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Section 2.4 Inverse Functions

In mathematics, an inverse is a function that serves to “undo” another function. That is, if \(f(x)\) produces \(y\text{,}\) then putting \(y\) into the inverse of \(f\) produces the output \(x\text{.}\) A function \(f\) that has an inverse is called invertible and the inverse is denoted by \(f^{-1}\text{.}\) It is best to illustrate inverses using an arrow diagram:

Notice how \(f\) maps \(1\) to \(a\text{,}\) and \(f^{-1}\) undoes this, that is, \(f^{-1}\) maps \(a\) back to \(1\text{.}\) Don't confuse \(f^{-1}(x)\) with exponentiation: the inverse \(f^{-1}\) is different from \(\frac{1}{f(x)}\text{.}\)

Not every function has an inverse. It is easy to see that if a function \(f(x)\) is going to have an inverse, then \(f(x)\) never takes on the same value twice. We give this property a special name.

A function \(f(x)\) is called one-to-one if every element of the range corresponds to exactly one element of the domain. Similar to the Vertical Line Test (VLT) for functions, we have the Horizontal Line Test (HLT) for the one-to-one property.

Example 2.17. Parabola is Not One-to-one.

The parabola \(f(x)=x^2\) it not one-to-one because it does not satisfy the Horizontal Line Test. For example, the horizontal line \(y=1\) intersects the parabola at two points, when \(x=-1\) and \(x=1\text{.}\)

We now formally define the inverse of a function.

Definition 2.18. Inverse of a Function.

Let \(f(x)\) and \(g(x)\) be two one-to-one functions. If \((f\circ g)(x)=x\) and \((g\circ f)(x)=x\) then we say that \(f(x)\) and \(g(x)\) are inverses of each other. We denote \(g(x)\) (the inverse of \(f(x)\)) by \(g(x)=f^{-1}(x)\text{.}\)

Thus, if \(f\) maps \(x\) to \(y\text{,}\) then \(f^{-1}\) maps \(y\) back to \(x\text{.}\) This gives rise to the cancellation formulas:

\begin{equation*} f^{-1}(f(x))=x,\mbox{ for every \(x\) in the domain of \(f(x)\)}\text{,} \end{equation*}
\begin{equation*} f(f^{-1}(x))=x,\mbox{ for every \(x\) in the domain of \(f^{-1}(x)\)}\text{.} \end{equation*}
Example 2.19. Finding the Inverse at Specific Values.

If \(f(x)=x^9+2x^7+x+1\text{,}\) find \(f^{-1}(5)\) and \(f^{-1}(1)\text{.}\)


Rather than trying to compute a formula for \(f^{-1}\) and then computing \(f^{-1}(5)\text{,}\) we can simply find a number \(c\) such that \(f\) evaluated at \(c\) gives \(5\text{.}\) Note that subbing in some simple values (\(x=-3,-2,1,0,1,2,3\)) and evaluating \(f(x)\) we eventually find that \(f(1)=1^9+2(1^7)+1+1=5\) and \(f(0)=1\text{.}\) Therefore, \(f^{-1}(5)=1\) and \(f^{-1}(1)=0\text{.}\)

To compute the equation of the inverse of a function we use the following guideline.

Guideline for Computing Inverses.
  1. Write down \(y=f(x)\text{.}\)

  2. Solve for \(x\) in terms of \(y\text{.}\)

  3. Switch the \(x\)'s and \(y\)'s.

  4. The result is \(y=f^{-1}(x)\text{.}\)

Example 2.20. Finding the Inverse Function.

We find the inverse of the function \(f(x)=2x^3+1\text{.}\)


Starting with \(y=2x^3+1\) we solve for \(x\) as follows:

\begin{equation*} y-1=2x^3\qquad\to\qquad \frac{y-1}{2}=x^3\qquad\to\qquad x=\sqrt[3]{\frac{y-1}{2}}\text{.} \end{equation*}

Therefore, \(\ds f^{-1}(x)=\sqrt[3]{\frac{x-1}{2}}\text{.}\)

This example shows how to find the inverse of a function algebraically. But what about finding the inverse of a function graphically? Step \(3\) (switching \(x\) and \(y\)) gives us a good graphical technique to find the inverse, namely, for each point \((a,b)\) where \(f(a)=b\text{,}\) sketch the point \((b,a)\) for the inverse. More formally, to obtain the graph of \(f^{-1}(x)\) reflect the graph of \(f(x)\) about the line \(y=x\text{.}\)

Exercises for Section 2.4.

Is the function \(f(x)=|x|\) one-to-one?


The graph of \(f(x) = |x|\) is shown below.

From this, we see that \(f(x)\) is not one-to-one because it does not satisfy the Horizontal Line Test.

If \(h(x)=e^x+x+1\text{,}\) find \(h^{-1}(2)\text{.}\)

\(h^{-1}(2) = 0 \)

Let us first notice that the function,

\begin{equation*} h(x)= e^{x} + x + 1\text{,} \end{equation*}

is one-to-one. Since we only need to find \(h^{-1}(2)\text{,}\) there is no need to solve for the general case (which would be difficult for this example). Instead, notice that

\begin{equation*} h(0) = 1 + 1 = 2\text{.} \end{equation*}

Thus, \(h^{-1}(2) = 0\text{.}\)

Find a formula for the inverse of the function \(\ds{f(x)=\frac{x+2}{x-2}}\text{.}\)

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


\begin{equation*} f(x) = \dfrac{x+2}{x-2} \end{equation*}

we can find \(f^{-1}(x)\) as follows. First, let \(y = f(x)\text{.}\) Then,

\begin{equation*} \begin{split} y \amp = \dfrac{x+2}{x-2} \\ y(x-2) \amp = x+2 \\ xy - x \amp = 2+2y\\ x(y-1)\amp = 2(1+y) \\ x \amp = \dfrac{2(y+1)}{(y-1)} \end{split} \end{equation*}

Therefore, \(f^{-1}(x) = \dfrac{2(x+1)}{(x-1)}\text{.}\)

Determine whether or not the following pairs are inverse functions.

  1. \(g(x) = \dfrac{ax+b}{cx-d}\) and \(f(x) = \dfrac{b+dx}{cx-a}\)


    If \(f(x)\) and \(g(x)\) are inverses of each other, then they must satisfy both \(f(g(x))=x\) and \(g(f(x))=x\text{.}\) Check:

    \begin{equation*} \begin{split} g(f(x)) \amp = g\left(\frac{b+dx}{cx-a}\right)\\ \amp = \frac{a\left(\frac{b+dx}{cx-a}\right) + b}{c\left(\frac{b+dx}{cx-a}\right) -d} \\ \amp =\frac{\frac{(ab+adx)+(bcx-ba)}{cx-a}}{\frac{(cb+cdx)-(dcx-da)}{cx-a}} \\ \amp =\frac{(ad+bc)x}{cb+da}\\ \amp =x \end{split} \end{equation*}
    \begin{equation*} \begin{split} f(g(x)) \amp = f\left(\frac{ax+b}{cx-d}\right)\\ \amp = \frac{b+d\left(\frac{ax+b}{cx-d}\right)}{c\frac{ax+b}{cx-d}-a} \\ \amp =\frac{\frac{(bcx-bd)+(dax+bd)}{cx-d}}{\frac{(cax+bc)-(acx-ad)}{cx-d}} \\ \amp =\frac{(bc+da)x}{bc+ad} \\ \amp =x \end{split} \end{equation*}

    Therefore, \(f(x)\) and \(g(x)\) are inverses of each other.

  2. \(g(x) = x^{2}\) and \(f(x) = \sqrt{x}\) for \(x \geq 0\)

  3. \(g(x) = \frac{1}{2}x^{5}\) and \(f(x) = \sqrt[5]{\frac{1}{2}x}\)

    Not inverses