Derivative of inverse that can't be expressed using elementary functions?

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R} $be defined as $f(x)=3x-\sin{x}$ $\\$

a) Prove $f$ is a bijection and its inverse is a differentiable function. $\\$

b) Determine, if they exist: $$(f^{-1})^{\prime}(3\pi)\\$$ $$\lim_{x\to\infty}({f^{-1})^{\prime}}(x)$$

I've successfully proven the a) part(I think so, at least) but I'm not quite sure what to do about $f^{-1}$ since it can't be written using elementary functions. How do I express its derivative then?

I'm not even sure what to do since everything I try gets me to that same expression involving $f^{-1}$. Any hint would be useful.

Answer 1

Hint:

The derivative of the inverse function theorem:

$$\left(f^{-1}(x)\right)'=\frac1{f'\left(f^{-1}(x)\right)}$$

Added on request: We easily have that $\;f'(x)=3-\cos x\;$ , so for any $\;x\in\Bbb R\;$ (as the given function is bijective, of course), we get that

$$\left(f^{-1}(x)\right)'=\frac1{f'\left(f^{-1}(x)\right)}=\frac1{\left[(3-\cos )\circ(f^{-1})\right](x)}$$

Let now $\;y\in\Bbb R\;$ be such that $\;f(y)=x\iff y=f^{-1}(x)\;$ , then

$$\left[(3-\cos )\circ(f^{-1})\right](x)=(3-\cos)(f^{-1}(x))=(3-\cos)(y)=3-\cos y$$

and also the upmost important relation $\;y\to\infty\iff x\to\infty\;$ .

For the case $\;x=3\pi\;$ , we get

$$\left(f^{-1}(3\pi)\right)'=\frac1{f'(f^{-1}(3\pi))}=\frac1{3-\cos y}$$

where $\;y\in\Bbb R\;$ is such that $\;f(y)=3\pi\iff y=\pi\;$

Answer 2

Let $f(x) = 3x - \sin(x) $ and $f'(x) = 3 - \cos(x)$

We need to find $f=3\pi$, which is also $f^{-1}(3\pi)$

$$ 3\pi = 3x - \sin(x) \implies x = \pi \implies f^{-1}(3\pi) = \pi $$

so using the inverse derivative formula

$$ \left(f^{-1}(3\pi)\right)' = \frac{1}{f'(f^{-1}(3\pi))} = \frac{1}{f'(\pi)} = \frac{1}{4}$$

In regards to the limit, you need to find:

$$ \underset{x \to \infty}{\lim} \frac{1}{f'(f^{-1}(x))} $$

and since $f'$ is periodic and $f^{-1}$ is defined everywhere in $\mathbb{R}$ then the limit does not exist.

Answer 3

Use the formula for the derivative of the inverse function: $$\bigl(f^{-1}\bigr)'(x)=\frac1{f'(y)}=\frac1{3-\cos y},\quad \text{where}\enspace y=f^{-1}(x).$$ Here $f^{-1}(3\pi)=\pi\;$ (and more generally $f^{-1}(3n\pi)=n\pi$, thus $$\bigl(f^{-1}\bigr)'(3\pi)=\frac1{4}.$$

As to the limit at infinity, the general formula shows $\;\dfrac1{y-\cos y}$ has no limit when $x$, hence $y$ tends to $+\infty$.