1. Functions, operations, domain
A function is any equation where if you input any value in for x (independent variable) you get 1 value for y (dependent value. What does this look like?
yes
Yes No
y = x + 3
y = √x + 5
x2 + y2 = 4
y = 1/(x + 2)
y2 = 3x - 5

2. Evaluating functions f(x) = F(0) = (02 – 1) / (0 + 4) = -1/4
F(-x) = (-x2 – 1) / (-x + 4) = (x2 -1)/(-x + 4)
-F(x) = (x2 – 1) / (x + 4) = (-x2 + 1)/(x +4)
F(x + 1) = ((x + 1)2 – 1) / (x ) = (x2 + 2x)/(x + 5)
F(2x) = ((2x)2 – 1) / (2x + 4) = (4x2 – 1)/(2x + 4)
F(x + h) = ((x + h)2 – 1) / (x + h + 4) = (x2 + 2xh + h2 – 1)/(x + h + 4)

3. Evaluating functions Practice: f(x) = |x| + 4 F(0) = |0| + 4 = 4
F(-x) = |-x| + 4 =|x| + 4
-F(x) = -(|x| + 4) = -|x| - 4
F(x + 1) = |x + 1| + 4
F(2x) = |2x| + 4 = 2|x| + 4
F(x + h) = |x + h| + 4

4. Domain:
Domain is the set of all x values that can be inputed into the equation without causing an undefined output.
Remember the domain starts off as (-∞,∞) then you take away any extraneous solutions or undefined out puts to two main things that cause this are:
Variables in a denominator can’t make the denominator equal 0
Variables inside a square root can not make the inside negative.
The domain of two functions together is the overlapping values of both functions. For example if the domain of f is [-3,6] and the domain of g is [0,9] the domain of f + g is [0,6] this is where those 2 domains overlap.

5. Domain:
Domain is the set of all x values that can be inputed into the equation without causing an undefined output.
For example lets say you have the equation: since there is an x in
the denominator if you plugged in 0 the answer would be undefined so 0 is not part of the domain.
Therefor x ≠ 0, your domain written in interval notation is (-∞,0) (0,∞). This is saying that the domain is from negative infinity to 0 but not including 0 and from 0 (not included to infinity.

6. Domain:
Example 2 lets say you have the equation: since there is an x in
the denominator it will create undefined x values, but what do we have to do first to find that excluded value or values?
FACTOR the denominator and see what values would create a 0 in the denominator by setting the factors equal to 0.
The factored form of this denominator is x(x+3). Therefore x ≠ 0 and x ≠ -3 the equation would be undefined so now lets write the domain in interval notation.
The domain written in interval notation is (-∞,-3) (-3,0) (0,∞).

7. You try
Find the excluded values from the domain then write the domain in interval notation.
Practice:
1 – x ≥ 0
-x ≥ -1
x ≤ 1

8. Function Operations
Function operation is a way to simplify 2 or more expressions together to make a new function.
f(x) + g(x) or (f + g)(x) means to add the two expressions of f and g together and simplify.
f(x) – g(x) or (f – g)(x) means to subtract the two expression of f and g together and simplify
f(x) ● g(x) or (f ● g)(x) means to multiply the two expressions of f and g together and simplify
f(x)/g(x) or (f ÷ g)(x) means to divide the two expressions of f and g together and simplify

9. Examples
Use the functions to find the following and find the domainf(x) = x - 1 and g(x) = 2x2
a) (f + g)(x) b) (f – g)(x) c) (f • g)(x) d) (f/g)(x)
x – 1 + 2x2
2x2 + x - 1
(-∞,∞)
b) x – 1 – 2x2
-2x2 + x - 1
(-∞,∞)
c) (x – 1)(2x2)
2x3 – 2x2
(-∞,∞)
d) (x-1)/(2x2)
x ≠ 0
(-∞,0) (0,∞)
Use the same functions to find the following.
a) (f + g)(3) b) (f – g)(4) c) (f • g)(2) d) (f/g)(1)
3 – 1 + 2(3)2
2 + 18 20
b) 4 – 1 – 2(4)2
3 – 32
-29
c) (2 – 1)(2(2)2)
(1)(8) 8
d) (1-1)/2(1)2
0/2

10. Practice
Use the functions to find the following and find the domain f(x) = √x and g(x) = 3x – 5
a) (f + g)(x) b) (f – g)(x) c) (f • g)(x) d) (f/g)(x)
d) √x / (3x – 5)
x ≠ 5/3
[0,5/3) (5/3,∞)
√x + 3x – 5
[0,∞)
b) √x – 3x + 5
[0,∞)
c) 3x√x – 5√x
[0,∞)
Use the same functions to find the following.
a) (f + g)(3) b) (f – g)(4) c) (f • g)(2) d) (f/g)(1)
√3 + 4
b) √4 – 7
- 5
c) 6√2 – 5√2 √2
d) √1 / (3•1 – 5)
1/-2
- ½

11. Other types of problems
Find the difference quotient of f; that is find for the function. Simplify.
f(x) = x2 – x + 4

12. Other types of problems
If f(x) = and f(2) = ½ , What is the value of B.
To do this we need to plug in 2 for both x’s and ½ for f(x)