Determine the domain and range of an inverse function

Figure 3. Domain and range of a function and its inverse

When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of [latex]f\left(x\right)=\sqrt{x}[/latex] is [latex]{f}^{-1}\left(x\right)={x}^{2}[/latex], because a square "undoes" a square root; but the square is only the inverse of the square root on the domain [latex]\left[0,\infty \right)[/latex], since that is the range of [latex]f\left(x\right)=\sqrt{x}[/latex].

We can look at this problem from the other side, starting with the square (toolkit quadratic) function [latex]f\left(x\right)={x}^{2}[/latex]. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and –3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the "inverse" is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function.

In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function [latex]f\left(x\right)={x}^{2}[/latex] with its range limited to [latex]\left[0,\infty \right)[/latex], which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

If [latex]f\left(x\right)={\left(x - 1\right)}^{2}[/latex] on [latex]\left[1,\infty \right)[/latex], then the inverse function is [latex]{f}^{-1}\left(x\right)=\sqrt{x}+1[/latex].

• The domain of [latex]f[/latex] = range of [latex]{f}^{-1}[/latex] = [latex]\left[1,\infty \right)[/latex].
• The domain of [latex]{f}^{-1}[/latex] = range of [latex]f[/latex] = [latex]\left[0,\infty \right)[/latex].

Example 4: Finding the Inverses of Toolkit Functions

Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed below. We restrict the domain in such a fashion that the function assumes all y-values exactly once.

Constant	Identity	Quadratic	Cubic	Reciprocal
[latex]f\left(x\right)=c[/latex]	[latex]f\left(x\right)=x[/latex]	[latex]f\left(x\right)={x}^{2}[/latex]	[latex]f\left(x\right)={x}^{3}[/latex]	[latex]f\left(x\right)=\frac{1}{x}[/latex]
Reciprocal squared	Cube root	Square root	Absolute value
[latex]f\left(x\right)=\frac{1}{{x}^{2}}[/latex]	[latex]f\left(x\right)=\sqrt[3]{x}[/latex]	[latex]f\left(x\right)=\sqrt{x}[/latex]	[latex]f\left(x\right)=|x|[/latex]

Solution

The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse.

The absolute value function can be restricted to the domain [latex]\left[0,\infty \right)[/latex], where it is equal to the identity function.

The reciprocal-squared function can be restricted to the domain [latex]\left(0,\infty \right)[/latex].

Analysis of the Solution

We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse.

Figure 4. (a) Absolute value (b) Reciprocal squared

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