Prove the inverse function theorem

[wany]

Homework Statement

Suppose that f' exists and is continuous on a nonempty, open interval (a,b) with [itex]f'(x) \neq 0[/itex] for all [itex]x \in (a,b) [/itex].
I already proved that f is 1-1 on (a,b) and takes (a,b) onto some open interval (c,d).
(i)Show that [itex]f^{-1}[/itex] is continuously differentiable on (c,d).
(ii)Using the function f(x)=x^3 show that (i) is false if the assumption [itex]f'(x) \neq 0[/itex] fails to hold for some [itex]x \in (a,b) [/itex].

Homework Equations

The Attempt at a Solution

Just to make sure to prove that f is 1-1 I did this through:
Proof by contradiction: suppose f is not 1-1 on (a,b)
therefore the slope must change signs or be 0 for some x in (a,b). So f' must be 0 or change signs. Since f' is continuous f' will =0 for some value. But this is a contradiction. So f has to be 1-1.
since f' exists on (a,b), f:(a,b) [itex]\rightarrow[/itex] (c,d).(i)By definition, since f' is continuous on (a,b) then f is continuously differentiable on (a,b).
The Inverse Function Theorem let's us say since f' exists and is continuous on (a,b), that is for any [itex]x \in (a,b) [/itex], f'(x)=y such that [itex]y \in (c,d) [/itex].
Then from this can we say that [itex]f^{-1}[/itex] is differentiable on (c,d) or is another step required, such that f is onto (c,d)?

(ii) so I am not sure how to do this but we must show that at some point the derivative of f=0 implies that the inverse of f will not be continuous.

 

[fzero]

You need to relate [tex](f^{-1})'[/tex] to [tex]f'[/tex] and use the properties of [tex]f'[/tex].

 

[wany]

So by the inverse function theorem we know that [itex](f^{-1})'(c)=\frac{1}{f'(a)}[/itex]
assuming that c=f(a)