4.4 Inverses of Combined Functions

Introduction

In the 1980's Hasbro changed the toy world by introducing a line of robot action figures known as Transformers. What made each Transformer special was that if you bent the legs, twisted the torso, and folded the arms, the robot would change into something else, like a car or an airplane. Undoing each of those steps in the reverse order would change it back into a robot again.

Converting a Transformer is very similar to the shoes-and-socks method of finding an inverse function that we introduced in. Unfortunately, that technique seldom works on combined functions. In this section we develop a new procedure that can be applied to any function.

Review the Shoes and Socks Method

Let's review the shoes-and-socks method for finding an inverse. We'll use $f(x)=\frac{x^{3}+1}{5}$ as an example. The first step in the process was to identify the operations performed by the function, following the standard order of operations. The next step was to list the inverse operations in the reverse order.

	Operations Performed by the Function		Operations Performed by the Inverse
1.	Cube	1.	Multiply by $5$
2.	Add $1$	2.	Subtract $1$
3.	Divide by $5$	3.	Cube root

Since the list on the right describes exactly what the inverse function does, we use it to build the equation for $f^{-1}$. This is the final step.

\begin{align} \small \color{#5fa2ce}{\text{Start with an $x$:}} &&& x \newline \small \color{#5fa2ce}{\text{Multiply by $5$:}} &&& 5x \newline \small \color{#5fa2ce}{\text{Subtract $1$:}} &&& 5x - 1 \newline \small \color{#5fa2ce}{\text{Cube root:}} &&& \sqrt[3]{5x-1} \newline \small \color{#5fa2ce}{\text{Write as a function:}} && f^{-1}(x) &= \sqrt[3]{5x-1}
\end{align}

Review Graphing Inverses

Now consider a function such as $f(x)=\frac{x-5}{x-4}$ where $x$ appears more than once in the equation. Since we cannot determine what happens to $x$ first, we cannot establish an order of operations and cannot use the shoes-and-socks method, despite the fact that this function is indeed one-to-one.

We can, however, find the graph of $f^{-1}(x)$ by reflecting points on the graph of $f(x)$ around the line $y=x$, as in the figure below (move the blue slider to see the reflection).

Notice that each point $\color{red}{(x, y)}$ on the graph of $f$ turns into the point $\color{black}{(y, x)}$ on the graph of the inverse. This may not seem like a significant fact, but it is the key to a new algebraic method for finding the equations of inverses.

Learn the Switch and Solve Method

We have seen that if $(x, y)$ is a point on the graph of a one-to-one function $f$, then the point $(y, x)$ is on the graph of $f^{-1}$. The significance of this that switching $x$ and $y$ in the equation of a one-to-one function creates an implicit equation for its inverse. All that remains is to solve that equation for $y$ so that we can write it with function notation. For obvious reasons, this will be called the switch-and-solve method.

The switch-and-solve method can be broken down into a four-step process.

$$ \begin{align} \text{Step 1:} &&& \small \color{#5fa2ce}{\text{Replace $f(x)$ with $y$}} \newline \text{Step 2:} &&& \small \color{#5fa2ce}{\text{Switch $x$ and $y$}} \newline \text{Step 3:} &&& \small \color{#5fa2ce}{\text{Solve for $y$}} \newline \text{Step 4:} &&& \small \color{#5fa2ce}{\text{Replace $y$ with $f^{-1}(x)$}} \end{align} $$

We will illustrate this process by returning to the same function $f(x)=\frac{x^{3}+1}{5}$ that we used earlier with the shoes-and-socks method.

$\text{Step 1: } \small \color{#5fa2ce}{\text{ Replace $f(x)$ with $y$. }}$
To find the inverse using the switch-and-solve method we start by rewriting the function as an equation with $y$ in place of $f(x)$.

$$ y=\frac{x^{3}+1}{5} $$

$\text{Step 2: } \small \color{#5fa2ce}{\text{ Switch $x$ and $y$. }}$
Then we turn every $x$ into a $y$ and vice versa.

$$ x = \frac{y^{3}+1}{5} $$

$\text{Step 3: } \small \color{#5fa2ce}{\text{ Solve for $y$. }}$
The next step is to solve this equation for $y$.

$$ \begin{align} 5x &= y^{3}+1 && \small \color{#5fa2ce}{\text{Multiply both sides by $5$}} \newline 5x-1 &= y^{3} && \small \color{#5fa2ce}{\text{Subtract $1$ from both sides}} \newline \sqrt[3]{5x-1} &= y && \small \color{#5fa2ce}{\text{Take the cube root}} \end{align} $$

$\text{Step 4: } \small \color{#5fa2ce}{\text{ Replace $y$ with $f^{-1}(x)$. }}$
Lastly we rewrite this equation using inverse function notation.

$$ f^{-1}(x) = \sqrt[3]{5x-1} $$

Notice that this is the same answer we obtained earlier with the shoes-and-socks method and that the steps needed to solve for $y$ are exactly the steps listed in the shoes-and-socks method.

Use the Switch and Solve Method

Since there are several solving techniques that may arise when doing the switch-and-solve method, we will explore a few more examples before looking at functions with more than one $x$.

Example 1

Here we will find the inverse of $f(x)=e^x-2$. Since this is an exponential function, after we switch variables we should expect the solving process to involve the natural logarithm.

$$ \begin{align} f(x) &= e^x-2 && \small \color{#5fa2ce}{\text{Original function}} \newline y &= e^x-2 && \small \color{#5fa2ce}{\text{Replace $f(x)$ with $y$}} \newline x &= e^y-2 && \small \color{#5fa2ce}{\text{Swap $x$ and $y$}} \newline x + 2 &= e^y && \small \color{#5fa2ce}{\text{Add $2$}} \newline \ln(x+2) &= \ln\left(e^y\right) && \small \color{#5fa2ce}{\text{Apply $\ln$ to both sides}} \newline \ln(x+2) &= y && \small \color{#5fa2ce}{\text{Simplify}} \newline f^{-1}(x) &= \ln(x+2) && \small \color{#5fa2ce}{\text{Replace $y$ with $f^{-1}(x)$}} \end{align} $$

We now know that the inverse of $f(x)=e^x-2$ is $f^{-1}(x)=\ln(x+2)$.

Example 2

In this next example we will use the switch-and-solve method to find the inverse of $f(x)=-5x+8$. The solving steps do not involve anything beyond basic inverse operations. Please try each step on your own and then check your answer.

$\text{Step 1: } \small \color{#5fa2ce}{\text{ Replace $f(x)$ with $y$. }}$

Answer

---

$$y=-5x+8$$

$\text{Step 2: } \small \color{#5fa2ce}{\text{ Switch $x$ and $y$. }}$

Answer

---

$$x=-5y+8$$

$\text{Step 3: } \small \color{#5fa2ce}{\text{ Solve for $y$. }}$

Answer

---

\begin{align} x&=-5y+8 \newline x-8 &= -5y && \small \color{#5fa2ce} {\text{subtract 8}} \newline \frac{x-8}{-5}&=y && \small \color{#5fa2ce} {\text{divide by -5}}\end{align}

$\text{Step 4: } \small \color{#5fa2ce}{\text{ Replace $y$ with $f^{-1}(x)$. }}$

Answer

---

$$f^{-1}(x)=\frac{x-8}{-5}$$

Example 3

Now let's look at a power function $f(x)=x^{2/3}+6$ and restrict its domain $x\leq0$ so that it is one-to-one. The steps below outline the complete process for finding the equation for $f^{-1}$.

$$ \begin{align} f(x) &= x^{2/3}+6 && \small \color{#5fa2ce} {\text{Original function}} \newline y &= x^{2/3}+6 && \small \color{#5fa2ce} {\text{Replace $f(x)$ with $y$}} \newline x &= y^{2/3}+6 && \small \color{#5fa2ce} {\text{Switch $x$ and $y$}} \newline x-6 &= y^{2/3} && \small \color{#5fa2ce} {\text{Subtract $6$}} \newline (x-6)^{3} &= y^{2} && \small \color{#5fa2ce} {\text{Apply the 3rd power}} \newline \pm\sqrt{(x-6)^{3}} &= y && \small \color{#5fa2ce} {\text{Apply the square root}} \newline y &= \pm(x-6)^{3/2} && \small \color{#5fa2ce} {\text{Rewrite}} \newline \end{align} $$

We now face a choice, should we pick $(x-6)^{3/2}$ or $-(x-6)^{3/2}$ as the inverse?

Since the inputs and outputs of a function and its inverse should switch places, comparing the domain and range of our original function against these two options should point us to the correct choice.

Both of our options have a domain of $x\geq6$, matching the range of $f$, but only $-(x-6)^{3/2}$ has a range that matches the domain of $f$, making it the correct choice for $f^{-1}$.

Another way to see this is to look at the graphs. The graphs of $f$ and $f^{-1}$ must be symmetric across the line $y=x$. In this case, our function has a y-intercept at $(0,6)$, so the inverse must have an x-intercept at $(6,0)$.

The takeaway is that whenever the equation for $f^{-1}$ is not one-to-one, as it was in this example, we need to look closely at the graph of the function and its domain and range to help us choose the correct inverse.

Inverses of Combined Functions

Let's return to the function $f(x)=\frac{x-5}{x-4}$ which we graphed earlier. Since $x$ occurs more than once, the switch-and-solve method is the best way to find the inverse. Here are all of the steps.

$$ \begin{align} f(x) &= \frac{x-5}{x-4} && \small \color{#5fa2ce} {\text{Original function}} \newline y &= \frac{x-5}{x-4} && \small \color{#5fa2ce} {\text{Replace $f(x)$ with $y$}} \newline x &= \frac{y-5}{y-4} && \small \color{#5fa2ce} {\text{Switch $x$ and $y$}} \newline x(y-4) &= y-5 && \small \color{#5fa2ce} {\text{Multiply by $y-4$}} \newline xy-4x &= y-5 && \small \color{#5fa2ce} {\text{Distribute $x$}} \newline xy-4x -y &= -5 && \small \color{#5fa2ce} {\text{Subtract $y$}} \newline xy-y &= 4x-5 && \small \color{#5fa2ce} {\text{Add $4x$}} \newline y(x-1) &= 4x-5 && \small \color{#5fa2ce} {\text{Factor out $y$}} \newline y &= \frac{4x-5}{x-1} && \small \color{#5fa2ce} {\text{Divide by $x-1$}} \newline f^{-1}(x) &= \frac{4x-5}{x-1} && \small \color{#5fa2ce}{\text{Replace $y$ with $f^{-1}(x)$}} \end{align} $$

Using function notation, we can now say that $f^{-1}=\frac{4x-5}{x-1}$ is the inverse of $f(x)=\frac{x-5}{x-4}$.

As you can see, finding a formula for an inverse function can be very difficult. In fact, sometimes it is impossible. For example, to find the inverse of $f(x)=xe^{x}$, $x\geq0$ we would need to solve $x=ye^y$ for $y$.

$$ \begin{align} x &= ye^y && \small \color{#5fa2ce} {\text{Original equation}} \newline \frac{x}{y} &= e^y && \small \color{#5fa2ce} {\text{Divide by } y} \newline \ln\left(\frac{x}{y}\right) &= y && \small \color{#5fa2ce} {\text{Apply the natural logarithm}} \end{align} $$

By removing the exponential we've freed up one $y$, but the other $y$ is now stuck inside of the logarithm. If we were to free the $y$ inside of the logarithm, then the other $y$ would be trapped back in an exponential. It's a loop we could never get out of.

Verify Inverse Functions

You'll remember that in Section 1.5 we introduced the idea that a function combined with its inverse should act like a round trip airplane flight, bringing you back to where you started. Now that we understand composition, we can state this rule more precisely.

Given a one-to-one function $f$ and its inverse $f^{-1}$, the following round trip properties hold

$$ (f \circ f^{-1})(x)=f(f^{-1}(x))=x $$

for all the $x$ in the domain of $f^{-1}$ and

$$ (f^{-1} \circ f)(x)=f^{-1}(f(x))=x $$

for all the $x$ in the domain of $f$. In other words, a function and its inverse should cancel each other, no matter which way they are composed together.

These properties are only true for inverse functions. As a consequence, they can be used to verify that two functions are inverses.

Earlier, in Example 2, we found that the inverse of $\color{red}{f(x)=-5x+8}$ was $\color{blue}{f^{-1}(x)=\frac{x-8}{-5}}$. Let's use the round trip properties to verify that these are indeed inverses.

First we'll check to see that $(f \circ f^{-1})(x) = x$.

$$ \begin{align} (f \circ f^{-1})(x) &= \color{red}{f \left( \color{blue}{f^{-1}(x)} \right) } && \small \color{#5fa2ce} {\text{Definition of composition}} \newline &= \color{red}{f \left( \color{blue}{\frac{x-8}{-5}} \right) } && \small \color{#5fa2ce} {\text{Use equation for } f^{-1}(x)} \newline &= \color{red}{-5 \left( \color{blue}{\frac{x-8}{-5}} \right) +8} && \small \color{#5fa2ce} {\text{Use equation for } f(x)} \newline &= \color{red}{ \left( \color{blue}{x-8} \right) +8} && \small \color{#5fa2ce} {\text{Multiplication and division cancel}} \newline &= x && \small \color{#5fa2ce} {\text{Simplify}} \end{align} $$

We also need to check composition in the other direction to see if $(f^{-1} \circ f)(x) = x$.

$$ \begin{align} (f^{-1} \circ f)(x) &= \color{blue}{f^{-1} \left( \color{red}{f(x)} \right) } && \small \color{#5fa2ce} {\text{Definition of composition}} \newline &= \color{blue}{f^{-1} \left( \color{red}{-5x + 8} \right) } && \small \color{#5fa2ce} {\text{Use equation for } f(x)} \newline &= \color{blue}{\frac{\color{red}{(-5x + 8)} -8}{-5}} && \small \color{#5fa2ce} {\text{Use equation for } f^{-1}(x)} \newline &= \color{blue}{\frac{\color{red}{-5x}}{-5}} && \small \color{#5fa2ce} {\text{Cancel +8 and -8}} \newline &= x && \small \color{#5fa2ce} {\text{Simplify}} \end{align} $$

Since both round trip properties hold, we have verified that these two functions are truly inverses of eachother.

Applications of Inverses

One of the main jobs of the ship L.M. Gould is to pick up containers of food and supplies from Punta Arenas, Chile, and deliver them to the scientists working at Palmer Station in Antarctica.

Suppose the amount of food, in pounds, that the station has on hand $x$ days after being re-supplied is given by $f(x)=11200-200x$. The inverse of this function is

$$ \begin{align} f(x) &= 11200-200x && \small \color{#5fa2ce} {\text{Original function}} \newline y &= 11200-200x && \small \color{#5fa2ce} {\text{Replace $f(x)$ with $y$}} \newline x &= 11200-200y && \small \color{#5fa2ce} {\text{Switch $x$ and $y$}} \newline y &= \frac{x - 11200}{-200} && \small \color{#5fa2ce} {\text{Solve for $y$}} \newline f^{-1}(x) &= \frac{x - 11200}{-200} && \small \color{#5fa2ce}{\text{Replace $y$ with $f^{-1}(x)$}} \end{align} $$

But how should we interpret the inverse, what would it mean in this situation? Our new switch-and-solve method not only gives the equation for the inverse, but can also help us see the context of the inverse as well.

The key is to recognize that the inputs and outputs are trading places. If the original function computes pounds of food based on the number of days since the supply ship arrived, then the inverse would allow the scientists to calculate how long it has been since the ship last visited by measuring the amount of food available.

Looking Ahead

In this chapter we've seen how to make new, more complicated functions by combining simple functions through the operations of function addition, subtraction, multiplication, division and composition.

We have also seen that as the functions become more complicated, their inverses, if they exist, are difficult or even impossible to find.

In the next chapter we will focus on the properties of polynomial and rational functions. Since their equations involve several terms and are not in general one-to-one, we will not spend time finding their inverses.