AP Calculus AB/BC

1. AP Calculus AB/BC 1.2 Functions, p. 12

2. D = Domain R = Range D = Domain R = Range 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. 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. 0 Notation: a Notation: Notation: a 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 Name: The set of numbers greater than a Name: The set of numbers greater than or equal to a

4. Notation: b Notation: b 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 numbers less than b Name: The set of numbers less than or equal b

5. a b Notation: Notation: a b 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: Open interval ab Closed at a and open at b

6. Notation: Notation: a a b b 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: Closed interval ab Open at a and closed at b

7. Example 2 For each function do the following: (a) Find the domain. (b) Find the range. (c) Draw the graph. (c) (a) Since the square root function can’t be negative, x > 1. Or, in interval notation D: [1,∞]. (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, yis always positive and y > 1. Or, in interval notation R: (1, ∞).

9. Even and Odd Functions If f(x) is a function, thenfor 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 −xinto 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 −xinto 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. The slope is and the y-intercept is . So Example 6 Write a piecewise formula for the graph. The first piece goes from 0, non-inclusive, to 2, inclusive. 2 (2, 1) The slope is −1 and the y-intercept is 2. So y = −x + 2 2 The second piece goes from 2, non-inclusive, to 5, inclusive. −x + 2, 2 < x≤5

14. is defined as: The Absolute Value Function 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 “fof 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