1.2-Basics of Functions

1.2-Basics of Functions

A relation is any set of ordered pairs. • The set of all first components of the ordered pairs is called the domain of the relation. • The set of all second components is called the range of the relation.

Find the domain and range of the relation: {(1994, 56.21), (1995, 51.00), (1996, 47.70), (1997, 42.78), (1998, 39.43)} Example 1 Solution The domain is the set of all first components. Thus, the domain is: {1994,1995,1996,1997,1998}. The range is the set of all second components. Thus, the range is: {56.21, 51.00, 47.70, 42.78, 39.43}.

Definition of a Function • A relation becomes a functionis a correspondence between two sets X and Y that assigns to each element xof set X exactly one element yof set Y.

(a) 1 2 3 4 6 8 9 Figure (a) shows that every element in the domain corresponds to exactly one element in the range. No two ordered pairs in the given relation have the same first component and different second components. YES- the relation is a function. Domain Range (b) Figure (b) shows that 6 corresponds to both 1 and 2. NO-This relation is not a function; two ordered pairs have the same first component and different second components. 1 2 3 4 6 8 9 Domain Range Determine whether each relation is a function. a. {(1, 6), (2, 6), (3, 8), (4, 9)} Solution The elements in the domain and in blue. The elements in the range are in yellow. Example 2 b. {(6,1),(6,2),(8,3),(9,4)}

Determine whether the equation defines y as a function of x: Example: Solve for y: Since each value of x corresponds to ONE value of y, it IS a funciton!

Determine whether the equation defines y as a function of x: Example: Solve for y: Since each value of x will correspond to TWO values of y, it Is NOT a function!

a. We find f (2) by substituting 2 for x in the equation. f (2) = 22 + 3 • 2 + 5 = 4 + 6 + 5 = 15 If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x) Functions are usually given as equations using “function” notation: Substitution: f(x) = x2 + 3x+ 5. Thus, f (2) = 15.

Equivalently, f (x + 3) = (x + 3)2 + 3(x + 3)+ 5 = x2 + 6x + 9 + 3x + 9 + 5 f (x + 3) = x2 + 9x + 23. Square x + 3 and distribute 3 throughout the parentheses. If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x) Text Example cont. Solution b.We find f (x + 3) by substituting x + 3 for x in the equation. f (x + 3) = (x + 3)2 + 3(x + 3)+ 5

Equivalently, f (-x) = (-x)2 + 3(-x)+ 5 f (-x) = x2 –3x + 5. If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x) Text Example cont. Solution c.We find f (-x) by substituting -x for x in the equation. f (-x) = (-x)2 + 3(-x)+ 5

“Piecewise” Functions are “split” into different pieces depending on the x-value Graph this function:

Graph this function:

Ex. 3 Find the domain of each function: a) Domain is the set of all real numbers. b) All real numbers except -5 c) Volume of a sphere: Since this is a radius, all r values must be positive. r > 0 d) Since what’s under the radical must be positive, solve the inequality:

Ex. Find the domain and range of the function using your graphing calculator:

Definition of a Difference Quotient • The expression: is called the difference quotient. This is used extensively in Calculus!

Example: Find and simplify the difference quotient for:

1.2-Basics of Functions