$\begingroup$Can anybody say anything about the inverse function of $y=1/x$ and plot it on a graph and then compare the graphs of the given function and it's inverse? Is $y=1/x$ invertible? If yes then do the graphs of $y=1/x$ and its inverse coincide?
$\endgroup$34 Answers
$\begingroup$For $x\ne0$, $1/(1/x)=x$, so the function is its own inverse on e.g. $\Bbb R\setminus\{0\}$.
$\endgroup$2$\begingroup$We have a function $$f(x)=\frac{1}{x}$$ The inverse function, $f^{-1}$ satisfies the equation $$f(f^{-1}(x))=x$$ Thus, $$\frac{1}{f^{-1}(x)}=x \implies f^{-1}(x)=\frac{1}{x}$$ $f$ is in fact its own inverse.
$\endgroup$$\begingroup$Solve the equation
$$y=\frac1x$$ for $x$. This immediately gives you
$$x=\frac1y$$ and is valid for $x,y\ne0$.
$\endgroup$$\begingroup$To get an inverse function you need to interchange $x$ and $y$ so here.. the relation remains unchanged, the equiangular hyperbola remains its own inverse.
For an implicit relation
$$ x^2 + y^2= \sin xy $$
the inverse function also coincides with the first function.
Graph of these functions appear as mirror images with respect to line $x=y$
As an exercise try to find the inverse function
$$ y= \pm \sqrt {(x+1)(3-x)} +1 $$
considering both signs of $\pm$ which also reflect about the line $x=y$ two parts of a circle of radius 2.
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