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During our study of pre-calculus and related subjects, we may be asked to find the inverse of a function. Finding the inverse of a function is an important procedure to learn since it's a foundational topic for more advanced mathematical subjects like calculus and real analysis. It also has many practical applications in other fields such as physics. Before we get started, it's wise to read up on related concepts like factors, multiplicative inverse, and inverse operations. A thorough understanding of these topics is valuable if we want to understand inverses.

When we take the inverse of a function, the notation is ${f}^{-1}\left(x\right)$ , which we refer to as "f inverse." Bear in mind that the -1 in ${f}^{-1}\left(x\right)$ doesn't indicate an exponent. Rather, it's just the notation we use to signify that we're taking the inverse of a function. Note that not all functions can be inverted. We can only take the inverse of a function when it's a one-to-one function. This means that each domain value must match up with a unique range value. In other words, we cannot have a recurring y-value in our range.

Therefore, before trying to find the inverse of a function, we must first determine whether the function is invertible in the first place. We can do this using the horizontal line test. If we can draw a horizontal line through any point on the graph and have it hit more than one point on the function, then it's not invertible.

If we have two functions that are inverses of each other, each ordered pair will be swapped. If we have a point (x, y) on a function, the corresponding ordered pair on its inverse will be (y, x). These two graphs will be mirror images of each other over the line $y=x$ . In order to find the inverse of a function, we simply swap out the x and y-variables. Let's say we have the function $f\left(x\right)=2x-1$ . This is a one-to-one function since it passes the horizontal line test on a graph.

In order to calculate ${f}^{-1}\left(x\right)$ , we need to switch the x and the y values like so:

$y=2x-1$

$x=2y-1$

Now we must solve for y to get the inverse. First, we'll add 1 to both sides.

$x+1=2y$

Now we'll isolate y by dividing both sides by 2.

$y=\frac{x+1}{2}$

This is our answer. ${f}^{-1}\left(x\right)=\frac{x+1}{2}$

If we graph these two functions, we'll see that they are indeed mirror images with respect to $y=x$ . Whenever solving these problems, it's always wise to plot each function out on graph paper. Not only does this provide an indication of whether a function is invertible, but it also gets us into the habit of thinking graphically.

Let's do another example. Suppose that we have a function defined as $g\left(x\right)={x}^{2}-2$ find ${g}^{-1}\left(x\right)$

If we plot this function on a graph, we'll see that it's a parabola, which means it fails the horizontal line test. Therefore, there is no inverse function without restricting the domain or range.

Let's try one more before moving on to some practice problems.

If $f\left(x\right)={x}^{3}$ find ${f}^{-1}\left(x\right)$

$y={x}^{3}$

$x={y}^{3}$

We take the cubed root of each side to get

$y=\sqrt[3]{x}$

Therefore ${f}^{-1}\left(x\right)=\sqrt[3]{x}$

1. Consider the function $f\left(x\right)=\frac{x+3}{5}$ . Is this function invertible? If so, find its inverse.

If we plot this function on a graph, we will see that it passes the horizontal line test. Since it's a one-to-one function, we can find an inverse function.

First, we'll switch out the x and y-variables.

$y=\frac{x+3}{5}$

$x=\frac{y+3}{5}$

First, we need to multiply each side by 5.

$5x=y+3$

Next, we'll subtract 3 from both sides.

$y=5x-3$

Therefore, ${f}^{-1}\left(x\right)=5x-3$ .

2. If we have function $g\left(x\right)=\frac{{x}^{2}}{2}$ , can we invert this function? If so, find the inverse.

Since this function has x-values that pair to multiply y-values, it's not a one-to-one function. If we were to graph it, we would see that it's a parabola, which means that it fails the horizontal line test. Therefore, this function has no inverse.

3. Let's say we have a function defined by $h\left(x\right)=\frac{3}{2}x+3$ . Can we invert this function? If so, find the inverse.

This is a one-to-one function. If we plot it on a graph, we see that it's a straight line, which means that it passes the horizontal line test. In other words, each x-value pairs with a distinct y-value. Therefore, we can invert it.

$y=\frac{3}{2}x+3$

$x=\frac{3}{2}y+3$

First, let's subtract both sides by 3.

$x-3=\frac{3}{2}y$

$y=\frac{x-3}{\frac{3}{2}}$

$y=\frac{2}{3}x-2$

Therefore, ${h}^{-1}\left(x\right)=\frac{2}{3}x-2$

4. Given the function $g\left(x\right)=\frac{3}{4}x$ , find the inverse if one exists.

This is a line on the graph, so it passes the horizontal line test and can be inverted.

$y=\frac{3}{4}x$

$x=\frac{3}{4}y$

We simply need to divide both sides by $\frac{3}{4}$ to get our answer.

$y=\frac{x}{\left(\frac{3}{4}\right)}$ $y=\frac{x}{\left(\frac{3}{4}\right)}$

$y=\frac{4}{3}x$

Therefore, ${g}^{-1}\left(x\right)=\frac{4}{3}x$

5. Given the function $f\left(x\right)=\frac{3x+1}{4}$ , find the inverse if one exists.

When we look at a graph of this function, it passes the horizontal line test.

First, we'll switch each variable.

$y=\frac{3x+1}{4}$

$x=\frac{3y+1}{4}$

We'll multiply both sides by 4.

$4x=3y+1$

Now we'll subtract 1 from both sides.

$3y=4x-1$

Now we divide both sides by 3.

${f}^{-1}\left(x\right)=\frac{4x-1}{3}$

6. Given the function $g\left(x\right)=\pi x$ , find the inverse if one exists.

This passes the horizontal line test on a graph, so it does have an inverse. We'll switch the two variables first.

$y=\pi x$

$x=\pi y$

Now we simply divide both sides by π.

$y=\frac{x}{\pi}$

${g}^{-1}\left(x\right)=\frac{x}{\pi}$

7. Given the function $h\left(x\right)=-3x-\frac{1}{3}$ , find the inverse if one exists.

This passes the horizontal line test, so an inverse exists.

$y=-3x-\frac{1}{3}$

$x=-3y-\frac{1}{3}$

We'll first add 1/3 to both sides.

$-3y=x+\frac{1}{3}$

Now we'll divide everything by -3.

$y=-\frac{1}{3}x-\frac{1}{9}$

If you're studying inverse functions and find that you're lost, it's a good idea to work with a tutor who will guide you through the topic. A tutor can create practice problems and help you if you become stuck. They can also design a learning plan tailored to your academic needs and periodically update it to reflect your progress. Furthermore, if you have a high school student who is struggling with algebra or pre-calculus material, a tutor will assist them as well. When you contact an Educational Director at Varsity Tutors, they will ask you questions about your situation and find a tutor appropriate for your needs. Learning math can be a struggle at times, which is why it's useful to study with the help of a tutor. Contact Varsity Tutors today to learn more.

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