Inverse Relations and Functions Section 7.7 Inverse Relations and Functions

Definition: The inverse of a function is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original function. Notation:   If  f  is a given function, then f -1 denotes the inverse of  f.

Definition of Inverse Function - Do Not Write- A function and its inverse can be described as the "DO" and the "UNDO" functions.  A function takes a starting value, performs some operation on this value, and creates an output answer.  The inverse of this function takes the output answer, performs some operation on it, and arrives back at the original function's starting value

Please Write in Notes This "DO" and "UNDO" process can be stated as a composition of functions.  If functions f and g are inverse functions,                                    .  A function composed with its inverse yields the original starting value.  Think of them as "undoing" one another and leaving you right where you started.

The inverse of a function may not always be a function! Remember The inverse of a function may not always be a function!  

Inverse Relations and Functions Lesson 7-7 Additional Examples a. Find the inverse of relation m. Relation m x –1 0 1 2 y –2 –1 –1 –2

Inverse Relations and Functions Lesson 7-7 Additional Examples a. Find the inverse of relation m. Relation m x –1 0 1 2 y –2 –1 –1 –2 Interchange the x and y columns. Inverse of Relation m x –2 –1 –1 –2 y –1 0 1 2

Inverse Relations and Functions Lesson 7-7 Additional Examples (continued) b. Graph m and its inverse on the same graph. Relation m

Inverse Relations and Functions Lesson 7-7 Additional Examples (continued) b. Graph m and its inverse on the same graph. Reversing the Ordered Pairs Relation m

Inverse Relations and Functions Lesson 7-7 Additional Examples (continued) b. Graph m and its inverse on the same graph. Reversing the Ordered Pairs Relation m Inverse of m

1. Set the function = y 2. Swap the x and y variables 3. Solve for y Example Solve algebraically:  Solving for an inverse algebraically is a three step process: 1.  Set the function = y 2.  Swap the x and y variables 3.  Solve for y

Inverse Relations and Functions Lesson 7-7 Additional Examples Find the inverse of y = x2 – 2. y = x2 – 2

Inverse Relations and Functions Lesson 7-7 Additional Examples Find the inverse of y = x2 – 2. y = x2 – 2 x = y2 – 2 Interchange x and y.

Inverse Relations and Functions Lesson 7-7 Additional Examples Find the inverse of y = x2 – 2. y = x2 – 2 x = y2 – 2 Interchange x and y. x + 2 = y2 Solve for y.

Inverse Relations and Functions Lesson 7-7 Additional Examples Find the inverse of y = x2 – 2. y = x2 – 2 x = y2 – 2 Interchange x and y. x + 2 = y2 Solve for y. ± x + 2 = y Find the square root of each side.

Inverse Relations and Functions Lesson 7-7 Additional Examples Graph y = –x2 – 2 and its inverse. The graph of y = –x2 – 2 is a parabola that opens downward with vertex (0, –2).

Inverse Relations and Functions Lesson 7-7 Additional Examples Graph y = –x2 – 2 and its inverse. The graph of y = –x2 – 2 is a parabola that opens downward with vertex (0, –2). The reflection of the parabola in the line x = y is the graph of the inverse.

Inverse Relations and Functions Lesson 7-7 Additional Examples Graph y = –x2 – 2 and its inverse. The graph of y = –x2 – 2 is a parabola that opens downward with vertex (0, –2). The reflection of the parabola in the line x = y is the graph of the inverse. You can also find points on the graph of the inverse by reversing the coordinates of points on y = –x2 – 2.

Inverse Relations and Functions Lesson 7-7 Additional Examples Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ.

Inverse Relations and Functions Lesson 7-7 Additional Examples Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ. Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1.

Inverse Relations and Functions Lesson 7-7 Additional Examples Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ. Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1. Since the principal square root is nonnegative, the range is the set of nonnegative numbers.

Inverse Relations and Functions Lesson 7-7 Additional Examples Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ. Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1. Since the principal square root is nonnegative, the range is the set of nonnegative numbers. b. Find ƒ –1

Inverse Relations and Functions Lesson 7-7 Additional Examples Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ. Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1. Since the principal square root is nonnegative, the range is the set of nonnegative numbers. b. Find ƒ –1 ƒ(x) = 2x + 2 y = 2x + 2 Rewrite the equation using y. x = 2y + 2 Interchange x and y.

Inverse Relations and Functions Lesson 7-7 Additional Examples Consider the function ƒ(x) = 2x + 2 . a. Find the domain and range of ƒ. Since the radicand cannot be negative, the domain is the set of numbers greater than or equal to –1. Since the principal square root is nonnegative, the range is the set of nonnegative numbers. b. Find ƒ –1 ƒ(x) = 2x + 2 y = 2x + 2 Rewrite the equation using y. x = 2y + 2 Interchange x and y. x2 = 2y + 2 Square both sides. y = x2 – 2 2 Solve for y. So, ƒ –1(x) = . x2 – 2 2

Inverse Relations and Functions Lesson 7-7 Additional Examples For the function ƒ(x) = x + 5, find (ƒ–1 ° ƒ)(652) and (ƒ ° ƒ–1)(– 86). 1 2

Inverse Relations and Functions Lesson 7-7 Additional Examples For the function ƒ(x) = x + 5, find (ƒ–1 ° ƒ)(652) and (ƒ ° ƒ–1)(– 86). 1 2 Since ƒ is a linear function, so is ƒ –1. Therefore ƒ –1 is a function.

Inverse Relations and Functions Lesson 7-7 Additional Examples For the function ƒ(x) = x + 5, find (ƒ–1 ° ƒ)(652) and (ƒ ° ƒ–1)(– 86). 1 2 Since ƒ is a linear function, so is ƒ –1. Therefore ƒ –1 is a function. So (ƒ –1 ° ƒ)(652) = 652 and (ƒ ° ƒ –1)(– 86) = – 86.

Inverse Relations and Functions Lesson 7-7 Checking For Understanding – Email your answers Find the inverse of y = x + 9. 2. If ƒ(x) = 5x + 8, find (ƒ–1 ° ƒ)(59) and (ƒ ° ƒ–1)(3001).