1.4 Inverse Functions

We use a function \( z \depends x \) to describe how one variable, \( z \), depends on another variable, \( x \). By inverting the function, we reverse the direction of dependence.

Definition.

Suppose \( b \mathrel{\overset{f}{\leftarrow}} a \) is a function, and suppose we can solve for the input variable, \( a \). Then

\[ \begin{align} a &\depends b \end{align} \]

is the inverse function for \( f \).

We'll avoid using the variables \( x \), \( y \), and, \( z \) when inverting a function, since these each come with a bias towards being used either for input or for output. See Appendix A. Are Variables Dummies? for a brief discussion of how we choose variable names.

Example.

Consider the function

\[ \begin{align} b &\mathrel{\overset{f}{\leftarrow}} a \\ b &= 2 a - 6. \end{align} \]

Let's solve for \( a \).

\[ \begin{align} &b = 2 a - 6 \\ &b + 6 = 2 a \\ &\frac{b + 6}{2} = a \end{align} \]

We've found the inverse of \( f \).

\[ \begin{align} a &\depends b \\ a &= \frac{b + 6}{2} \end{align} \]

We may cancel a function by applying its inverse.

Important.

If two functions

\[ \begin{align} b &\mathrel{\overset{f}{\leftarrow}} a & &\andSpaced & a &\mathrel{\overset{g}{\leftarrow}} b \end{align} \]

are inverse, then we can make cancellations,

\[ \begin{align} g \of f \of a &= a \\ f \of g \of b &= b. \end{align} \]

The cube root function is defined as the inverse of the cubing function.

\[ \begin{align} b &\depends a & & & a &\depends b \\ b &= a^3 & &\andSpaced & a &= \sqrt[3]{b} \end{align} \]

Let's see how the cube root cancels with cubing.

\[ \begin{align} &\sqrt[3]{3^3} = \sqrt[3]{27} = 3 \\ &\bigl(\sqrt[3]{-8}\bigr)^3 = (-2)^3 = -8 \end{align} \]

If we cube a number and then take the cube root, then we get back the number we started with. And the same is true working in the other order.

Example.

Let's find an inverse for the function

\[ \begin{align} b &\depends a \\ b &= (a + 1)^3 - 4. \end{align} \]

It is our job to solve for \( a \).

\[ \begin{align} &b = (a + 1)^3 - 4 \\ &b + 4 = (a + 1)^3 \end{align} \]

We'll take cube roots of both sides of this equation so that we may make a cancellation.

\[ \begin{align} &\sqrt[3]{b + 4} = \sqrt[3]{(a + 1)^3} \\ &\sqrt[3]{b + 4} = a + 1 \\ &\sqrt[3]{b + 4} - 1 = a \end{align} \]

We've found the inverse.

\[ \begin{align} a &\depends b \\ a &= \sqrt[3]{b + 4} - 1 \end{align} \]

We would like to define the square root as the inverse of the squaring function,

\[ \begin{align} b &\depends a & & & a &\depends b \\ b &= a^2 & &\andSpaced & a &= \sqrt[2]{b}, \end{align} \]

but we run into trouble!

Any positive \( b \)-value, say \( b = 9 \), corresponds to two \( a \)-values: \( a = 3 \) and \( a = -3 \).
Any negative \( b \)-value, say \( b = -9 \), corresponds to zero \( a \)-values.

For the square root to make any sense, we must have a single \( a \)-value for each \( b \)-value. To remedy these issues, we'll use restrictions:

\[ \begin{align} a &\ge 0 & &\andSpaced & b &\ge 0. \end{align} \]

We may now define the square root as the inverse to the squaring function. These restrictions have real consequences, however. We'll refuse to take the square root of a negative number. And the cancellation rules

\[ \begin{align} &\sqrt[2]{a^2} = a \\ &\bigl(\sqrt[2]{b}\bigr)^2 = b \end{align} \]

only hold for positive \( a \)-values and positive \( b \)-values.