Find a constant L for Lipschitz condition There is a function $f(x) = x^3\sin(\frac{1}{x})$ and I need to do two things:
a) find a constant $L$ that satisfies Lipschitz condition for $x\in[\frac{1}{\pi},\frac{2}{\pi}]$
b) prove that there exists an inverse function to $f: (\frac{1}{\pi},\frac{2}{\pi}) \rightarrow \mathbb R$ and find domain of the inverse function
My attempts:
a) Lipschitz condition claims $\forall_{{x,y} \in (a,b)}$ $|\frac{f(x)-f(y)}{x-y}| \leq L$ . Of course this function is continuously differentiable (on the given interval), so by MVT we know that  $\exists_{c \in (a,b):}\frac{f(a)-f(b)}{a-b} = f'(c)$ so basically I need to find a maximum slope of the function $f$ on the given interval. The first derivative is $x(3x\sin(\frac{1}{x})-\cos(\frac{1}{x})$) and there are two ways: if $f'$ is monotonic on the given interval I only need to check values at the ends of the interval (Weierstrass theorem), and if $f'$ is not monotonic I need to find $c \in (a,b): f'(c) = \max$ so basically look for extrema of $f'$. Neither of these two approaches seems to be easy enough to apply them during the test (I would rather not use wolframalpha or any other drawing tool).
b) of course there exists an inverse function if the given function $f$ is both injective and surjective. But... how to check if this function is injective? I mean, the most obvious condition $f(x) = f(y) <=> x = y$ is not the easiest one to check. 
 A: (a)$f'(x)=x(3xsin(\dfrac{1}{x})-cos\dfrac{1}{x})$,so $f'(x)$is continous on $[\dfrac{1}{\pi},\dfrac{2}{\pi}]$and is bound.$\lvert f'(x)\rvert\le \dfrac{2}{\pi}(3\dfrac{2}{\pi}+1)$.Therefore $L$ can be $\dfrac{2}{\pi}(3\dfrac{2}{\pi}+1)$.
(b)By theorem of inverse function,we only need to find the interval of $f'(x)\ne 0$.$f'(x)=0\Leftrightarrow 3xsin\dfrac{1}{x}=cos\dfrac{1}{x}$.It's easy to know that $cos\dfrac{1}{x}\le 0,3xsin\frac{1}{x}\ge 0$.so $f'(x)\ne 0$ in $[\dfrac{1}{\pi},\dfrac{2}{\pi}].$so there exists a inverse function of $f$on interval $[\dfrac{1}{\pi},\dfrac{2}{\pi}].$
A: The way the question is phrased, you're asked to find any constant $L$ that works, not necessarily the optimal one. For $x\in[1/\pi,2/\pi]$, we have
$$\begin{align}
|f'(x)|&=\left|3x^2\sin\left(\frac 1x\right)-x\cos\left(\frac 1x\right)\right| \\
&\leq 3x^2\left|\sin\left(\frac 1x\right)\right|+x\left|\cos\left(\frac 1x\right)\right| \\
&\leq 3x^2+x\leq 4
\end{align} $$
so $L=4$ will work. For part (b), $f'(x)>0$ since $\cos(1/x)<0$, $\sin(1/x)>0$, and $x>0$ when $x\in(1/\pi,2/\pi)$. Hence, $f$ is strictly increasing and injective. Note that $f$ is not surjective on $\mathbb{R}$. You're asked to find the domain of the inverse function which is $f\left[\left(\frac 1\pi,\frac 2\pi\right)\right]=\left(f\left(\frac 1\pi\right),f\left(\frac 2\pi\right)\right)$ since $f$ is increasing and continuous.
