Determine the Domain and Range of an Inverse Function
The outputs of the function [latex]f[/latex] are the inputs to [latex]{f}^{-1}[/latex], so the range of [latex]f[/latex] is also the domain of [latex]{f}^{-1}[/latex]. Likewise, because the inputs to [latex]f[/latex] are the outputs of [latex]{f}^{-1}[/latex], the domain of [latex]f[/latex] is the range of [latex]{f}^{-1}[/latex]. We can visualize the situation.

domain and range of inverse functions
The range of a function [latex]f\left(x\right)[/latex] is the domain of the inverse function [latex]{f}^{-1}\left(x\right)[/latex].
The domain of [latex]f\left(x\right)[/latex] is the range of [latex]{f}^{-1}\left(x\right)[/latex].
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.
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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).
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!
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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.
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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].
No. If two supposedly different functions, say, [latex]g[/latex] and [latex]h[/latex], both meet the definition of being inverses of another function [latex]f[/latex], then you can prove that [latex]g=h[/latex]. We have just seen that some functions only have inverse functions if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverse functions. However, on any one domain, the original function still has only one unique inverse function.
- If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse.
- If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function.
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Fora refresher of the tool kit functions, expand the review below.