Inverses of Some Standard Functions

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Lets try to define an inverse for some of the functions that we have seen up till now.


\(f(x) = x\,\,\,\,\,\,\mathbb{R} \to \mathbb{R}\)

We see that this function is one-one and onto. The inverse exists.

\({f^{ - 1}}(x) = x\,\,\,\,\,\,\mathbb{R} \to \mathbb{R}.\)



\(f(x) = {x^2}\,\,\,\,\,\,\mathbb{R} \to \mathbb{R}\)

This function is many-one and into. (why?). To define the inverse, we first need to make \(f\) one-one and onto.

We see that to make the function one-one, we can select the domain as only [\(0,\;\infty \)) (instead of \(\mathbb{R} \)).

In this domain, the function is one-one, as is clear from the second diagram above.

Also, the range is [\(0,\;\infty \)). Therefore, redefine this function:

\[f(x) = {x^2}\mathop {[0,\infty )}\limits_A {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \to {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop {[0,\infty )}\limits_B {\rm{ }}\]

This function is now one-one and onto, and hence invertible

\[{f^{ - 1}}(x){\rm{ }} = {\rm{}}\sqrt x {\rm{ }}\mathop {[0,\infty )}\limits_B {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  \to {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop {[0,\infty )}\limits_A \]

Additionally, note that if we draw the graph of \(\sqrt x \) and \(x^2\) on the same axis, they are mirror images of each other in the line \(x=y.\)

(As we saw in the case of \(\,f(x) = \log x\)  and \({f^{ - 1}}(x) = {2^x}).\)

A little thought will show why \(f\) and  \({f^{ - 1}}\) should be mirror images in the mirror \(y=x.\)

In the equation  \(y = f(x),\) \(x\) can be treated as the independent and y the dependent variable.

In the equation \(x = {f^{ - 1}}(y),\) we can reverse the roles. We can treat \(y\text{-axis}\) as the independent variable axis

By convention, the independent variable is taken on the horizontal axis. Therefore we convert this horizontal view (the second diagram above) into a conventional vertical view. How? By taking the reflection in \(y = x.\)

(c)   \(f (x)=\text{sin}x\;\)\(\mathbb{R}\; \)\( \rightarrow \; \)\([–1, 1]\)

This function is onto but not one-one. Lets make it one-one first. Select that interval as the domain, in which \(\text{sin}x\) remains one-one.

In other words, we redefine \(f\) as \(\begin{align}f(x) = \sin x\,\,\,\,\,\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right] \to [ - 1,1]\end{align}\) (we could have taken any other interval in which \(\text{sin}x\) remains one-one, to define the inverse. But conventionally, we take the interval closest to the origin; \(\begin{align}\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\end{align}\) \(\) in this case, which is centred on the origin). The inverse can now be defined as:

\[{f^{ - 1}}(x){\rm{}} = {\rm{}}{\sin ^{ - 1}}x\quad [ - 1,1] \to {\rm{}}\left[ { - \frac{\pi }{2},\frac{\pi }{2}} \right]\]

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