
g(x)
(x2 - 2x + 4)
=
7x2 + 13x - 3
Domain: All Real Numbers
Distribute the negative
Now try multiplication…
Let g(x) = (2x – 6)
Let f(x) = (x + 5)
Find f(x)
(x + 5)


g(x)
(2x – 6)
=
2x2 + 4x - 30
Domain: All Real Numbers
And what about division…
Let f(x) = (x3 + 4x2 – 55x - 30)
Find f(x)
Let g(x) = (x – 6)
 g(x)
x2 + 10x + 5
(x – 6)
(x3 + 4x2 – 55x - 30)
x3 – 6x2
10x2 - 55x
x3 + 4x2 – 55x – 30
x–6
Domain: x = 6
10x2 – 60x
5x - 30
5x – 30
0
Some Domain Considerations …
The domain of the solution consists of the x values
that are in the domains of both f and g.
Typical domain restrictions involve square roots (no
negatives) and denominators (no zeros).
Let f(x) = (x + 5)-1
Let g(x) = x½
The domain in f(x) is all Real Numbers, x = -5
The domain in g(x) is all Real Numbers, x > 0
And what about division…
Let f(x) = 8x5/3
Let g(x) = 6x1/2
Domain: x  0
Domain: All Real Numbers
Find f(x)
f(x)
=
g(x)
 g(x)
4x7/6
8x5/3
=
1/2
3
6x
Domain: x > 0
Domain: Only positive numbers (no zero, no negatives)
And now something new…
f
g
g
f
Is just another way to substitute.
f(x) = x + 7
f
g(x) = 2x + 3 Find f
g
“f of g”
g is another way of saying f( g(x) ).
f(g(x)) = (2x + 3) + 7
= 2x + 10
Domain: All Real Numbers
Let’s try…
g(x) = x + 4
f(x) = 3x – 1
Find f
g
replace
f(g(x)) = f(x + 4)
= 3(x + 4) - 1
= 3x + 12 - 1
Find g
f
= 3x + 11
Domain: All Real Numbers
g(f(x)) = g(3x – 1) = (3x – 1) + 4
= 3x - 1 + 4
Domain: All Real Numbers
= 3x + 3
Another example …
g(x) = 5x - 2
f(x) = 2x-1
Find f
g
f(g(x)) = 2(5x – 2)-1
=
2
5x - 2
Find g
Domain: x = 2/5
f
g(f(x)) = 5(2x-1 ) - 2
= 10x-1 – 2 =
10
x
Domain: x = 0
-2
And some more …
f(x) = 3x-½
g(x) = 6x - 8
Find f(g(4))
Work from the inside out. Find g(4) first.
g(4) = 6(4) – 8 = 16
Now find f(16).
f(16) = 3(16) =
-½

 1
3 1
 2
 16




1
= 3 
 4
=¾
On Your Own …
g(x) = x2 – 2x + 1
f(x) = 6x-½
Find f
g
f(g(x)) = 6(x2 – 2x + 1) -½
(x2 – 2x + 1) is a perfect square!
(x – 1)2
So, the square root of (x – 1)2 is (x – 1)
= 6(x – 1)
Find g
f
=
6x - 6
On Your Own …
g(x) = x2 – 2x + 1
f(x) = 6x-½
Find f
g
f(g(x)) = 6(x2 – 2x + 1) -½
(x2 – 2x + 1) is a perfect square!
(x – 1)2
So, the square root of (x – 1)2 is (x – 1)
= 6(x – 1)
Find g
f
=
6x - 6