Inverse Functions
Inverses are functions that reverse or undo the operations of their base functions. They can be expressed as:
The input of the base function will result in the output of the inverse function and vice versa.
When you switch the input and output values of a function, they end up undoing each other. For example, let's say that we have the given function:
This would mean that its inverse would be:
Let's see what happens when we take the inverse of the function output:
\(= f^{-1} (f(4))\)
\(= f^{-1}(5)\)
\(=4\)
We can see that taking the inverse of the function results in the original input (\(\boldsymbol{4}\)).
Steps for Calculating Inverses
When solving for \(f^{-1}(x)\):
- Write the original function as \(y=f(x)\)
- Swap \(x\) with \(y\) and vice versa
- Isolate \(y\) by reversing the operations
- Substitute \(y\) for \(f⁻¹(x)\)
NOTE: Reversing operations of the original function works in reverse order of BEDMAS (ie addition and subtraction come before multiplication and division).
Example
Determine the inverse of \(f(x)=2x+1\).
First, we can rewrite the function as such:
\(y=2x+1\)
Next, we can switch \(x\) and \(y\):
\(x=2y+1\)
Now, we can find the inverse by isolating \(y\):
\(2y = x - 1\)
\(\cfrac{\cancel{2}y}{\cancel{2}} = \cfrac{x - 1}{2}\)
\(y = \cfrac{x-1}{2}\)
Finally, we can change \(y\) to \(f^{-1}(x):\)
\(f^{-1}(x)= \cfrac{x-1}{2}\)
Therefore, we can determine that \(\boldsymbol{f(x)^{-1} = \cfrac{x-1}{2}}\).
\(f(x)=5x^2\)
First, we can rewrite the function as:
\(y=5x^2\)
Next, we can switch \(x\) and \(y\):
\(x=5y^2\)
Now, we can find the inverse by isolating \(y\):
\(\cfrac{x}{5} = \cfrac{\cancel{5}y^2}{\cancel{5}}\)
\(\sqrt{\cfrac{x}{5}} = \sqrt{y^2}\)
\(y = \sqrt{\cfrac{x}{5}}\)
Finally, we can change \(y\) to \(f^{-1}(x):\)
\(f^{-1}(x) = \sqrt{\cfrac{x}{5}}\)
Therefore, we can determine that \(\boldsymbol{f^{-1}(x) = \sqrt{\cfrac{x}{5}}}\).
\(g(x)= \cfrac{x+5}{3}\)
First, we can rewrite the function as:
\(y=\cfrac{x+5}{3}\)
Next, we can switch \(x\) and \(y\):
\(x=\cfrac{y+5}{3}\)
Now, we can find the inverse by isolating \(y\):
\(3(x)= \cancel{3}\left(\cfrac{y+5}{\cancel{3}}\right)\)
\(3x=y+5\)
\(y = 3x-5\)
Finally, we can change \(y\) to \(g^{-1}(x):\)
\(g^{-1}(x)=3x-5\)
Therefore, we can determine that \(\boldsymbol{g^{-1}(x) = 3x-5}\).
\(h(x)=4(17-6x)^2-3\)
First, we can rewrite the function as:
\(y=4(17-6x)^2 - 3\)
Next, we can switch \(x\) and \(y\):
\(x=4(17-6y)^2 - 3\)
Now, we can find the inverse by isolating \(y\):
\(x+3=4(17-6y)^2\)
\(\cfrac{x+3}{4} = \cfrac{\cancel{4}(17+6x)^2}{\cancel{4}}\)
\(\sqrt{\cfrac{x+3}{4}} = \sqrt{(17+6x)^2}\)
\(\sqrt{\cfrac{x+3}{4}} -17 = 6x\)
\(\cfrac{\sqrt{\cfrac{x+3}{4}}-17}{6} = \cfrac{\cancel{6}x}{\cancel{6}}\)
\(y= \cfrac{\sqrt{\cfrac{x+3}{4}}-17}{6}\)
Finally, we can change \(y\) to \(h^{-1}(x):\)
\(h^{-1}(x)= \cfrac{\sqrt{\cfrac{x+3}{4}}-17}{6}\)
Therefore, we can determine that \(\boldsymbol{h^{-1}(x) = \cfrac{\sqrt{\cfrac{x+3}{4}}-17}{6}}\).
Graphing Inverses
When graphing an inverse, even if you haven't calculated the actual function, reverse the \(x\) and \(y\)-coordinates of each point. Inverse functions are reflected along the line \(y = x\). In addition:
- The domain of the original function is the range of the inverse function
- The range of the original function is the domain of the inverse function
The graph below shows the function \(2x+1\) displayed against its inverse, \(\cfrac{x-1}{2}\):
The black dotted line shows how these two functions are reflected about the line \(y = x\).
You can identify if an inverse is a function if you use the Horizontal Line Test on the the graph of the original function. If the horizontal line touches the graph more than once, it proves that the inverse isn't a function.
Another way of determining if the inverse is a function is by performing the Vertical Line Test on the inverse graph. As with the HLT, if it touches the graph only once, then the inverse is considered a function.
Example
For the following graph:
- Determine the graph, domain, and range for its inverse
- Identify if its inverse is a function
i. We can create the inverse graph by first creating a table of values:
| x Values | -5 | 0 | 3 | 4 | 3 | 0 | -5 |
|---|---|---|---|---|---|---|---|
| y Values | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
Using this table of values, we can draw our inverse graph by inverting the \(x\) and \(y\) -values:
From the inverse equation, we can identify the Domain as \(\boldsymbol{\{x\in\mathbb{R} | x \le 4\}}\) and the Range as \(\boldsymbol{\{y\in\mathbb{R}\}}\).
ii. We can determine whether the inverse equation is a function by performing the horizontal line test on the original function:
As you can observe, the inverse is NOT a function since it passes through more than one point on the graph.
In order to make the inverse of a function pass the Horizontal Line Test, we can limit its domain. Let's test this on the function, \(x^2\):
In this instance, we're limiting its domain to \(\{x\in\mathbb{R} | 0 \le x\}\). When we perform the Horizontal Line Test, it will only go through one point at a time:
We can graph the inverse of this function to perform the Vertical Line Test. Since the line only goes through one point at a time, we can confirm that the inverse is a function as well:
Given the following graph:
- Determine the graph, domain, and range for its inverse
- Identify if its inverse is a function
i. We can create the inverse graph by first creating a table of values:
| x Values | 1 | 0 | -1 | -2 | -1 | 0 | 1 |
|---|---|---|---|---|---|---|---|
| y Values | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
Using this table of values, we can draw our inverse graph by inverting the \(x\) and \(y\)-values:
From this inverse graph, we can determine the Domain as \(\boldsymbol{\{x \in \mathbb{R}\}}\) and the Range as \(\boldsymbol{\{y \in \mathbb{R} | y \geq -2\}}\).
ii. We can determine that the inverse equation is a function by performing the horizontal line test on the original function:
As you can observe, the inverse is a function since the horizontal line will pass through only one point at any location.
Domain and Range
As stated above, when solving for an inverse equation, the original Domain becomes the new Range and vice versa. In several cases, finding the inverse equation helps us find either set.
Using inverses, we can find the Domain and Range of \(g(x) = \cfrac{-2x}{6-3x} + 4\).
First, we can rewrite the function as such:
\(y= \cfrac{-2x}{6-3x} + 4\)
Next, we can switch \(x\) and \(y\):
\(x = \cfrac{-2y}{6-3y} + 4\)
Then, we can determine the inverse by isolating \(y\):
\(\cfrac{-2y}{6-3y} = x-4\)
\((\cancel{6-3x})\left(\cfrac{-2y}{\cancel{6-3x}}\right) = (6-3y)(x-4)\)
\(-2y = 6x-24-3xy+12y\)
\(-2y-12y+3xy = 6x-24\)
\(-14y+3xy = 6x+24\)
\(y(-14+3x) = 6x+24\)
\(\cfrac{y(\cancel{-14+3x})}{\cancel{-14+3x}} = \cfrac{6x+24}{-14+3x}\)
\(y = \cfrac{6x+24}{-14+3x}\)
\(f^{-1}(x)=\cfrac{6x+24}{-14+3x}\)
We can evaluate the denominator of the original function to determine any potential limits to the Domain:
\(6-3x = 0\)
\(6 = 3x\)
\(\cfrac{6}{3} = \cfrac{3x}{3}\)
\(x = 2\)
We can evaluate the denominator of the inverse function to determine any potential limits to the Range:
\(-14+3y = 0\)
\(3y = 14\)
\(\cfrac{\cancel{3}y}{\cancel{3}} = \cfrac{-14}{3}\)
\(y = \cfrac{-14}{3}\)
From the original function, we are able to determine that \(\mathbf{D} = \boldysymbol{\{x\in\mathbb{R} | x \ne 2\}}\). From the inverse function, we are able to determine that \(\mathbf{R} = \boldsymbol{\{y\in\mathbb{R} | y \ne \cfrac{-14}{3}\}}\).
First, we can rewrite the function as such:
\(y = \sqrt{16-0.5x} +2\)
Next, we can switch \(x\) and \(y\):
\(x = \sqrt{16-0.5y} +2\)
Then, we can find the inverse by isolating \(y\):
\(\sqrt{16-0.5y} = x-2\)
\((\sqrt{16-0.5y})^2 = (x-2)^2\)
\(-0.5y = (x-2)^2 - 16\)
\(\cfrac{\cancel{-0.5}y}{\cancel{-0.5}} = \cfrac{(x-2)^2-16}{-0.5}\)
\(y = \cfrac{(x-2)^2-16}{-0.5}\)
\(f^{-1}(x) = \cfrac{(x-2)^2-16}{-0.5}\)
After, we can evaluate the original function to determine any potential limits to the Domain:
\(\sqrt{16-0.5x}\)
First, we can evaluate the original function at \(x = 32\):
\(= \sqrt{16-0.5(32)}\)
\(= \sqrt{16-16}\)
\(\sqrt{0} = 0\)
Next, we can evaluate the original function at \(x=34\):
\(= \sqrt{16-0.5(34)}\)
\(= \sqrt{16-17}\)
\(\sqrt{-1} = \text{ERROR!!}\)
Finally, we can evaluate the quadratic in the numerator of the inverse function to determine any potential limits to the Range:
\(= (y-2)^2\)
\(= (2-2)^2\)
\((0)^2 = 0\)
From the original function, we are able to determine that \(\mathbf{D} = \boldsymbol{\{x\in\mathbb{R} | x \le 32\}}\). From the inverse function, we are able to determine that \(\mathbf{R} = \boldsymbol{\{y\in\mathbb{R} | 0 \le y\}}\).
Enter in coefficients for the quadratic expression or click on the button to generate new values. This tool will calculate the inverse of the quadratic expression.