1. 7-7 Inverse Relations & Functions
M11.D.1.1.3: Identify the domain, range, or inverse of a relation

2. Objectives
The Inverse of a Function

3. Finding the Inverse of a Relation
a. Find the inverse of relation m.
Relation m
x –
y –2 –1 –1 –2
Interchange the x and y columns.
Inverse of Relation m
x –2 –1 –1 –2
y –

4. Continued b. Graph m and its inverse on the same graph.
Reversing the Ordered Pairs
Relation m
Inverse of m

5. Interchanging x and y 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.

6. Graphing a Relation and Its Inverse
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.

7. Finding an Inverse Function
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

8. Continued c. Find the domain and range of ƒ –1.
The domain of ƒ –1 equals the range of ƒ, which is the set of nonnegative numbers.
Note that the range of ƒ–1 is the same as the domain of ƒ.
Since x , –1. Thus the range of ƒ–1 is the set of numbers greater than or equal to –1.
x2 – 2 2
> –
d. Is ƒ –1 a function? Explain.
For each x in the domain of ƒ–1, there is only one value of ƒ –1(x). So ƒ –1 is a function.

9. Real-World Example
The function d = 16t 2 models the distance d in feet that an object falls in t seconds. Find the inverse function. Use the inverse to estimate the time it takes an object to fall 50 feet.
d = 16t 2
t 2 = d 16
Solve for t. Do not interchange variables.
t = d 4
Quantity of time must be positive.
t = 1 4
The time the object falls is 1.77 seconds.

10. Vocabulary
If and are inverse functions, then
and

11. Composition of Inverse Functions
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) = –

12. Homework
p 410 #1,2,5,6,14,15,23,24,31,32