If a function maps every “input” to exactly one “output,” an inverse of that function maps every “output” to exactly one “input.” We need a more formal definition to actually say anything with rigor.

Let \(f\) be a function with domain \(A\) and range \(B\):

Let \(g\) be a function with domain \(B\) and range \(A\):

We say that \(f\) and \(g\) are inverses of each other if \(f(g(b)) = b\) for all \(b\) in \(B\), and also \(g(f(a)) = a\) for all \(a\) in \(A\). Sometimes we write \(g = f^{-1}\) in this case.

and

So, we could rephrase these conditions as

\[ f(f^{-1}(x)) = x\qquad \text {and}\qquad f^{-1}(f(x)) = x. \]

These two simple equations are somewhat more subtle than they initially appear.

Consider the graph of \(y=f(x)\) below

Is \(f(x)\) invertible at \(x=1\)?

yes no

\[ f^{-1}(1) = \answer {1.5} \]

So far, we’ve only dealt with abstract examples. Let’s see if we can ground this in a real-life context.

We have examined several functions in order to determine their inverse functions, but there is still more to this story. Not every function has an inverse, so we must learn how to check for this situation.

Look again at the last question. If two different inputs for a function have the same output, there is no hope of that function having an inverse function. Why? This is because the inverse function must also be a function, and a function can only have one output for each input. More specifically, we have the next definition.

You may recall that a plot gives \(y\) as a function of \(x\) if every vertical line crosses the plot at most once, and we called this the vertical line test. Similarly, a function is one-to-one if every horizontal line crosses the plot at most once, and we call this the horizontal line test.

Below, we give a graph of \(f(x)=-5x^2+30x+60\). While this graph passes the vertical line test, and hence represents \(y\) as a function of \(x\), it does not pass the horizontal line test, so the function is not one-to-one.

Although the line \(y=20\) intersects the curve \(y=f(x)\) exactly once, the function \(f\) is not one-to-one on its domain, since the line \(y=80\) intersects the curve twice.

As we have discussed, we can only find an inverse of a function when it is one-to-one. If a function is not one-to-one, but we still want an inverse, we must restrict the domain. Let’s see what this means in our next examples.

Consider the graph of the function \(f\) below:

On which of the following intervals is \(f\) one-to-one?

\([A,B]\) \([A,C]\) \([B,D]\) \([C,E]\) \([C,D]\)

This idea of restricting the domain is critical for understanding functions like \(f(x) = \sqrt {x}\).

Consider the function

\[ f(x) = x^2. \]

Does \(f\) have an inverse? If so, what is it? If not, attempt to restrict the domain of \(f\) and find an inverse on the restricted domain.

In this case \(f\) is not one-to-one. However, it is one-to-one on the interval \([0,\infty )\). Hence we can find an inverse of \(f(x)=x^2\) on this interval. We plug \(f^{-1}(x)\) into \(f\) and write \begin{align*} f(f^{-1}(x)) &= \left (f^{-1}(x)\right )^2\\ x &= \left (f^{-1}(x)\right )^2. \end{align*}

Since the domain of \(f\) is \([0, \infty )\), we know that \(x\) is positive. This means we can take the square-root of each side of the equation to find that

\[ \sqrt {x} = f^{-1}(x). \]

Consider the graph of \(y=f(x)\) below

Is the function \(f\) one-to-one on the interval \([1,2]\)?

yes no

Let \(f^{-1}\) be the inverse function of \(f\) restricted to the domain \([1,2]\).

(a)

Find the domain of \(f^{-1}\).

\[ \text {The domain of $f^{-1}$ is } \left [\answer {0}, \answer {2}\right ] \]

(b)

Find the value of \(f\left ( f^{-1}(0.5)\right )\).

\[ f\left ( f^{-1}(0.5)\right ) = \answer {0.5} \]

(c)

Choose the correct answer regarding \(f^{-1}(0.5)\).

\(f^{-1}(0.5)>1.5\) \(f^{-1}(0.5)\le 1.5\) \(f^{-1}(0.5)\) is not defined

Consider the function

\[ f(x) = x^3. \]

Does the function \(f\) have an inverse? If so, what is it? If not, attempt to restrict the domain of \(f\) and find an inverse on the restricted domain.

In this case \(f\) is one-to-one. We may write \begin{align*} f(f^{-1}(x)) &= \left (f^{-1}(x)\right )^3\\ x &= \left (f^{-1}(x)\right )^3\\ \sqrt [3]{x} &= f^{-1}(x). \end{align*}

For your viewing pleasure we give a graph of \(y=f(x)=x^3\) and \(y=f^{-1}(x)= \sqrt [3]{x}\). Note, the graph of \(f^{-1}\) is the image of \(f\) after being flipped over the line \(y=x\).