2. Using basic algebraic functions, what limitations are there
when working with real numbers? A) You CANNOT 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 CANNOT 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.
3. Example Find the domain: There is an x under an even radical AND
x terms in the denominator, so we must consider both of these as
possible limitations to our domain. 65 2 2 xx x }3,2:{: 3,2,0)2)(3(
065 2,02 2 xxxDomain xxx xx xx 4. Find 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 1 1 1 5 4 4 1 1 5 5 0 1 ( ) 5
f x x 5. Find the domain of the function Solution: 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 = 4 2 2 3 10 8 ( ) 3 28 x x
g x x x The domain consists of the set of all real numbers except
x= 4 and x= 7 or {x | x 4 and x 7}. , 4 ( 4,7) (7, ) 6. Rational
Functions 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] 7. Visualizing
Domain and Range Keep the following in mind regarding the graph of
a function: Domain = the set of a functions inputs; found on the
x-axis (horizontal). The domain of a function is the set of all
first coordinates of the ordered pairs of a relation Range = the
set of a functions outputs; found on the y-axis (vertical). The
range of a function is the set of all second coordinates of the
ordered pairs of a relation. 8. Example Graph the function. Then
estimate the domain and range. (Note: Square root function moved
one unit right) Domain = [1, ) Range = [0, ) ( ) 1f x x ( ) 1f x x
9. Algebra of functions (f + g)(x) = f(x) + g(x) (f - g)(x) = f(x)
g(x) (fg)(x) = f(x)g(x) 0)(, )( )( )( xg xg xf x g f 10. Example
Find each function and state its domain: f + g f g f g f /g ;1 1g
xf x x x ;1 : 11x Domainf xx xg x ;1 : 11x Domainf xx xg x 2 1; :1
1 1x x Domaing xx xf x 1 ; : 1 1 x D x f omain x xg x 11. BA 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. Their difference 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 If
f and g are functions with domains A and B: 12. Their product f g
is the function given by BA The domain of f g consists of the
numbers x that are in the domain of f and in the domain of g. Their
quotient f /g is the function given by (f / g ) (x) = f(x) / g(x)
where g(x) 0; (f g)(x) = f(x) g(x) If f and g are functions with
domains A and B: The domain of f / g consists of the numbers x for
which g(x) 0 that are in the domain of f and in the domain of g.
0)(xgBA 13. 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) 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 (f +
g)(5) = 3(5) + 7 = 22or 14. 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) 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 (f - g)(5) = -(5) - 3 = -8or 15.
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) 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 16. ( ) ( ) f x g x 2 3 16 x x The domain
of f / g is {x | x > 3, x 4}. ( ) 3f x x 2 ( ) 16g x x Given the
functions below, find (f/g)(x) and give the domain. ( / )( )f g x
The radicand x 3 cannot be negative. Solving x 3 0 gives x 3 We
must exclude x = -4 and x = 4 from the domain since g(x) = 0 when x
= 4. 17. Composition of functions Composition of functions is the
successive application of the functions in a specific order. Given
two functions f and g, the composite function is defined by and is
read f of g of x. The domain of is the set of elements x in the
domain of g such that g(x) is in the domain of f. Another way to
say that is to say that the range of function g must be in the
domain of function f. Composition of functions means the output
from the inner function becomes the input of the outer function. f
g f g x f g x f g 18. 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 f g x g(x) f(g(x)) domain of g range of f range of
g domain of f g f 19. f g x f g x 1 2 f x 1 2x 1 2x . gf x xgxxf
Find 2 1 )(and)(Suppose Suppose f x x( ) and g x x ( ) 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 1 0 2x 2x The domain of f g is {x | x > -2}. 20.
2 2 2 1 3 2 4x xf g x 2 2 2 2 1 2 6 9 1 2 12 18 1 3g x x x f x x x
Example Evaluate and : f g x g f x 3f x x 2 2 1g x x 2 2 4 (you
check) f g x x 2 2 12 17g f x x x You can see that function
composition is not commutative. NOTE: This is not a formal proof of
the statement. 21. (Since a radicand cant be negative in the set of
real numbers, x must be greater than or equal to zero.) Example
Find the domain of and :f g x g f x 1f x x g x x 1 : 0f g x x
Domain x x 1 : 1g f x x Domain x x (Since a radicand cant be
negative in the set of real numbers, x 1 must be greater than or
equal to zero.) 22. Solution to Previous Example : Determine a
function that gives the cost of producing the helmets in terms of
the number of hours the assembly line is functioning on a given
day. Cost C n C P t 2 75 2C t t 2 14 525 100 2 40 $5 8C t t 2 75 2n
P t t t 7 1000C n n 23. 1. Suppose that and2 ( ) 1f x x ( ) 3g x x
( ) ?g f x ( ) ( ( ))g f x g f x 2 ( 1)g x 2 3( 1)x 2 3 3x 2.
Suppose that and2 ( ) 1f x x ( ) 3g x x (2) ?g f (2) 2g f g f 2 (2
1)g (3)g (3)(3) 9 24. The End Call us for more Information:
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