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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 $\text{\hspace{0.17em}}f(x)=\sqrt{x}\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}{f}^{-1}(x)={x}^{2},\text{\hspace{0.17em}}$ because a square “undoes” a square root; but the square is only the inverse of the square root on the domain $\text{\hspace{0.17em}}\left[0,\infty \right),\text{\hspace{0.17em}}$ since that is the range of $\text{\hspace{0.17em}}f(x)=\sqrt{x}.$
We can look at this problem from the other side, starting with the square (toolkit quadratic) function $\text{\hspace{0.17em}}f(x)={x}^{2}.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}f(x)={x}^{2}\text{\hspace{0.17em}}$ with its range limited to $\text{\hspace{0.17em}}\left[0,\infty \right),\text{\hspace{0.17em}}$ which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).
If $\text{\hspace{0.17em}}f(x)={\left(x-1\right)}^{2}\text{\hspace{0.17em}}$ on $\text{\hspace{0.17em}}\left[1,\infty \right),\text{\hspace{0.17em}}$ then the inverse function is $\text{\hspace{0.17em}}{f}^{-1}(x)=\sqrt{x}+1.$
Is it possible for a function to have more than one inverse?
No. If two supposedly different functions, say, $\text{\hspace{0.17em}}g\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}h,\text{\hspace{0.17em}}$ both meet the definition of being inverses of another function $\text{\hspace{0.17em}}f,\text{\hspace{0.17em}}$ then you can prove that $\text{\hspace{0.17em}}g=h.\text{\hspace{0.17em}}$ We have just seen that some functions only have inverses 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 inverses. However, on any one domain, the original function still has only one unique inverse.
The range of a function $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ is the domain of the inverse function $\text{\hspace{0.17em}}{f}^{-1}(x).$
The domain of $\text{\hspace{0.17em}}f(x)\text{\hspace{0.17em}}$ is the range of $\text{\hspace{0.17em}}{f}^{-1}(x).$
Given a function, find the domain and range of its inverse.
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 in [link] . We restrict the domain in such a fashion that the function assumes all y -values exactly once.
Constant | Identity | Quadratic | Cubic | Reciprocal |
$f(x)=c$ | $f(x)=x$ | $f(x)={x}^{2}$ | $f(x)={x}^{3}$ | $f(x)=\frac{1}{x}$ |
Reciprocal squared | Cube root | Square root | Absolute value | |
$f(x)=\frac{1}{{x}^{2}}$ | $f(x)=\sqrt[3]{x}$ | $f(x)=\sqrt{x}$ | $f(x)=\left|x\right|$ |
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 $\text{\hspace{0.17em}}\left[0,\infty \right),$ where it is equal to the identity function.
The reciprocal-squared function can be restricted to the domain $\text{\hspace{0.17em}}\left(0,\infty \right).$
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