An inverse is an undoing operation.

When I tie my shoelaces, I have performed an operation. When I untie my shoelaces, I am performing an inverse operation.

Some places measure temperatures in degrees Fahrenheit, \(^{\circ}\mbox F\), while other places measure temperatures in degrees Celsius, \(^{\circ}\mbox C\). It is not uncommon for some people to convert back and forth.

To convert from Celsius to Fahrenheit, the formula \(F=\dfrac{9}{5}C + 32\) can be used, where \(F\) is the temperature measured in Fahrenheit and \(C\) is the temperature measured in Celsius.

To convert from Fahrenheit to Celsius, the formula \(C=\dfrac{5(F-32)}{9}\) can be used.

To further illustrate the idea of an inverse, let's convert \(15^{\circ}\mbox{C}\) to \(^{\circ}\mbox F\) by substituting \(15\) for \(C\) in the first formula. Then, \(F=\dfrac{9}{5}(15) + 32 = 27+32=59\).

Now, take the output from the first formula and plug it into the second formula. That is, substitute \(F=59\) into the second formula. The result should be \(15\).

We can check by substituting and indeed, \(C=\dfrac{5(59-32)}{9}=\dfrac{135}{9}=15\), as expected.

The second formula undoes the work of the first formula, and the first formula also undoes the work of the second formula. The two functions are inverses of one another.

In earlier modules, we introduced the notion of a function by using a machine.

This machine took an input and cubed it to produce an output. We defined the function \(f(x)=x^3\). What function would undo the operation of cubing?

You probably have encountered the cube root in past experiences. The function \(g(x)=\sqrt[3]{x}\) will undo the work of the function \(f(x)=x^3\), and the function \(f(x)=x^3\) will undo the work of the function \(g(x)=\sqrt[3]{x}\).

Referring to the two tables of values, you will notice that the input of \(f(x)=x^3\) is the output of \(g(x)=\sqrt[3]{x}\) and the output of \(f(x)=x^3\) is the input of \(g(x)=\sqrt[3]{x}\).

That is, it appears that the domain of \(f(x)\) is the range of \(g(x)\) and the range of \(f(x)\) is the domain of \(g(x)\).

The inverse of a function is a relation which undoes the work of the function. The inverse of a function may or may not be a function (that is, it may or may not pass the vertical line test).

When the inverse of a function, \(f(x)\), is also a function, it is denoted as \(f^{-1}(x)\).

Note: \(f^{-1}(x)\neq \dfrac{1}{f(x)}\)

When you see this notation, it is referring to the inverse of a function.

To find the inverse of a function (or relation), we interchange (or switch) \(x\) and \(y\).

Let \(f = \left\{(-2,4),(-1,1),(0,0),(1,1),(2,4)\right\}\).

a. Determine the inverse of \(f\).

b. State the domain and range of \(f\) and its inverse.

c. Is the inverse of \(f\) a function?

Solution

a. By switching the \(x\) and \(y\) coordinates in each ordered pair,

the inverse of \(f = \left\{(4,-2),(1,-1),(0,0),(1,1),(4,2)\right\}\)

b.

Notice that the domain of \(f\) equals the range of the inverse of \(f\) and the range of \(f\) equals the domain of the
inverse of \(f\).

c. The inverse of \(f\) is not a function. There are two ordered pairs in the set which have \(x\)-coordinate \(4\) and
different \(y\)-coordinates. Such a relation would fail the vertical line test.

a. Graph \(y=x^2\) and its inverse.

b. State the domain and range of \(y=x^2\) and the inverse of \(y=x^2\).

c. Is the inverse of \(y=x^2\) a function? Justify your answer.

a. Graph \(y=x^2\) and its inverse.

b. State the domain and range of \(y=x^2\) and the inverse of \(y=x^2\).

c. Is the inverse of \(y=x^2\) a function? Justify your answer.

Solution

b.

Notice again that the domain of \(y=x^2\) is the range of the inverse of \(y=x^2\) and the range of \(y=x^2\) is the domain of the inverse of \(y = x^2\).

c. The inverse of \(y=x^2\) is not a function since it fails the vertical line test.