How To Find The Inverse Of An Equation

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March ane, 2022

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Section one-ii : Inverse Functions

In the last example from the previous section we looked at the two functions $f\left( x \correct) = 3x - ii$ and $grand\left( ten \right) = \frac{10}{3} + \frac{2}{3}$ and saw that

\[\left( {f \circ g} \right)\left( x \right) = \left( {m \circ f} \right)\left( x \right) = x\]

and every bit noted in that department this means that there is a nice human relationship between these ii functions. Allow'south see just what that relationship is. Consider the following evaluations.

\[\crave{colour} \begin{align*}f\left( { \color{PineGreen}- 1} \correct) & = three\left( { - 1} \correct) - ii = {\color{Red}- 5} \hspace{0.5in} \Rightarrow \hspace{0.25in} & 1000\left( {\color{Scarlet} - 5} \right) & = \frac{{ - 5}}{three} + \frac{2}{3} = \frac{{ - iii}}{3} = {\colour{PineGreen}- 1}\\ & & & \\ g\left( {\color{PineGreen}ii} \right) & = \frac{2}{3} + \frac{2}{3} = {\color{Reddish}\frac{4}{3}}\hspace{0.5in} \Rightarrow \hspace{0.25in} & f\left( {\color{Red}\frac{4}{3}} \right) & = iii\left( {\frac{four}{3}} \right) - 2 = 4 - ii = {\color{PineGreen}ii}\cease{align*}\]

In the first instance we plugged $x = - 1$ into $f\left( x \right)$ and got a value of $-v$. Nosotros then turned around and plugged $x = - 5$ into $g\left( 10 \correct)$ and got a value of -1, the number that we started off with.

In the 2d case we did something like. Here we plugged $ten = ii$ into $k\left( x \right)$ and got a value of$\frac{iv}{3}$, we turned around and plugged this into $f\left( x \right)$ and got a value of 2, which is once again the number that we started with.

Note that we really are doing some function limerick here. The starting time case is really,

\[\left( {g \circ f} \right)\left( { - i} \right) = k\left[ {f\left( { - 1} \right)} \right] = yard\left[ { - five} \right] = - i\]

and the 2d case is really,

\[\left( {f \circ grand} \right)\left( 2 \correct) = f\left[ {m\left( two \right)} \right] = f\left[ {\frac{4}{3}} \right] = 2\]

Note too that these both concur with the formula for the compositions that we found in the previous section. We get back out of the function evaluation the number that we originally plugged into the composition.

So, just what is going on here? In some mode we can think of these two functions equally undoing what the other did to a number. In the first case nosotros plugged $x = - 1$ into $f\left( ten \right)$ and and then plugged the result from this role evaluation back into $g\left( 10 \right)$ and in some way $yard\left( x \right)$ undid what $f\left( ten \right)$ had done to $x = - 1$ and gave u.s.a. dorsum the original $10$ that we started with.

Function pairs that exhibit this behavior are chosen inverse functions. Before formally defining inverse functions and the notation that nosotros're going to employ for them nosotros need to get a definition out of the way.

A function is called ane-to-ane if no two values of $x$ produce the same $y$. Mathematically this is the same equally maxim,

\[f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right)\hspace{0.25in}\hspace{0.25in}{\rm{whenever}}\hspace{0.25in}\,\,\,\,\,\,{x_1} \ne {x_2}\]

Then, a function is one-to-one if whenever we plug different values into the function we get different function values.

Sometimes it is easier to empathise this definition if nosotros see a function that isn't one-to-1. Let's take a look at a function that isn't one-to-ane. The function $f\left( x \correct) = {10^2}$ is not one-to-i considering both $f\left( { - 2} \right) = iv$ and $f\left( ii \correct) = 4$. In other words, there are two different values of $x$ that produce the same value of $y$. Note that we can turn $f\left( x \right) = {x^two}$ into a one-to-1 office if we restrict ourselves to $0 \le ten < \infty $. This can sometimes be done with functions.

Showing that a function is one-to-one is often boring and/or difficult. For the almost part we are going to presume that the functions that we're going to be dealing with in this form are either one-to-i or nosotros accept restricted the domain of the function to go information technology to be a i-to-one function.

At present, let's formally define just what changed functions are. Given two ane-to-one functions $f\left( x \correct)$ and $thou\left( x \right)$ if

\[\left( {f \circ g} \right)\left( 10 \correct) = ten\hspace{0.25in}\hspace{0.25in}{\rm{AND}}\hspace{0.25in}\hspace{0.25in}\left( {chiliad \circ f} \right)\left( x \correct) = ten\]

then we say that $f\left( x \right)$ and $one thousand\left( 10 \right)$ are inverses of each other. More than specifically we will say that $g\left( x \right)$ is the inverse of $f\left( x \correct)$ and denote it by

\[g\left( ten \correct) = {f^{ - i}}\left( x \right)\]

Likewise, nosotros could too say that $f\left( x \right)$ is the changed of $g\left( ten \right)$ and denote it by

\[f\left( x \right) = {m^{ - ane}}\left( 10 \right)\]

The notation that we use really depends upon the problem. In about cases either is acceptable.

For the two functions that nosotros started off this department with nosotros could write either of the post-obit two sets of notation.

\[\begin{align*}f\left( x \right) & = 3x - 2\hspace{0.25in}\hspace{0.25in} & {f^{ - i}}\left( x \right) & = \frac{x}{3} + \frac{2}{three}\\ & & & \\ yard\left( x \right) & = \frac{ten}{3} + \frac{2}{3}\hspace{0.25in}\hspace{0.25in}& {1000^{ - 1}}\left( x \correct) & = 3x - 2\end{align*}\]

Now, exist careful with the notation for inverses. The "-i" is Non an exponent despite the fact that it sure does wait like ane! When dealing with changed functions we've got to recall that

\[{f^{ - 1}}\left( x \right) \ne \frac{1}{{f\left( x \right)}}\]

This is one of the more common mistakes that students brand when offset studying inverse functions.

The procedure for finding the inverse of a part is a fairly elementary 1 although there are a couple of steps that tin can on occasion be somewhat messy. Here is the process

Finding the Inverse of a Function

Given the role $f\left( x \correct)$ we want to find the inverse office, ${f^{ - one}}\left( x \right)$.

First, replace $f\left( x \correct)$ with $y$. This is done to make the rest of the process easier.
Supplant every $x$ with a $y$ and supervene upon every $y$ with an $x$.
Solve the equation from Step 2 for $y$. This is the stride where mistakes are most frequently made then exist careful with this footstep.
Replace $y$ with ${f^{ - i}}\left( x \correct)$. In other words, we've managed to observe the inverse at this point!
Verify your work by checking that \[\left( {f \circ {f^{ - 1}}} \right)\left( ten \right) = x\] and \[\left( {{f^{ - 1}} \circ f} \correct)\left( x \right) = x\] are both true. This work tin sometimes be messy making it easy to make mistakes and then again be careful.

That'southward the process. Most of the steps are not all that bad only as mentioned in the process in that location are a couple of steps that nosotros really need to exist careful with since it is easy to make mistakes in those steps.

In the verification step we technically really do need to check that both $\left( {f \circ {f^{ - 1}}} \correct)\left( x \right) = 10$ and $\left( {{f^{ - ane}} \circ f} \right)\left( x \right) = x$ are true. For all the functions that we are going to exist looking at in this course if one is true then the other will as well exist truthful. All the same, at that place are functions (they are beyond the scope of this course nevertheless) for which it is possible for only one of these to be truthful. This is brought up because in all the issues here we will be just checking one of them. We just need to ever retrieve that technically nosotros should bank check both.

Let's work some examples.

Example i Given $f\left( 10 \right) = 3x - 2$ find ${f^{ - 1}}\left( x \right)$.

Evidence Solution

Now, we already know what the changed to this function is as we've already washed some work with information technology. Nonetheless, it would be nice to really get-go with this since we know what we should go. This volition work as a nice verification of the process.

So, let'due south become started. Nosotros'll showtime replace $f\left( x \right)$ with $y$.

\[y = 3x - 2\]

Next, supplant all $ten$'s with $y$and all $y$'southward with $x$.

\[x = 3y - 2\]

At present, solve for $y$.

\[\begin{marshal*}x + two & = 3y\\ \frac{1}{3}\left( {x + two} \correct) & = y\\ \frac{x}{3} + \frac{two}{3} & = y\end{align*}\]

Finally replace $y$ with ${f^{ - 1}}\left( 10 \right)$.

\[{f^{ - ane}}\left( x \correct) = \frac{x}{iii} + \frac{2}{three}\]

Now, nosotros need to verify the results. We already took care of this in the previous section, however, we really should follow the process so we'll practice that here. It doesn't thing which of the two that nosotros bank check we merely need to check one of them. This time nosotros'll bank check that $\left( {f \circ {f^{ - 1}}} \right)\left( 10 \right) = x$ is true.

\[\begin{align*}\left( {f \circ {f^{ - 1}}} \right)\left( x \right) & = f\left[ {{f^{ - one}}\left( ten \correct)} \right]\\ & = f\left[ {\frac{x}{3} + \frac{ii}{iii}} \right]\\ & = 3\left( {\frac{10}{3} + \frac{2}{3}} \right) - 2\\ & = x + two - 2\\ & = x\end{align*}\]

Case two Given $g\left( x \right) = \sqrt {x - 3} $ find ${k^{ - one}}\left( x \correct)$.

Testify Solution

The fact that nosotros're using $g\left( ten \correct)$ instead of $f\left( x \correct)$ doesn't change how the process works. Here are the first few steps.

\[y = \sqrt {x - 3} \hspace{0.25in}\,\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,\,\,\,\,\,x = \sqrt {y - iii} \]

At present, to solve for $y$we will need to starting time square both sides and then proceed equally normal.

\[\begin{align*}x & = \sqrt {y - 3} \\ {x^ii} & = y - 3\\ {x^2} + 3 & = y\end{marshal*}\]

This inverse is then,

\[{thousand^{ - 1}}\left( x \right) = {x^2} + 3\]

Finally permit'south verify and this time we'll use the other ane just then we can say that we've gotten both down somewhere in an example.

\[\begin{align*}\left( {{g^{ - i}} \circ g} \right)\left( x \right) & = {g^{ - i}}\left[ {g\left( ten \right)} \right]\\ & = {g^{ - 1}}\left( {\sqrt {x - 3} } \right)\\ & = {\left( {\sqrt {x - 3} } \right)^2} + 3\\ & = x - iii + iii\\ & = 10\end{align*}\]

So, nosotros did the work correctly and we do indeed take the inverse.

The next example can exist a fiddling messy then exist careful with the work hither.

Example 3 Given $\displaystyle h\left( ten \right) = \frac{{ten + 4}}{{2x - 5}}$ discover ${h^{ - one}}\left( x \right)$.

Show Solution

The first couple of steps are pretty much the same equally the previous examples so here they are,

\[y = \frac{{x + iv}}{{2x - five}}\hspace{0.25in}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,ten = \frac{{y + 4}}{{2y - 5}}\]

Now, exist careful with the solution footstep. With this kind of problem information technology is very easy to make a mistake here.

\[\brainstorm{align*}x\left( {2y - 5} \right) & = y + 4\\ 2xy - 5x & = y + four\\ 2xy - y & = 4 + 5x\\ \left( {2x - ane} \right)y & = 4 + 5x\\ y & = \frac{{4 + 5x}}{{2x - 1}}\end{align*}\]

So, if we've done all of our work correctly the inverse should be,

\[{h^{ - 1}}\left( x \correct) = \frac{{4 + 5x}}{{2x - 1}}\]

Finally, nosotros'll need to do the verification. This is besides a fairly messy process and information technology doesn't actually affair which ane nosotros piece of work with.

\[\brainstorm{align*}\left( {h \circ {h^{ - one}}} \right)\left( x \correct) & = h\left[ {{h^{ - one}}\left( x \right)} \right]\\ & = h\left[ {\frac{{4 + 5x}}{{2x - 1}}} \right]\\ & = \frac{{\frac{{4 + 5x}}{{2x - 1}} + four}}{{ii\left( {\frac{{iv + 5x}}{{2x - 1}}} \right) - 5}}\end{marshal*}\]

Okay, this is a mess. Allow's simplify things up a little fleck by multiplying the numerator and denominator by $2x - 1$.

\[\begin{marshal*}\left( {h \circ {h^{ - 1}}} \right)\left( 10 \right) & = \frac{{2x - i}}{{2x - 1}}\,\,\frac{{\frac{{4 + 5x}}{{2x - 1}} + 4}}{{2\left( {\frac{{4 + 5x}}{{2x - 1}}} \correct) - v}}\\ & = \frac{{\left( {2x - 1} \correct)\left( {\frac{{4 + 5x}}{{2x - 1}} + four} \right)}}{{\left( {2x - 1} \correct)\left( {two\left( {\frac{{4 + 5x}}{{2x - i}}} \right) - 5} \correct)}}\\ & = \frac{{4 + 5x + 4\left( {2x - 1} \right)}}{{ii\left( {4 + 5x} \right) - 5\left( {2x - 1} \right)}}\\ & = \frac{{4 + 5x + 8x - 4}}{{8 + 10x - 10x + 5}}\\ & = \frac{{13x}}{{13}} = 10\end{align*}\]

Wow. That was a lot of work, but it all worked out in the terminate. We did all of our work correctly and we do in fact accept the inverse.

In that location is one final topic that nosotros need to address quickly before we go out this department. There is an interesting relationship betwixt the graph of a part and the graph of its inverse.

Here is the graph of the part and inverse from the first ii examples.

$Two graphs. The graph on the left is for Example 1 and the graph on the right is for Example 2. Each graph has a dashed line representing $y=x$ as well as the graph of $f\left(x\right)$ and ${{f}^{-1}}\left( x \right)$. The graph of the function and the inverse are reflections of each other about the line $y=x$. In the Example 1 graph $f\left(x\right)$ is a line that starts at approximately (0.6,0) and ends at approximately (2.5,6) and ${{f}^{-1}}\left( x \right)$ is a line that starts at approximately (0,0.6) and ends at approximately (4,2). In the Example s graph $f\left(x\right)$ starts at (3,0) and curves upwards to the right and ends at approximately (6, 1.7) and ${{f}^{-1}}\left( x \right)$ starts at (0,3) and curves upwards and ends at approximately (2.23, 8).$

In both cases we can see that the graph of the inverse is a reflection of the actual function well-nigh the line $y = 10$. This will e'er exist the case with the graphs of a function and its inverse.