Inverse of a Function

How to Verify if Functions are Inverses?  - How it Works - Video

Guidelines for Finding the Inverse 

Guidelines for Finding the Inverse:

1 - Verify the function is one-to-one. Look at the domain. Is it increasing/decreasing throughout? Does it pass the horizontal line test?

2 - If yes. Then solve x in terms in y. 

3 - Verify if f(f-1(x)) and f-1(f(x)) = x. 

Looking for more inverse functions facts.

Example 1

Example 1:

Here we have our function, f(x) = 2x - 3.

Our first step - Guideline 1 - is determine if our function is one-to-one. There are multiple ways of doing that including determining the domain and range, where is it increasing/decreasing?, or its graph. 

We used the horizontal line test to figure out if the function is one-to-one. Since it only touches the graph one time, the inverse of the functions is also a function. Let's solve for x in terms of y - Guideline 2.

We have our inverse function, f-1(x) = (x + 3)/2. Notice that we don't have f-1(y). After solving for x in terms of y, we change the y's to x's. The reason for that is x is standard input value.

Here we have our function, f(x) = 2x - 3, and our inverse function, f-1(x) = (x + 3)/2.

To check if we did it correctly, we need to substitute f(x) into f-1(x) and vice versa - Guideline 3.

Now since both f(f-1(x)) and f-1(f(x)) = x. We know that we did it correctly.

Now, let's use that graph again. We graph the inverse function, f-1(x) = (x + 3)/2. We have also added y = x as a dashed line. Each corresponding point is the same distance from the dashed line.

Let's take a closer look at the point (4, 5) in the original function. It becomes (5, 4). This goes off conditions 1 and 2 on the theorem on inverse functions. So we can flip flop x and y points in our ordered pairs to graph the inverse function.

Mathematically that would be the point (x, y) is on the graph of f if and only if (y, x) is on the graph of f-1.

Example 2

Example 2:

Here we have our function, f(x) = 3x2 - 5. Notice we also have restricted our domain to x ≥ 0. If we do not do that, then our parabolic equation would not pass the horizontal test, which would mean that the inverse of f(x) = 3x2 - 5 would not be a function. That is why we restricted it to just the nonnegative numbers.

Our first step - Guideline 1 - is determine if our function is one-to-one. There are multiple ways of doing that including determining the domain and range, where is it increasing/decreasing?, or its graph. 

We used the horizontal line test to figure out if the function is one-to-one. Since it only touches the graph one time, the inverse of the functions is also a function. Let's solve for x in terms of y - Guideline 2.

We have our inverse function, f-1(x) = sqrt[ (x + 5) / 3 ]. Notice that we don't have f-1(y). After solving for x in terms of y, we change the y's to x's. The reason for that is x is standard input value.

Here we have our function, f(x) =3x2 - 5, and our inverse function, f-1(x) = sqrt[ (x + 5) / 3 ].

To check if we did it correctly, we need to substitute f(x) into f-1(x) and vice versa - Guideline 3.

Now since both f(f-1(x)) and f-1(f(x)) = x. We know that we did it correctly.

Now, let's use that graph again. We graph the inverse function, f-1(x) = sqrt[ (x + 5) / 3 ]. We have also added y = x as a dashed line. Each corresponding point is the same distance from the dashed line.

Let's take a closer look at the point (0, -5) in the original function. It becomes (-5, 0). This goes off conditions 1 and 2 on the theorem on inverse functions. So we can flip flop x and y points in our ordered pairs to graph the inverse function.

Mathematically that would be the point (x, y) is on the graph of f if and only if (y, x) is on the graph of f-1.

Example 3

Example 3:

Here we have our function, f(x) = (2x + 3) / (3x - 2). 

Our first step - Guideline 1 - is determine if our function is one-to-one. There are multiple ways of doing that including determining the domain and range, where is it increasing/decreasing?, or its graph. 

We used the horizontal line test to figure out if the function is one-to-one. Since it only touches the graph one time, the inverse of the functions is also a function. Let's solve for x in terms of y - Guideline 2.

We have our inverse function, f-1(x) = (2x + 3) / (3x -2). Notice that we don't have f-1(y). After solving for x in terms of y, we change the y's to x's. The reason for that is x is standard input value. 

Woh! We have the inverse is the same as our original function. Sometimes quirks like that happens in math.

Here we have our function, f(x) = (2x + 3) / (3x - 2), and our inverse function, f-1(x) = (2x + 3) / (3x - 2). 

To check if we did it correctly, we need to substitute f(x) into f-1(x) and vice versa - Guideline 3.

1st = sign - Substituted f(x) into  f-1(f(x)).

2nd = sign - Distributed the coefficients 2 and 3. Created same denominator to combine terms.

3rd = sign - Distributed the coefficients 3 and -2.

4th = sign - Combined fractions.

5th = sign - Canceled the denominators of each fraction. Rearranged the terms.

6th = sign - Combined like terms.

7th = sign - Divided by 13.

Now since both f(f-1(x)) and f-1(f(x)) = x. We know that we did it correctly.

Now, let's use that graph again. We graph the inverse function, f-1(x) = (2x + 3) / (3x - 2). We have also added y = x as a dashed line. Each corresponding point is the same distance from the dashed line. Remember the original function is underneath the inverse function since they are the same.

Let's take a closer look at the point (-2, 1/8) in the original function. It becomes (1/8, -2). This goes off conditions 1 and 2 on the theorem on inverse functions. So we can flip flop x and y points in our ordered pairs to graph the inverse function.

Mathematically that would be the point (x, y) is on the graph of f if and only if (y, x) is on the graph of f-1.

Live Worksheet

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