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**Definition of a Relation**

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, and the set of all second components is called the range of the relation.

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**Example:Analyzing U.S. Mobile-Phone Bills as a Relation**

Find the domain and range of the relation {(1994, 56.21), (1995, 51.00), (1996, 47.70), (1997, 42.78), (1998, 39.43)} 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}.

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**Definition of a Function**

A function is a correspondence between two sets X and Y that assigns to each element x of set X exactly one element y of set Y. For each element x in X, the corresponding element y in Y is called the value of the function at x. The set X is called the domain of the function, and the set of all function values, Y, is called the range of the function.

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**Example: Determining Whether a Relation is a Function**

Determine whether each relation is a function. a. {(1, 6), (2, 6), (3, 8), (4, 9)} b. {(6,1),(6,2),(8,3),(9,4)} Solution We begin by making a figure for each relation that shows set X, the domain, and set Y, the range, shown below. 1 2 3 4 6 8 9 Domain Range (a) 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. Thus, the relation is a function. 6 8 9 1 2 3 4 Domain Range (b) Figure (b) shows that 6 corresponds to both 1 and 2. This relation is not a function; two ordered pairs have the same first component and different second components.

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**When is a relation a function?**

Determine whether each relation is a function. S = {(1,2), (3,4),(5,6),(7,8)} Each first component is unique, therefore S is a function T = {(1,2), (3,4),(6,5),(1,5)} Note that the first component in the first pair is the same as the first component in the second pair, therefore T is not a function.

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Function Notation When an equation represents a function, the function is often named by a letter such as f, g, h, F, G, or H. Any letter can be used to name a function. Suppose that f names a function. Think of the domain as the set of the function's inputs and the range as the set of the function's outputs. The input is represented by x and the output by f (x). The special notation f(x), read "f of x" or "f at x," represents the value of the function at the number x. If a function is named f and x represents the independent variable, the notation f (x) corresponds to the y-value for a given x. Thus, f (x) = 4 - x2 and y = 4 - x2 define the same function. This function may be written as y = f (x) = 4 - x2.

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**Example: Evaluating a Function**

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x) Solution We substitute 2, x + 3, and -x for x in the definition of f. When replacing x with a variable or an algebraic expression, you might find it helpful to think of the function's equation as f (x) = x2 + 3x + 5. a. We find f (2) by substituting 2 for x in the equation. f (2) = • = = 15 Thus, f (2) = 15. more

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**Example: Evaluating a Function**

If f (x) = x2 + 3x + 5, evaluate: a. f (2) b. f (x + 3) c. f (-x) 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 + 3) = (x + 3)2 + 3(x + 3) + 5 = x2 + 6x x = x2 + 9x + 23. Square x + 3 and distribute 3 throughout the parentheses. more

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**Example: Evaluating a Function**

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

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**Finding a Function’s Domain**

If a function f does not model data or verbal conditions, its domain is the largest set of real numbers for which the value of f (x) is a real number. Exclude from a function's domain real numbers that cause division by zero and real numbers that result in an even root of a negative number.

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**Example: Finding the Domain of a Function**

Normally it is safe to say the domain of a function is all real numbers. However, there are 2 conditions which must be considered: 1)division by zero and 2)even roots of negative numbers. Consider the following functions and find the domain of each function: Solution a. The function f (x) = x2 – 7x contains neither division nor an even root. The domain of f is the set of all real numbers. b. The function contains division. Because division by 0 is undefined, we must exclude from the domain values of x that cause x2 – 9 to be 0. Thus, x cannot equal –3 or 3. The domain of function g is {x | x = -3, x = 3}. / more

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**Example: Finding the Domain of a Function**

Continuing… Solution c. The function contains an even root. Because only nonnegative numbers have real square roots, the quantity under the radical sign, 3x + 12, must be greater than or equal to 0. 3x + 12 > 0 3x > -12 x > -4 The domain of h is { x | x > -4} or the interval [-4, oo).

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**Problems Evaluate each function for the given values. F(x) = 3x + 7**

F(4) b. F(x+1) c. F(-x) F(x) = f(16) b. f(-24) c.f(25-2x) Exercises page 160, numbers 1-7 odd, 21 – 31 odd, 51 – 71 odd.

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**Definition of the Difference Quotient**

The expression is called the difference quotient.

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**Example: Evaluating the Difference Quotient**

For , find and simplify: b. Repeat for Exercises pg 160, numbers odd.

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**Piecewise-Defined Functions**

Functions that are made up of different pieces based upon different domains are called piecewise-defined functions.

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**Example: Piecewise-defined function**

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Section 7.6 Functions Math in Our World. Learning Objectives Identify functions. Write functions in function notation. Evaluate functions. Find.

Section 7.6 Functions Math in Our World. Learning Objectives Identify functions. Write functions in function notation. Evaluate functions. Find.

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