Inverse Functions Lesson

We previously learned that a function is a relation where each x-value corresponds to one and only one y-value. A function is allowed to have a y-value that corresponds to different x-values.
{(3,1),(6,5),(9,1),(2,3)}
In our function above, the y-value of 1 is linked up with two different x-values (3 and 9). This does not violate the definition of a function since each x-value is associated with only one y-value.

One-to-One Function

A one-to-one function has a stricter definition. For a function to be a one-to-one function, each x-value can correspond to only one y-value and each y-value can correspond to only one x-value.
If we look at our function from above:
{(3,1),(6,5),(9,1),(2,3)} This function is not a one-to-one function. For each y-value, we do not have one x-value. The y-value of 1 is associated with two different x-values (3 and 9). Let's look at an example.
Example 1: Determine if the function is one-to-one.
{(7,5),(3,8),(2,6),(-7,-1)}
Since each x-value corresponds to one y-value and each y-value corresponds to one x-value, we have a one-to-one function.

Inverse Functions

When we have a one-to-one function, we can create a function known as the inverse by interchanging x and y. Let's suppose we have a function F:
F = {(4,-1),(2,3),(7,8),(13,12)}
The inverse of F is found by interchanging the x and y values. Let's call this function G:
G = {(-1,4),(3,2),(8,7),(12,13)}
We can say that F and G are inverses of each other.
When we work with inverses, we have a special notation. The inverse of f is notated as f^-1. This is not f to the power of negative one. It is read as "f inverse" or "the inverse of f".
f^-1 consists of all ordered pairs (y,x) where (x,y) are the components of f. This basically means that all ordered pairs in the function f are reversed in the inverse. The y-values of the function f become the x-values in the inverse and the x-values in the function f become the y-values in the inverse. This means the domain of f^-1 is the range of f and the range of f^-1 is the domain of f. Let's look at an example.
Example 2: Find the inverse of the function, state the domain and range for the inverse.
F = {(9,7),(3,4),(-2,1),(-8,11)}
To find the inverse of F, we interchange the x and y values for each ordered pair.
F^-1 = {(7,9),(4,3),(1,-2),(11,-8)}
The domain for the inverse is the range from F. Similarly, the range for the inverse is the domain from F.
domain: {7,4,1,11}
range: {9,3,-2,-8}

Horizontal Line Test

When studying functions, we developed the vertical line test to determine if the graph of a relation is a function. For a relation to be a function, no vertical line should intersect the graph in more than one location. If it does, this means an x-value corresponds to more than one y-value, therefore, we do not have a function. Similarly, to determine if we have a one-to-one function, we can use a horizontal line test. The horizontal line test tells us if any horizontal line intersects the graph in more than one location, it is not the graph of a one-to-one function. Since a horizontal line always has the same y-coordinate, if it impacts the graph in more than one location, this tells us the same y-value is associated with two different x-values, therefore, the function is not a one-to-one function. Let's look at an example.
Example 3: Use the horizontal line test to determine if the function is one-to-one. If we look at our graph, we can see that it fails the horizontal line test. A horizontal line would impact the graph in more than one location, therefore, this is not the graph of a one-to-one function.

Finding the Equation of the Inverse of y = f(x)

For a one-to-one function defined by an equation y = f(x), we can find the inverse f^-1(x) using the following steps:

• Replace f(x) with y
• Interchange x and y in the equation
• Solve for y
• Replace y with f^-1(x)

Let's look at an example.
Example 4: Find the inverse of each function. $$f(x)=(x - 1)^3 + 2$$ Step 1) Replace f(x) with y. $$y=(x - 1)^3 + 2$$ Step 2) Interchange x and y in the equation. $$x=(y - 1)^3 + 2$$ Step 3) Solve for y. $$x - 2=(y - 1)^3$$ $$\sqrt[3]{x - 2}=y - 1$$ $$y=\sqrt[3]{x - 2}+ 1$$ Step 4) Replace y with f^-1(x) $$f^{-1}(x)=\sqrt[3]{x - 2}+ 1$$

How to Determine if Two Functions are Inverses

How can we determine if two functions are inverses of each other? We previously stated that the domain of a function becomes the range of its inverse and the range of a function becomes the domain of its inverse. Let's take a simple function such as: $$f(x)=2x$$ This function maps an x-value of 5 to a y-value of 10. $$f(5)=2(5)=10$$ If we find the inverse of the function, it will map an x-value of 10 to a y-value of 5. $$f^{-1}(x)=\frac{x}{2}$$ $$f^{-1}(10)=\frac{10}{2}=5$$ We can think about the inverse as reversing the procedure of the original function. In the original function, we multiply 2 by an unknown number (x). In the inverse, we take the unknown number (x) and divide by 2. We know that multiplying by 2 and dividing by 2 are opposite operations and will cancel each other out. What happens if we plug in f^-1(x) for x in the original function? $$f(f^{-1}(x))=2\left(\frac{x}{2}\right)$$ $$\require{cancel}f(f^{-1}(x))=\cancel{2}\left(\frac{x}{\cancel{2}}\right)=x$$ Based on our example, we can state the following rule:
For inverses f and f^-1: $$f(f^{-1}(x))=x$$ $$f^{-1}(f(x))=x$$ In other words, we can prove that two functions are inverses by plugging the inverse function in for x in the original function. If they are inverses, the results should just be x. We also need to check the other scenario. We will plug in the original function for x in the inverse function. Again, if they are inverses, the result should just be x.
In other words, if f(x) and g(x) are inverses:
f(g(x)) = x
g(f(x)) = x
Let's look at an example.
Example 5: Determine if the functions f and g are inverses. $$f(x)=\sqrt[5]{x + 1}$$ $$g(x)=x^5 - 1$$ To determine if these two functions are inverses, let's plug g(x) in for x in f(x). $$f(g(x))=\sqrt[5]{(x^5 - 1) + 1}$$ $$f(g(x))=\sqrt[5]{x^5}$$ $$f(g(x))=x$$ Now we will plug f(x) in for x in g(x). $$g(f(x))=(\sqrt[5]{x + 1})^5 - 1$$ $$g(f(x))=x + 1 - 1$$ $$g(f(x))=x$$ In each case, we get x as a result. We can state that these two functions are inverses.

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