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I'm quite new to calculus and frankly speaking I'm an autodidact. Could you tell me how to solve this:

$$f: [\frac{\pi}{2}, \pi) \ni x \rightarrow \frac{1}{\sin x} \in \mathbb{R}$$

Prove that there exists a inverese of $f$. In what set is $f^{-1}$ differentiable? Calculate $f^{-1'}$

I would really appreciate a thorough explanation. Thank you.

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up vote 2 down vote accepted

For the existence, you may want to prove $f$ is monotone by examining $f^{\prime}$.

Find the range of $f$ (that will be the domain of $f^{-1}$) and examine where $f^{\prime}(x)\neq 0$. At the points $y=f(x)$ where $f^{\prime}(x)\neq 0$, $f^{-1}$ will be differentiable. In addition, $$(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}$$

Proof of the last equality:

Since $f$ is continuous and 1-1 so without loss of generality we can suppose that $f$ is increasing. So for $c\in (a,b)=D_f$ so that $f^{\prime}(c)\neq 0$, \begin{equation}0\neq f^{\prime}(c)= \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\Rightarrow \lim_{x\to c}\frac{x-c}{f(x)-f(c)}=\frac{1}{f^{\prime}(c)}\end{equation} Let $\epsilon>0$. Then, \begin{equation}\exists \delta_0>0: 0<\left|x-c\right|<\delta_0\Rightarrow \left|\frac{x-c}{f(x)-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\end{equation} In addition, $\exists \delta_1>0:[c-\delta_1,c+\delta_1]\subset (a,b)$. Let $\delta=\min{\left\{\delta_0,\delta_1\right\}}$

Let $y_1=f(c-\delta),y_2=f(c+\delta)$. Then since $f$ is increasing, \begin{equation}y_1<f(c)<y_2\Rightarrow \exists \delta_2>0:y_1<f(c)-\delta_2<f(c)<f(c)+\delta_2<y_2\end{equation} If $y\in (f(c)-\delta_2,f(c)+\delta_2)\subset (y_1,y_2)$ then by the Intermediate Value Theorem $\exists x\in (c-\delta,c+\delta):f(x)=y$. Thus, \begin{gather} 0<\left|y-f(c)\right|<\delta_2\Rightarrow \exists x\in (a,b):f(x)=y\text{ and } 0<\left|x-c\right|<\delta \Rightarrow \left|\frac{x-c}{f(x)-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\Rightarrow \notag\\ \left|\frac{f^{-1}(y)-f^{-1}(c)}{y-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\Rightarrow \lim_{y\to f(c)}\frac{f^{-1}(y)-f^{-1}(c)}{y-f(c)}=\frac{1}{f^{\prime}(c)} \end{gather} and so, $f^{-1}$ is differentiable at $f(c)$ and $(f^{-1})^{\prime}(f(c))=\frac{1}{f^{\prime}(c)}$

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Thanks a lot. I need to find the points where f'(x)=0 using Lagrange theorem. But how do I prove the last equality? – Bilbo Dec 18 '12 at 17:53
By Lagrange's theorem, you mean the Mean Value Theorem? – Nameless Dec 18 '12 at 17:54
Right. Thank you. – Bilbo Dec 18 '12 at 18:03

One approach is the one suggested by Nameless. Even more easily, you could notice that $x \mapsto \sin x$ strictly decreases from $1$ to $0$ as $x$ ranges from $\pi/2$ to $\pi$. Hence $x \mapsto 1/\sin x$ strictly increases, and it is injective.

Alternatively, given $y \in \mathbb{R}$, the equation $$\frac{1}{\sin x}=y$$ can be easily discussed for $x \in \left[ \frac{\pi}{2}, \pi \right)$. The quantity $1/y$ must lie in the interval $(0,1]$, i.e. $y \geq 1$.

The question about the differentiability of the inverse follows from the theorem of differentiation of the inverse function, as suggested by Nameless.

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