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0.3 The Algebra of Functions
It is often possible to view function as the
combination of other functions.
Suppose we have two functions
f(x) = x + 10 and
g(x) = 4x – 7
Find f(x) + g(x), f(x) – g(x), f(x)g(x), and
f(x)/g(x)
f(x) + g(x) = (x + 10) + (4x - 7)
= x + 10 + 4x – 7
= 5x + 3
f(x) – g(x) = (x + 10) - (4x - 7)
= x + 10 – 4x + 7
= -3x + 17
f(x)g(x) = (x + 10) (4x - 7)
= 4x2 –7x + 40x –70
= 4x2 + 33x – 70
f ( x) x  10

g ( x) 4 x  7
Composition of Functions
Another important way of combining two
functions f(x) and g(x) is to substitite g(x) for
every occurrence of x in f(x). The resulting
function is called the composition of f(x) and
g(x).
The composition of f(x) and g(x) is written f(g(x))
and is read
“f of g of x”
Find f(g(x)) when
f(x) = x2 – 2x + 1 and
g(x) = x + 2
f ( g ( x))  ( g ( x))  2( g ( x))  1
2
 ( x  2)  2( x  2)  1
2
 ( x  4 x  4)  (2 x  4)  1
2
 x  4x  4  2x  4 1
2
 x  2x 1
2
0.4 Zeros of Functions – The Quadratic
Formula and Factoring
A zero of a function f(x) is a value of x for which
f(x) = 0
The Quadratic Formula
Recall that a quadratic function has the form
f(x) = ax2 + bx + c
where a cannot equal zero
The zeros of this function are exactly the solutions
of the quadratic equation
ax2 + bx + c = 0
One way of solving this equation and
finding the zeros is by using the quadratic
formula
 b  b  4ac
x
2a
2
The quadratic formula will provide two
solutions if b2 – 4ac is positive
 b  b  4ac
x
2a
 b  b  4ac
x
2a
2
2
and
one solution if b2 – 4ac is zero
b
x
2a
and no solutions if b2 – 4ac is negative.
Solve the quadratic equation
3x2 – 6x + 2 = 0
a = 3, b = -6, c = 2
so
b  4ac  (6)  4(3)(2)  36  24
2
2
 12  (4)(3)  4 3  2 3
is positive and there are two solutions
 b  b  4ac  (6)  2 3 6  2 3
x


2a
2(3)
6
2
6 2 3
3
 
 1
6
6
3
or
3
1
3
and
3
1
3
The graph appears as
Factoring
If f(x) is a polynomial, we can often write f(x) as
the product of linear factors (ax+b).
If a polynomial can be rewritten in this manner,
then the zeros of f(x) can be determined by setting
each linear factor equal to zero and solving for x.
Factor the quadratic function x2 – 4x – 12
x2 – 4x – 12 = (x – 6)(x + 2)
Factor 3x2 – 21x + 30
3x2 – 21x + 30
= 3(x2 – 7x + 10)
= 3(x – 5)(x – 2)
Solve the equation x2 – 2x – 15 = 0
Start by factoring x2 – 2x – 15
x2 – 2x – 15 = (x – 5)(x + 3)
Then we have (x – 5)(x + 3) = 0
Either (x – 5) = 0 or (x + 3) = 0 will make this a
true statement
For (x – 5) = 0, x = 5
For (x + 3) = 0, x = -3
0.5 Exponents and Power Functions
For any nonzero number b and any positive
integer n, we have the following definitions
b  b  b  ...  b
n
where b is multiplied by itself n times, and
b
n
1
 n
b
Find 33
33 = (3)(3)(3) = (9)(3) = 27
Find 4-2
4
2
1
1
 2 
16
4
Next, let us consider numbers of the form
b1/n. For values of b that are zero or positive,
b1/n is zero or the number whose nth power is
b.
21/2 is the positive number whose square is 2
or
1/ 2
2
 2
21/3 is the positive number whose cube is 2 or
1/ 3
2
 2
3
21/4 is the positive number whose fourth
power is 2 or
1/ 4
2
etc.
 2
4
If n is even, there is no number whose nth
power is b if b is negative. However, if n is
odd then b is permitted to be negative.
(8)
1/ 3
 2
since (-2)(-2)(-2) = -8
Now let us consider number of the form bm/n
where m and n are positive integers.
We define bm/n = (b1/n)m whenever b1/n is
defined.
Find 272/3
27
2/3
 (27
)  (3)  9
1/ 3 2
2
We have the following laws for working with
exponents
Calculate (21/2)(81/2)
1/ 2
2
8
1/ 2
 (2  8)
1/ 2
 16
1/ 2
4