Download Thinking Mathematically by Robert Blitzer

COMBINATIONS OF
FUNCTIONS; COMPOSITE
FUNCTIONS
DOMAIN OF A FUNCTION

Exclude any number that causes division by zero
x 1
f ( x) 
x 3


Domain: (-∞, 3)  (3, ∞)
Exclude any number that results in the square
root of a negative number
g ( x)  3 x  12

Domain: [-4, ∞)
Do checkpoint #1 page 211
DEFINITIONS: SUM, DIFFERENCE, PRODUCT,
AND QUOTIENT OF FUNCTIONS
Let f and g be two functions. The sum of f + g,
the difference f – g, the product fg, and the
quotient f /g are functions whose domains are
the set of all real numbers common to the
domains of f and g Df ∩ Dg, defined as follows:
Sum:
Difference:
Product:
Quotient:
(f + g)(x) = f (x)+g(x)
(f – g)(x) = f (x) – g(x)
(f • g)(x) = f (x) • g(x)
(f / g)(x) = f (x)/g(x), provided
g(x) does not equal 0
EXAMPLE
Let f(x) = 2x + 1 and g(x) = x2 – 2. Find f + g, f - g, fg,
and f/g.
Solution:
f + g = 2x + 1 + x2 – 2 = x2 + 2x – 1
f – g = (2x + 1) - (x2 – 2)= -x2 + 2x + 3
fg = (2x + 1)(x2 – 2) = 2x3+ x2 – 4x - 2
f/g = (2x + 1)/(x2 – 2)
Do checkpoint 2, page 214
EXAMPLE 3
Let f(x) = x  3 and g(x) = x  2 .
a) Find f + g
b) The domain of f + g
Solution: x  3  x  2
a) f + g =
Domain of f: [-3, ∞)
Domain of g: [2, ∞)
b) Domain of f + g is [2, ∞)
Do checkpoint #3 page 215
THE COMPOSITION OF FUNCTIONS
 The
composition of the function f with g
is denoted by f o g and is defined by the
equation
 (f o g)(x) = f (g(x)).
 The domain of the composite function f o g
is the set of all x such that
 x is in the domain of g and
 g(x) is in the domain of f.
 (see visual on page 216)
TEXT EXAMPLE
Given f (x) = 3x – 4 and g(x) = x2 + 6, find:
a. (f o g)(x) b. (g o f)(x)
Solution
a. We begin with (f o g)(x), the composition of f with g. Because (f o g)(x)
means f (g(x)), we must replace each occurrence of x in the equation for
f by g(x).
This is the given equation for f.
f (x) = 3x – 4
(f o g)(x) = f (g(x)) = 3g(x) – 4
= 3(x2 + 6) – 4
= 3x2 + 18 – 4
= 3x2 + 14
Thus, (f o g)(x) = 3x2 + 14.
Replace g(x) with x2 + 6.
Use the distributive property.
Simplify.
EXAMPLE
Given f (x) = 3x – 4 and g(x) = x2 + 6, find:
a. (f o g)(x) b. (g o f)(x)
Solution
b. Next, we find (g o f )(x), the composition of g with f. Because (g o f )(x)
means g(f (x)), we must replace each occurrence of x in the equation for
g by f (x).
This is the given equation for g.
g(x) = x2 + 6
(g o f )(x) = g(f (x)) = (f (x))2 + 6
= (3x – 4)2 + 6
= 9x2 – 24x + 16 + 6
= 9x2 – 24x + 22
Replace f (x) with 3x – 4.
Square the binomial, 3x – 4.
Simplify.
Thus, (g o f )(x) = 9x2 – 24x + 22. Notice that (f o g)(x) is not the same as
(g o f )(x).
HOMEWORK

Page 219 #13 – 71 every other odd
# 81 – 92
 Bring your textbooks tomorrow
