MHB Continuity of the Inverse Function

[bsmithysmith]

I just started Calculus 1, a summer quarter that's compressed and I'm having trouble understanding a theorem that state continuity of the inverse function. Within my textbook, it mentions "If f(x) is continuous on an interval I with range R, and if inverse f(x) exists, then the inverse f(x) is continuos with domain R".

Not understanding it too much, but from what I can derive from it is that when a function has an inverse, it follows the same rules of the function; if that makes sense.

Lets say:

$$y=\frac{-2}{x-5}$$
and the inverse
$$y=\frac{5x-2}{x}$$

Then they both are continuous with their domains R. Maybe I'm not getting it, or can't put it in words, but it's a little confusing to me. They are the same equations, although inverses, but follow the same properties of continuity.

I need simplification because I also don't understand what I'm talking about, myself.

 

[Prove It]

I just started Calculus 1, a summer quarter that's compressed and I'm having trouble understanding a theorem that state continuity of the inverse function. Within my textbook, it mentions "If f(x) is continuous on an interval I with range R, and if inverse f(x) exists, then the inverse f(x) is continuos with domain R".

Not understanding it too much, but from what I can derive from it is that when a function has an inverse, it follows the same rules of the function; if that makes sense.

Lets say:

$$y=\frac{-2}{x-5}$$
and the inverse
$$y=\frac{5x-2}{x}$$

Then they both are continuous with their domains R. Maybe I'm not getting it, or can't put it in words, but it's a little confusing to me. They are the same equations, although inverses, but follow the same properties of continuity.

I need simplification because I also don't understand what I'm talking about, myself.

Well first of all, it should be obvious that NEITHER of those functions has a domain of $\displaystyle \begin{align*} \mathbf{R} \end{align*}$.

Anyway, all that the theorem you're referring to is saying is that provided your original function is one-to-one on its domain (which is a requirement to have an inverse function), if your original function is continuous, so is the inverse function.