What is the Inverse of a function?
If the function itself is considered a "DO" action, then the inverse is the "UNDO"
What about the domain and Range?
Definition: The inverse of a function is when the domain and the range trade places. All elements of the domain become the range, and all elements of the range become the domain.
Therefore, the inverse of a function is equivalent to what kind of transformation
Reflection over the line y = x
| Original function f(x) | Inverse of function or f-1(x) |
|---|---|
{ (0,3 ) , (1,4) , (2, 5) } |
{ (3, 0 ) , (4,1) , (5, 2) } |
Is the inverse of a function always a function?
To answer that question let's look at the function in Diagram 1. In that graph, you can see
the original function and its inverse
Is the inverse of that function , also a function?
As you can see, the inverse of that function does not pass the vertical line test and, therefore , is not a function.
Conclusion: The inverse of a function is not necessarily a function.
So...when is the inverse of a function also a function?
f-1(x), the inverse, is itself a function only when f(x), the original function, is a one-to-one function.
Remember: 1 to 1 functions must pass the horizontal line test!
Example 1In this case, f(x) is a function, but f-1(x) is not a function.
Function #1, f(x) is not a one to one function because it does not pass the horizontal line test .
f-1(x) is not a function
In this case, f(x) is a function and f-1(x) is also a function.
This function passes the horizontal line test so, its inverse will also be a function.
Practice Problems Part I
Remember: $$ f^{-1} (x) $$ means 'the inverse of f(x) '
$$ f^{-1} (x) $$ = { (2,1) , (4,3) , (6,5) }
Remember: $$ f^{-1} (x) $$ means 'the inverse of f(x) '
$$ f^{-1} (x) $$ = { (14,33) , (15, 23) , (12,11) , (14, 13) }}
Remember: $$ f^{-1} (x) $$ means 'the inverse of f(x) '
$$ f^{-1} (x) $$ = { (12, -1) , (114, 13) , (61,15) , ( 12,1) }
y = 3 is a horizontal line. In a horizontal line all elements of the range are 3.
So, it looks something $$ f^{-1} (x) $$ ={ ( -1,3 ) , (0,3) , (1,3)}
Therefore the inverse of this function will be whatever line has 3 for all elements in its domain.
Therefore the inverse of y = 3 is the line x = 3
Solving Algebraically for the Inverse
What is the inverse of f(x) = x + 1?
Just like in our prior examples, we need to switch the domain and range. In an equation , the domain is represented by the x variable and the range by the y variable
f(x): took an element from the domain and added 1 to arrive at the corresponding element in the range.
f-1(x) : took an element from the domain and subtracted 1 to arrive at the corresponding element in the range.
Practice Problems Part II
Problem 1
What is the inverse of the function f(x) = x + 22?
| replace f(x) with y |
y = x + 22 |
| switch x and y |
x = y + 22 |
| solve for new 'y' |
 |
| replace 'y' with f-1(x) |
f-1(x) = x - 22 |
Problem 2
What is the inverse of the function f(x) = 2x?
| replace f(x) with y |
y = 2x |
| switch x and y |
x = 2y  |
| solve for new 'y' |
$$ \frac{1}{2} x = \frac{1}{2} \cdot 2y \\ \frac{1}{2} x = y $$ |
| replace 'y' with f-1(x) |
$$ f^{-1 } (x) =\frac{1}{2} x $$ |
Problem 3
What is the inverse of the function $$ f(x) = \frac{1}{2}x + 3 $$?
| replace f(x) with y |
$$ y = \frac{1}{2}x + 3 $$ |
| switch x and y |
$$ x = \frac{1}{2}y + 3 $$ |
| solve for new 'y' |
$$ \color{Red}{2 \cdot } x = \color{Red}{2 \cdot } \big ( \frac{1}{2}y + 3 \big) \\ 2 \cdot x = \frac{2}{2}\cdot y + 2 \cdot 3 \\ 2x = y + 6 \\ 2x -6 = y $$ |
| replace 'y' with f-1(x) |
$$ f^{-1 } (x) = 2x -6 $$ |
Problem 4
What is the inverse of the function $$ f(x) = x^2 + 3 $$?
| replace f(x) with y |
$$ y = x^2 + 3 $$ |
| switch x and y |
$$ x = y^2 + 3 $$ |
| solve for new 'y' |
$$ x \color{Red}{ -3} = y^2 + 3 \color{Red}{ -3} \\ x -3 = y^2 \\ \sqrt{x-3} = \sqrt{y^2} \\ \sqrt{x-3} = y $$ |
| replace 'y' with f-1(x) |
$$ f^{-1 } (x) = \sqrt{x-3} $$ |
Problem 5
What is the inverse of the function $$ f(x) = 5 \sqrt{ x } $$ ?
| replace f(x) with y |
$$ y = 5 \sqrt{ x } $$ |
| switch x and y |
$$ x = 5 \sqrt{ y } $$ |
| solve for new 'y' |
$$ x{ \color{Red}{^2}} = \big( 5 \sqrt{ y } \big) { \color{Red}{^2}} \\ x^2 = 5{ \color{Red}{^2}} \cdot ( \sqrt{ y } ){ \color{Red}{^2}} \\ x^2 = 25y \\ \frac{1}{25} \cdot x^2 = \frac{1}{25} \cdot 25y \\ \frac{1}{25} \cdot x^2 = y $$ |
| replace 'y' with f-1(x) |
$$ f^{-1 } (x) = \frac{1}{25} \cdot x^2 $$ |
