If $f:\mathbb{R}\to \mathbb{R}$ is an invertible function, is it necessary that the function has to be strictly monotonic without any additional condition?
For invertibility to hold, we have to ensure that it's a bijective function on $\mathbb{R}$.
Now, let's say, a function is continuous and has a convex up form starting from $-\infty$ and is finally asymptotic at $y=5$. At point $x=3$, it has a jump discontinuity such that $(x,y)=(3,8)$. Can such a function satisfy the conditions of $f$ in question? I don't think so as the codomain is $\mathbb{R}$ and therefore, it has to be surjective on the codomain I guess. Like, we won't be able to find $f^{-1}(12)$. Right? Or, there's no relationship with the range and codomain here?
Can this be a suitable example to the function in question? (Please see the picture below)
In the picture, the green coloured circles represent open intervals and the filled-blue circles represent closed intervals.
Best Answer
Your function is clearly injective and surjective, hence is bijective, hence invertible, so everything is OK.
The strictly monotone bit is needed when your function is continuous by IVT.