1.7 Inverse functions  (Page 3/10)

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 $f(x)=\sqrt{x}$ is $f^{-1}(x)=x^2,$ because a square “undoes” a square root; but the square is only the inverse of the square root on the domain $\left[0,\infty \right),$ since that is the range of $f(x)=\sqrt{x}.$

We can look at this problem from the other side, starting with the square (toolkit quadratic) function $f(x)=x^2.$ 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 $f(x)=x^2$ with its range limited to $\left[0,\infty \right),$ which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function).

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

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