Inverse Functions

Section 6.8 Inverse Functions

Example 6.30.

Let \(f(x) = 2 x^3 +4\text{.}\) Consider the following two contrasting problems.

Evaluate \(f\) at \(x=-1\text{.}\)
The inverse problem: Find \(x\text{,}\) if \(f(x) = 2\text{.}\)

For the first, we simply evaluate \(f\)

\begin{equation*} f(-1) = 2(-1)^3 + 4 = -2+4 = 2. \end{equation*}

For the inverse problem we need to solve the equation \(f(x) = 2\) for \(x\text{.}\)

\begin{align*} f(x) &= 2\\ 2x^3 + 4 &= 2\\ 2x^3 &= -2\\ x^3 &= -1\\ x &= \sqrt[3]{-1} = -1 \end{align*}

In the inverse problem, we are told the output \(y = f(x)\) of the function and find the corresponding input \(x\text{.}\) We could have solved this problem regardless of the given \(y\)-value. Regardless of wherever I choose the output \(y\) on the graph, I can always trace back to find a unique corresponding input \(x\text{.}\)

Figure 6.31. Graph of \(y=2x^3+4\text{.}\)

Example 6.32.

Let \(f(x) = 2 x^3 +4\text{.}\) Find \(x\text{,}\) if \(f(x) = y\text{.}\)

We must solve the equation \(f(x) = y\) for \(x\text{.}\)

\begin{align*} f(x) &= y\\ 2x^3 + 4 &= y\\ 2x^3 &= y-4\\ x^3 &= \frac{y-4}{2}\\ x &= \boxed{\sqrt[3]{\frac{y-4}{2}}} \end{align*}

To illustrate, if \(y = 2\text{,}\) then the corresponding \(x\) is

\begin{equation*} x = \sqrt[3]{\frac{2-4}{2}} = \sqrt[3]{\frac{-2}{2}} = \sqrt[3]{-1} = -1, \end{equation*}

so that \(f(-1) = 2\text{.}\)

The solution for \(x\) in terms of \(y\) above determines a new function which has a special relationship with \(f\text{.}\)

Definition 6.33.

Suppose \(f\) and \(g\) are two functions satisfying the following conditions:

\((g\circ f)(x) = x\text{,}\) for all \(x\) in the domain of \(f\text{;}\) and
\((f\circ g)(y) = y\text{,}\) for all \(y\) in the domain of \(g\text{.}\)

The we say \(g\) is an inverse function of \(f\text{;}\) and reflexively, \(f\) is an inverse function of \(g\text{.}\)

Example 6.34.

Let \(f(x) = 2x^3 + 4\) and

\begin{equation*} g(y) = \sqrt[3]{\frac{y-4}{2}}\text{.} \end{equation*}

Show that \(f\) and \(g\) are inverse functions of each other by confirming that

\((g\circ f)(x) = x\text{,}\) for all real numbers \(x\text{;}\) and
\((f\circ g)(y) = y\text{,}\) for all real numbers \(y\text{.}\)

Solution.

First,

\begin{align*} (g\circ f)(x) &= g(f(x)) \\ &= g\left(2x^3+4\right) \\ &= \sqrt[3]{\frac{\left(2x^3+4\right)-4}{2}} \\ &= \sqrt[3]{\frac{2x^3}{2}} \\ &= \sqrt[3]{x^3} \\ &= x \end{align*}

Second,

\begin{align*} (f\circ g)(y) &= f(g(y)) \\ &= f\left(\sqrt[3]{\frac{y-4}{2}}\right) \\ &= 2\left(\sqrt[3]{\frac{y-4}{2}}\right)^3 + 4 \\ &= 2\left(\frac{y-4}{2}\right) + 4 \\ &= y-4 + 4 \\ &= y \end{align*}

Example 6.35.

Let \(f(x) = 3x - 4\text{.}\) Find an inverse function for \(f\text{,}\) if possible.

Solution.

Suppose \(f(x) = y\text{.}\) We need to solve this equation for \(x\text{.}\)

\begin{align*} f(x) &= y\\ 3x-4 &= y\\ 3x &= y+4\\ x &= \frac{y+4}{3} \end{align*}

So, let \(g(y) = \frac{y+4}{3}\text{.}\) Then \(g\) is an inverse function of \(f\text{.}\) You can confirm this with

\begin{equation*} (g\circ f)(x) = g(f(x)) = g(3x - 4) = \frac{3x-4+4}{3} = \frac{3x}{3} = x \end{equation*}

and

\begin{equation*} (f\circ g)(y) = f(g(y)) = f\left(\frac{y+4}{3}\right) = 3\left(\frac{y+4}{3}\right) - 4 = y+4 - 4 = y. \end{equation*}

Not all functions have inverse functions, an issue we address in Section 6.9. An inverse function, if it exists, is in fact unique. As such, it deserves its own special notation introduce below.

Theorem 6.36.

Suppose \(f\) has an inverse function \(g\text{.}\) Then \(g\) is the unique such function and will be denoted by \(f^{-1}\) from now on. It satisfies,

\((f^{-1}\circ f)(x) = x\text{,}\) for all \(x\) in the domain of \(f\text{;}\) and
\((f\circ f^{-1})(y) = y\text{,}\) for all \(y\) in the domain of \(f^{-1}\text{.}\)

There is a nice relation between the graph of a function and its inverse function (when there is one).

Example 6.37.

Let \(f(x) = 2 x^3 +4\text{.}\) Then we saw in Example 6.34 that \(g(y) = \sqrt[3]{\frac{y-4}{2}}\) is an inverse function of \(f\text{.}\) As such, it is the only inverse function of \(f\) and we may now write

\begin{equation*} f^{-1} (y) = \sqrt[3]{\frac{y-4}{2}}. \end{equation*}

I personally like to write \(f^{-1}(y)\) emphasizing that the inverse function accepts a \(y\)-value from the graph of \(f\) as its input. However, it is customary to use \(x\) as the input to a function so we usually rewrite this as

\begin{equation*} f^{-1} (x) = \sqrt[3]{\frac{x-4}{2}}. \end{equation*}

We can now plot \(y=f^{-1}(x)\) and compare it to the graph of \(y=f(x)\text{.}\)

Observe the labeled points. For instance, \(f(1)=6\) so that \((1,6)\) is a point on the graph of \(f\text{.}\) Then, \(f^{-1}(6) = 1\) so that \((6,1)\) is a point on the graph of its inverse function. Repeat this observation for the other marked points on the two graphs.

Principle 6.40. Graphs of Inverse Function.

Suppose \(f\) has an inverse function. Then,

\((a,b)\) is a point on the graph of \(f\) if and only \((b,a)\) is a point on the graph of \(f^{-1}\text{.}\)
The graph of \(y=f^{-1}(x)\) is obtained by reflecting the graph of \(y=f(x)\) over the line \(y=x\text{.}\)
The domain of \(f\) is the range of \(f^{-1}\text{.}\)
The range of \(f\) is the domain of \(f^{-1}\text{.}\)

Example 6.41. Evaluating the Inverse Function Graphically.

Below is the graph of a function \(f\) which has an inverse function.

Find \(f(6)\text{.}\)
Find \(f^{-1}(6)\text{.}\)
Find the real number zeros of \(f^{-1}\)
Find the domain and range of \(f^{-1}\)
Sketch the graph of \(y=f^{-1}(x)\)

Solution.

\(f(6) = -1\) because \((6,-1)\) is on the graph of \(f\text{.}\)
\(f^{-1}(6) = -2\) because \((-2,6)\) is on the graph of \(f\text{.}\)
\(f(0)=2\) so that \(f^{-1}(2) = 0\) and \(x=2\) is the only zero of \(f^{-1}\)
The domain of \(f\) is \([-5,6]\) which is also the range of \(f^{-1}\text{.}\) The range of \(f\) is \([-1,8]\) which is also the domain of \(f^{-1}\text{.}\)
Take the key points on the graph and interchange the coordinates. Plot these and connect the dots.

Warning 6.42. Many Kinds of Inverse.

Be careful with the interpretation of the exponent \(-1\text{.}\) For instance, \(2^{-1} = 1/2\) refers to the multiplicative inverse of the real number \(2\) satisfying

\begin{equation*} 2\times 2^{-1} = 1. \end{equation*}

If \(f\) is a function that has an inverse function, then \(f^{-1}\) refers to the functional inverse satisfying

\begin{equation*} (f\circ f^{-1})(x) = x. \end{equation*}

In general,

\begin{equation*} f^{-1}(x) \neq \frac{1}{f(x)}. \end{equation*}

However,

\begin{equation*} [f(x)]^{-1} \neq \frac{1}{f(x)}, \end{equation*}

because \(f(x)\) is a real number obtained by evaluating \(f\text{.}\) (We are assuming \(f(x)\neq 0\text{.}\))

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