1. Unit 1 Lesson 6
Inverse Functions

2. What is an inverse of a function?
(Informal definition): The inverse of 𝑓 𝑥 , denoted 𝑓 −1 𝑥 , is the function that reverses the effect of 𝑓 𝑥 .
Reading this graph, you should see that values in the domain are entered into 𝑓(𝑥), generating values in the range.
Values in the range of values are entered into 𝑓 −1 (𝑥), generating values that were in the original domain

3. What is an inverse of a function? (cont.)
For example, the inverse of 𝑓 𝑥 = 𝑥 3 is 𝑓 −1 𝑥 = 𝑥 1 3
Looking at the table below, we can see this relationship between the domain and range when using an inverse.

4. Formal definition of an inverse
Definition Inverse Let 𝑓 𝑥 have domain 𝐷 and range 𝑅. If there is a function 𝑔 𝑥 with domain 𝑅 such that 𝑔 𝑓 𝑥 =𝑥 for 𝑥∈𝐷 and 𝑓 𝑔 𝑥 =𝑥 for 𝑥∈𝑅 then 𝑓 𝑥 is said to be invertible. The function 𝑔 𝑥 is called the inverse function and is denoted 𝑓 −1 𝑥

5. How do we tell if an inverse exists?
Show that 𝑓 𝑥 =2𝑥−18 is invertible. What are the domain and range of 𝑓 −1 𝑥 ?
Step 1: Solve the equation 𝑦=𝑓 𝑥 for 𝑥 in terms of 𝑦.
Step 2: Prefer to have the inverse in terms of x. So we interchange the x and y.
𝑦=2𝑥−18
2𝑥=𝑦+18
𝑓 −1 𝑥 = 1 2 𝑥+9
𝑥= 1 2 𝑦+9
It is usually a good idea to check our answer. Next slide please!
Note: This gives us the inverse of the variable 𝑦: 𝑓 −1 𝑦 = 1 2 𝑦+9

6. Checking our last answer
𝑓 −1 𝑥 = 1 2 𝑥+9
Checking our last answer
𝑓 𝑥 =2𝑥−18
To check our last answer we have to verify that 𝑓 −1 𝑓 𝑥 =𝑥 and 𝑓 𝑓 −1 𝑥 =𝑥
1 2 2𝑥−18 +9=
𝑓 −1 𝑓 𝑥 =
𝑓 −1 2𝑥−18 =
𝑥−9 +9=𝑥
𝑓 1 2 𝑥+9 =
𝑥+9 −18=
𝑓 𝑓 −1 𝑥 =
𝑥−18 +18=𝑥
Because 𝑓 −1 is a linear function, its domain and range are ℝ.

7. To reiterate the concept of an inverse
Functions 𝑓 and 𝑔 are inverses iff 𝑓 𝑔 𝑥 =𝑥 AND 𝑔 𝑓 𝑥 =𝑥
If 𝑓 2 =4, and 𝑔 𝑥 is the inverse of 𝑓(𝑥), then 𝑔 4 =2
However, there is a problem with this particular example. What is it?

8. The problem with 𝑓 𝑥 = 𝑥 2
The problem is that every positive number occurs twice as an output of 𝑓 𝑥 = 𝑥 2 .
That means that our inverse does not give us a one-to-one mapping between the domain and range!

9. Definition of a One-to-One Function
Definition One-to-One Function A function 𝑓 𝑥 is one-to-one on the domain D if, for every value c, the equation 𝑓 𝑥 =𝑐 has at most one solution for 𝑥∈𝐷. Or, equivalently, if for all 𝑎,𝑏∈𝐷, 𝑓 𝑎 ≠𝑓 𝑏 𝑢𝑛𝑙𝑒𝑠𝑠 𝑎=𝑏

10. Theorem: Existence of Inverses
Theorem Existence of Inverses The inverse function 𝑓 −1 𝑥 exists if and only if 𝑓 𝑥 is one-to-one on its domain D. Furthermore,
Domain of 𝑓= range of 𝑓 −1
Range of 𝑓= domain of 𝑓 −1

11. Checking for one-to-one with the Horizontal Line Test
Graphically, we can determine if a function is one-to-one, which guarantees that an inverse exists!
Horizontal Line Test A function 𝑓 𝑥 is one-to-one if every horizontal line intersects the graph of 𝑓 𝑥 in at most one point.

12. Another Informal Test for Invertibility
From our last slide, we can easily justify that any function that is monotonically increasing or monotonically decreasing has an inverse.
Why?
Because such functions always pass the horizontal line test, so they are always one-to-one!

13. Inverses are reflections over the line 𝑦=𝑥
Another way of graphically looking at an inverse is to see it as a reflection over the line 𝑦=𝑥
When we create an inverse for 𝑓 𝑥 = 𝑥 2 +1 using our “steps”, we encounter a problem with its inverse. What is the problem?
The inverse is not a function!!!

14. Fix for Some Invertibility Problems
For some functions that do not pass the Horizontal Line Test, we can still produce an inverse
We can do this by either
Creating a piecewise function for the function and its inverse
Restricting the domain of the original function, only looking at part of the “picture”. (This is the most often used option!)
𝑓 𝑥 = 𝑥 2 , 𝑥≤0 & 𝑥 2 , 𝑥>0
𝑓 −1 𝑥 = − 𝑥 , 𝑥≤0 & 𝑥 , 𝑥>0

15. Restricting a domain to help create an inverse
Find a domain on which 𝑓 𝑥 = 𝑥 2 is one-to-one and determine its inverse on this domain.
If we restrict the domain such that 𝐷= 𝑥:𝑥≥0 , then we can create an inverse
𝑓 −1 𝑥 = 𝑥
Not one-to-one!
One-to-one!

16. Inverse Trig Functions Notation
The actual “inverse” trigonometric functions can be written in one of two equivalent ways.
sin −1 𝑥 = arcsin 𝑥
csc −1 𝑥 = arccsc 𝑥
cos −1 𝑥 = arccos 𝑥
sec −1 𝑥 = arcsec 𝑥
tan −1 𝑥 = arctan 𝑥
cot −1 𝑥 = arccot 𝑥

17. Manipulating Inverse Trig Functions
The formula for inverse trig functions is read this way:
sin −1 𝑥 =𝑎𝑛𝑔𝑙𝑒  “the arcsin of x is the angle whose sine is x”
This means the equation can be rewritten as sin (𝑎𝑛𝑔𝑙𝑒) =𝑥
Example: Rewrite sin −1 −1 =− 𝜋 2
sin − 𝜋 2 =−1

18. Creating Inverse Trig Functions
For a trig function to have an inverse, it must pass the horizontal line test.
Since all of the trig functions are periodic, they all fail the horizontal line test.

19. Creating Inverse Trig Functions
Before we can create the inverse of a trig function we have to restrict the domains of each trig function so that they can pass the horizontal line test!
restrict sine to − 𝜋 2 , 𝜋 2
restrict cosine to 0,𝜋

20. Restricted Domains and Ranges of Trig Functions and Their Inverses
Inv Trig Func
𝐲= 𝐬𝐢𝐧 𝐱
− 𝜋 2 ≤𝑥≤ 𝜋 2
−1≤𝑦≤1
𝐲= 𝐬𝐢𝐧 −𝟏 𝐱
−1≤𝑥≤1
− 𝜋 2 ≤𝑦≤ 𝜋 2
𝐲= 𝐜𝐨𝐬 𝐱
0≤𝑥≤𝜋
𝐲= 𝐜𝐨𝐬 −𝟏 𝐱
0≤𝑦≤𝜋
𝐲= 𝐭𝐚𝐧 𝐱
𝑦∈ℝ
𝐲= 𝐭𝐚𝐧 −𝟏 𝐱
𝑥∈ℝ
arcsin
arccos
arctan

21. Graphs of Inverse Trig Tunctions
The other 3 look like …
arccsc
arcsec
arccot

22. Inverse trig functions
What do all the “restrictions” on domain mean?
Our answers for arcsin, arctan, and arccsc will come from Quad I and IV
Our answers for arccos, arcsec, and arccot will come from Quad I and II

23. The Art of Finding Trig and Inverse Trig Exact Values
Step 1 – Know what is being asked of you!
Looking for the sine ratio for the given angle
sin 𝜋 3
sin −
Looking for the angle measure for the given ratio
sin cos −
Looking for the sine ratio of the angle whose cosine is the given ratio
Looking for the angle measure using the ratio given by the tangent function using the given angle
sin −1 tan 𝜋 4

24. The Art of Finding Trig and Inverse Trig Exact Values
Step 2 – Pick the right tools to help get the answer!
Unit Circle
4 points from the Unit Circle 0, 𝜋 2 ,𝜋, 3𝜋 2
Co-terminal Angles
Reference Angles
Right triangles (if using non-unit circle angles/ratios)
Special right triangles
Knowing the quadrants where the ratios are positive (ASTC)

25. Find the Values (Remember ASTC!)
cos − =
𝜋 4
𝜋 4
tan − =
cos −1 − =
− 𝜋 2
5𝜋 6
sin −1 −1 =
cos − =
sin −1 − =
𝜋 6
− 𝜋 4

26. What does sin −1 𝑥 “look” like?
Since we are looking at the inverse of sine, 𝑥 represents a ratio 𝑜𝑝𝑝 ℎ𝑦𝑝
Using SOHCAHTOA, we’ll say that 𝑜𝑝𝑝=𝑥 and ℎ𝑦𝑝=1
With the help of the Pythagorean Theorem, we can say that 3rd side = 1− 𝑥 2
Why did I stick cos −1 𝑥 here?
𝑥= 𝑜𝑝𝑝 ℎ𝑦𝑝 = 𝑥 1
cos −1 𝑥 𝑥 1
sin −1 𝑥

27. Use the triangle and basic right triangle relationships to answer the following:
Given sin −1 𝑥
𝜋 2
sin −1 𝑥 + cos −1 𝑥 =
cos −1 𝑥 1 𝑥
1− 𝑥 2
cos sin −1 𝑥 =
sin −1 𝑥
1− 𝑥 2
sin cos −1 𝑥 =
𝑥 1− 𝑥 2
tan sin −1 𝑥 =

28. Find the following: 1+ 𝑥 2 sec tan −1 𝑥 = 𝑥 2 −1 𝑥 sin sec −1 𝑥 =
The basic rule with these problems is to “stick” the inverse trig in the angle at the bottom left, and use SOHCAHTOA and Pythag Thm to establish the side-values
1+ 𝑥 2 𝑥
1+ 𝑥 2
sec tan −1 𝑥 =
tan −1 𝑥 1 𝑥
𝑥 2 −1 𝑥
𝑥 2 −1
sin sec −1 𝑥 =
sec −1 𝑥 1

29. Find the sinθ, cosθ and tanθ if:
sin 𝜃 =
𝜃= tan −
313 12
cos 𝜃 =
𝜃= tan −1 𝑥
12 13 13
tan 𝜃 =
3 2
𝜃= cos −
sin 𝜃 = 2 3
1 2
cos 𝜃 =
𝜃= cos −1 𝑥 1
tan 𝜃 = 3
What about csc 𝜃 , sec 𝜃 , cot 𝜃 ?

30. Compositions of functions
Recall 𝑓 𝑓 −1 𝑥 =𝑥 for any function.
This is true for trig functions as long as x is within the defined interval:
If example has arc on the outside of parentheses then look at range
If example has arc on the inside then look at domain.
Don’t forget that you can use coterminal angles (±2π), but only do this with sin, cos, tan, NOT WITH arcsin, arccos, arctan

31. Examples sin −1 sin − 𝜋 3 =− 𝜋 3 tan arctan −5 =−5
Find the exact value of each:
tan tan −1 − b. sin −1 sin 5𝜋 3
c. cos arccos 𝜋
“arc” on inside, looking at domain.
“arc” on outside, looking at range.
Domain of arctan is ℝ, so -5 is valid
Range of arcsin is − 𝜋 2 ≤𝑦≤ 𝜋 2
tan arctan −5 =−5
5𝜋 3 is outside range. Subtract 2𝜋 to get back into range. 5𝜋 3 −2𝜋=− 𝜋 3
sin −1 sin − 𝜋 =− 𝜋 3
“arc” on inside, looking at domain.
Domain of arccos −1,1 , π>1, so this is undefined.

32. Evaluating compositions of functions using triangles
Evaluate each using triangles:
a. cos arcsin − b. tan arccos 2 3
The cosine of the angle whose sine is -3/5. Triangle is in Quad IV
The tangent of the angle whose cosine is 2/3. Triangle is in Quad I
So the answer is 4 5
So the answer is

33. Find the following:
sin −1 sin 𝜋 7 =
𝜋 7
Why?
sin −1 sin 𝜋 =
sin −1 sin 𝜋 ≠𝜋
sin −1 sin 5𝜋 7 =
2𝜋 7

34. Homework Chapter 1 Precalculus Review Packet
Pgs , #2, 3, 4, 9, 13, 16, 19, 23 – 31 odd, 39, 43