6.1 Introduction to Combinations of Functions (#1-4)

• Look at the temperature of a liquid place in a refrigerator problem on P(404-405) in the text.

• Join your team and work on problems 1-4 at the end of section 6.1 in the text.

Functions ReviewFind the indicated function values and determine whether the given values are in the domain of the function.

f(1) and f(5), for

f(1) =

Since f(1) is defined, 1 is in the domain of f.

f(5) = Since division by 0 is not defined, the number 5 is not in the domain of f.

1( )

5f x

x

1 1 1

1 5 4 4

1 1

5 5 0

Find the domain of the functionSolution:

We can substitute any real number in the numerator, but we must avoid inputs that make the denominator 0.

Solve x2 3x 28 = 0. (x 7)(x + 4) = 0

x 7 = 0 or x + 4 = 0 x = 7 or x = 4The domain consists of the set of all real numbers except 4 and 7 or {x|x 4 and x 7}.

2

2

3 10 8( )

3 28

x xg x

x x

, 4 ( 4,7) (7, )

To find the domain of a function that has a variable in the

denominator, set the denominator equal to zero and solve the equation. All solutions

to that equation are then removed from consideration for

the domain.

Find the domain:

Since the radical is defined only for values that are greater than

or equal to zero, solve the inequality

( ) 5f x x

5 0x 5x 5x ( ,5]

Visualizing Domain and Range

Keep the following in mind regarding the graph of a function:

• Domain = the set of a function’s inputs, found on the x-axis (horizontal).

• Range = the set of a function’s outputs, found on the y-axis (vertical).

ExampleGraph the function. Then estimate the domain and range.

Domain = [1, )

Range = [0, )

( ) 1f x x ( ) 1f x x

The domain of a function is normally all real numbers but there

are some exceptions:

• A) You can not divide by zero.– Any values that would result in a zero denominator

are NOT allowed, therefore the domain of the function (possible x values) would be limited.

B) You can not take the square root (or any even root) of a negative number.

Any values that would result in negatives under an even radical (such as square roots) result in a domain restriction.

Example

• Find the domain

• There are x’s under an even radical AND x’s in the denominator, so we must consider both of these as possible limitations to our domain.

65

22

xx

x

2

2 0, 2

5 6 0

( 3)( 2) 0, 2,3

:{ : 2, 3}

x x

x x

x x x

Domain x x x

6.3 Algebra of Functions (#1,2,3,5) 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 )

The domain of f + g consists of the numbers x that are in the domain of f and in the domain of g.

BA

Their difference f - g is thefunction given by

( f - g ) (x ) = f (x ) - g (x )

The domain of f - g consists ofthe numbers x that are in thedomain of f and in the domain ofg. BA

Their product f g is the function given by

The domain of fg consists of the numbers x that are in the domain of f and in the domain of g.

BA

(f g)(x) = f(x)g(x)

Their quotient f / g is thefunction given by

( f / g )(x ) = f(x ) / g (x ) , g (x ) 0

The domain of f / g consists ofthe numbers x for which g(x) 0that are in the domain of f and inthe domain of g. 0)( xgBA

Example• Given that f(x) = x + 2 and g(x) = 2x + 5, find

each of the following. a) (f + g)(x) b) (f + g)(5) Solution: a)

( )( ) ( ) ( )

2 2 5

3 7

f g x f x g x

x x

x

b) We can find (f + g)(5) provided 5 is in the domain of each function. This is true.

f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15(f + g)(5) = f(5) + g(5) = 7 + 15 = 22 or

(f + g)(5) = 3(5) + 7 = 22

Example• Given that f(x) = x + 2 and g(x) = 2x + 5, find

each of the following. a) (f - g)(x) b) (f - g)(5) Solution: a)

( )( ) ( ) ( )

2 (2 5)

2 2 5

3

f g x f x g x

x x

x x

x

b) We can find (f - g)(5) provided 5 is in the domain of each function. This is true.

f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15(f - g)(5) = f(5) - g(5) = 7 - 15 = -8 or

(f - g)(5) = -(5) - 3 = -8

Example• Given that f(x) = x + 2 and g(x) = 2x + 5, find

each of the following. a) (f g)(x) b) (f g)(5) Solution: a)

2

( )( ) ( ) ( )

( 2)(2 5)

2 9 10

fg x f x g x

x x

x x

b) We can find (f g)(5) provided 5 is in the domain of each function. This is true.

f(5) = 5 + 2 = 7 g(5) = 2(5) + 5 = 15(f g)(5) = f(5)g(5) = 7 (15) = 105 or

(f g)(5) = 2(25) + 9(5) + 10 = 105

( / )( )f g x

( )

( )

f x

g x

2

3

16

x

x

We must exclude x = -4 and x = 4 from the domain since g(x) = 0 when x = 4.

The domain of f / g is {x | x > 3, x 4}.

( ) 3f x x 2( ) 16g x x

Given the functions below, findand give the domain.

( / )( )f g x

The radicand x – 3 cannot be negative.Solving gives 3 0 3x x

Composition of functions

• Composition of functions means the output from the inner function becomes the input of the outer function.

• f(g(3)) means you evaluate function g at x=3, then plug that value into function f in place of the x.

• Notation for composition:

))(())(( xgfxgf

Given two functions f and g, thecomposite function, denoted byf g (read as “f composed withg”), is defined by

f g x f g x

The domain of f g is the set ofall numbers x in the domain ofg such that g( x) is in the domainof f.

f g x f g x 1

2f

x

1

2x

1

2x

Suppose f x x( ) and g xx

( )1

2. Findf g .

Suppose f x x( ) and g xx

( ) 1

2. Find

the domain of f g .

The domain of f g consists of those x in the domain of g, thus x = -2 is not in the domain of f g .

In addition, g(x) > 0, so

10

2x

2x

The domain of f g is {x | x > -2}.

Suppose that andfind

1)( 2 xxfxxg 3)(

xfg

xfgxfg

12 xg

13 2 x33 2 x

Suppose that andfind

1)( 2 xxfxxg 3)(

2g f

2 2g f g f

2(2) 1g 3g

(3)(3) 9