1. AP Calculus AB/BC
1.2 Functions

2. Engineers use functions to optimize designs in both form and function.
Definition: Function
A function is a rule (usually an equation) that assigns a unique value in the Range, R, to each value in the Domain, D.
D = Domain
R = Range
D = Domain
R = Range
Function
Not a Function
A function is written using the notation y = f(x). (Read “f of x”.) This giving different functions different names.

3. Sometimes, the domain and range of a function is stated, especially when the domain has to be restricted. Intervals may be open, closed or half-open, and finite or infinite.
Name: The set of all real numbers
Notation: a
Name: The set of numbers greater than a
Notation: a
Name: The set of numbers greater than or equal to a
Notation:

4. Sometimes, the domain and range of a function is stated, especially when the domain has to be restricted. Intervals may be open, closed or half-open, and finite or infinite. b
Name: The set of numbers less than b
Notation: b
Name: The set of numbers less than or equal b
Notation:

5. Sometimes, the domain and range of a function is stated, especially when the domain has to be restricted. Intervals may be open, closed or half-open, and finite or infinite. b a
Name: Open interval ab
Notation: b a
Closed at a and open at b
Notation:

6. Sometimes, the domain and range of a function is stated, especially when the domain has to be restricted. Intervals may be open, closed or half-open, and finite or infinite. b a
Name: Closed interval ab
Notation: b a
Open at a and closed at b
Notation:

7. Example 2
For each function do the following:
(a) Find the domain.
(b) Find the range.
(c) Draw the graph.
(a) Since the square root function can’t be negative, x ≥ 1. Or, in interval notation D: [1,∞).
(c)
(b) Since the domain has to start at 0 or greater, the range has to start at 2 and increases. Or interval notation R: [2,∞).

8. Example 2
For each function do the following:
(a) Find the domain.
(b) Find the range.
(c) Draw the graph.
(c)
(a) Since x is in the denominator, x ≠ 0. Or, in interval notation D: (−∞, 0) U (0, ∞).
(b) Since x2 is always positive, y is always positive and y > 1. Or, in interval notation R: (1, ∞).

9. Even and Odd Functions
If f(x) is a function, then for every x in the domain of f(x), f(x) is an:
Even function when f(−x) = f(x)
For example, If f(x) = x2, then f(−x) = (−x)2 = x2 = x2.
Odd function when f(−x) = −f(x)
For example, If f(x) = x3, then f(−x) = (−x)3 = −x3.

10. Example 4
Determine whether each function is even, odd, or neither.
Find f(−x) by substituting −x into f(x).
Since f(−x) = f(x), f(x) is even.

11. Example 4
Determine whether each function is even, odd, or neither.
Find f(−x) by substituting −x into f(x).
Since f(−x) ≠ f(x) ≠ −f(x) , f(x) is neither even nor odd.

12. Example 5
Piecewise Functions
Graph the piecewise defined function.
First graph y = 3 – x with x = 1 as the endpoint.
When x = 1, y = 3 – 1 = 2. Also, the y-intercept is 3.
And, since x ≤ 1, the line is drawn in the negative direction.
Next, graph y = 2x with x = 1 as the endpoint.
When x = 1, y = 2 ∙ 1 = 2. Also, the slope is 2.
And, since x > 1, the line is drawn in the positive direction.

13. Write a piecewise formula for the graph.
Example 6
Write a piecewise formula for the graph.
The first piece goes from 0, non-inclusive, to 2, inclusive. 2
The slope is −1 and the y-intercept is 2. So y = −x + 2
(2, 1) 2
The second piece goes from 2, non-inclusive, to 5, inclusive.
−x + 2,
The slope is and the y-intercept is So
2 < x ≤ 5

14. The Absolute Value Function
is defined as:
The graph of│x│ is made up of two straight lines.
When x < 0, the function starts at 0 and goes up with a slope of −1.
When x > 0, the function starts at 0 and goes up with a slope of 1.

15. Composite Functions
We say that the function f(g(x)), read as “f of g of x”, is the composite of g and f.
The usual notation is f ○ g, which of read “f of g.
Thus, the value of f ○ g at x is (f ○ g )(x) = f(g(x)).

16. Example 8
Let f(x) = x + 5 and g(x) = x2 – 3. Find f(g(x)).
f(g(x)) is found by substituting the expression for g(x) into the x inside of f(x).
f(g(x)) = ( ) + 5
x2 – 3
= x2 + 2

17. Now it’s your turn.
Let f(x) = x + 5 and g(x) = x2 – 3. Find g(f(x)).
g(f(x)) is found by substituting the expression for f(x) into the x inside of g(x).
f(g(x)) = ( )2 – 3
x + 5
= x2 + 10x + 25 – 3
= x2 + 10x + 22 p