Show that `f: R-[0]vecR0[0]` given by `f(x)=3/x` is invertible and it is inverse of itself.

Explanation

`f(n)= frac{3}{x} `


`f^{prime}(n)= 3 cdot(-1) x^{-2} `


`=frac{-3}{x^{2}} quad<0 quad forall x in R `


` =(text { one-one })`



now, `y=frac{3}{x} Rightarrow x=frac{3}{y}`.


`

`therefore` onto ie. `f(r) rightarrow` bijective invertible