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Section 6.2 Finding a Formula for the Inverse

In the last section, we learned what an inverse function is and how to evaluate it. Sometimes, however, we want to be able to evaluate the inverse function at a lot of different values. In that case, it would be helpful to have a formula for the inverse function. That is what we are going to learn how to build in this section.

Remember: the key to inverse functions is that they swap inputs and outputs. So, our steps for finding the formula for the inverse function will require some swapping:

Example 6.8.

Suppose \(f(x)=3-7x\) and we want to find a formula for \(f^{-1}(x)\text{.}\) Let's follow the steps above:

  1. We begin by writing \(y=3-7x\text{.}\)

  2. Now we swap \(x\) and \(y\) to get \(x = 3-7y\text{.}\)

  3. Finally, we solve for \(y\text{:}\)

    \begin{align*} x \amp = 3-7y\\ x-3 \amp = -7y\\ \frac{x-3}{-7} \amp = y \end{align*}

Therefore, our answer is \(f^{-1}(x)=\frac{x-3}{-7}\text{.}\)

Checkpoint 6.9.

Suppose \(g(x)=x^5+4\text{.}\) Find a formula for \(g^{-1}(x)\text{.}\)

Answer.

\(g^{-1}(x)=\sqrt[5]{x-4}\)

Solution.

To find a formula for the inverse, we start by writing \(y=x^5+4\text{,}\) then swap \(x\) and \(y\) to get \(x=y^5+4\text{.}\) All that is left is to solve for \(y\text{:}\)

\begin{align*} x \amp = y^5+4\\ x-4 \amp = y^5\\ \sqrt[5]{x-4} \amp = y \end{align*}

Therefore, our final answer is \(g^{-1}(x)=\sqrt[5]{x-4}\text{.}\)

Checkpoint 6.10.

Suppose \(r(x)=\dfrac{3x-1}{2-x}\text{.}\) Find a formula for \(r^{-1}(x)\text{.}\)

Answer.

\(r^{-1}(x)=\dfrac{2x+1}{3+x}\)

Solution.

To find a formula for the inverse, we start by writing \(y=\dfrac{3x-1}{2-x}\text{,}\) then swap \(x\) and \(y\) to get \(x=\dfrac{3y-1}{2-y}\text{.}\) All that is left is to solve for \(y\text{.}\) There are a lot of moving parts to solving this equation, so I will highlight the variable \(y\) in red so that it is easier to keep track of:

\begin{align} x \amp = \dfrac{3{\color{red}{y}}-1}{2-{\color{red}{y}}}\label{eqtna}\tag{1}\\ x(2-{\color{red}{y}})\amp = 3{\color{red}{y}}-1\label{eqtnb}\tag{2}\\ 2x-x{\color{red}{y}} \amp = 3{\color{red}{y}}-1\label{eqtnc}\tag{3}\\ 2x \amp = 3{\color{red}{y}}+x{\color{red}{y}}-1\label{eqtnd}\tag{4}\\ 2x+1 \amp = 3{\color{red}{y}}+x{\color{red}{y}}\label{eqtne}\tag{5}\\ 2x+1 \amp = {\color{red}{y}}(3+x)\label{eqtnf}\tag{6}\\ \dfrac{2x+1}{3+x} \amp = {\color{red}{y}}\label{eqtng}\tag{7} \end{align}

Therefore, our final answer is \(r^{-1}(x)=\dfrac{2x+1}{3+x}\text{.}\)

The algebra on this one can be hard to follow, so we'll take a moment here to walk through the steps of what happened:

  1. In equation (1), we have two problems we need to deal with in order to solve for \(y\text{:}\) we have two \(y\)'s in our equation, and one of them is in the denominator of a fraction. We need to deal with the second problem first.

  2. To get from equation (1) to (2), we multiplied both sides by \(2-y\text{,}\) which canceled the denominator on the right-hand side.

  3. To get from equation (2) to (3), we just distributed the \(x\text{.}\) The reason we wanted to do this is that we are trying to solve for \(y\text{,}\) but in equation (2), the \(y\) is buried inside of parentheses. Distributing the \(x\) on the outside gets the \(y\) that we need out of those parentheses.

  4. At this point, we no longer have a \(y\) in the denominator of a fraction, nor is it buried inside of parentheses. So, we need to deal with the other problem: there are two \(y\)'s floating around, so we need to combine them into just one \(y\) that we can solve for. To do this, we need to get everything with a \(y\) on one side and everything without a \(y\) on the other side.

  5. To get from equation (3) to (4), we moved the \(xy\) to the right-hand side so that all of the \(y\)'s are on the same side of the equal sign.

  6. To get from equation (4) to (5), we moved the \(1\) to the left-hand side so that everything without a \(y\) is on the same side of the equal sign.

  7. This step is the one that trips students up the most: to get from equation (5) to (6), we factored a \(y\) out of the right-hand side. A helpful way to think about this is that it's just reverse-distribution. It's the same thing we did to get to equation (3), but backwards. Notice that if we went backwards and distributed out the \(y(3+x)\) in equation (6), we would get the \(3y+xy\) in equation (5).

  8. Now that we finally have only one \(y\) in our equation, we just need to get it by itself. So, to get from equation (6) to (7), we divide both sides by \(3+x\) to get \(y\) by itself and get our final answer.