Show that it's image is R and prove that it is an injective and find the tangent of the opposite function Q1.p1:
To show that this function is injective and that it`s imgae is R.
$$f\left(x\right)=x^3+3x+1$$
My solution: let's look at it's derivative: $f'\left(x\right)=3x^2+3\:>\:0$ 
and that's why it is monotonic, and from that we conclude that it is injecyive. About it's image: $\lim _{x\to \infty \:}f\left(x\right)\:=\:\infty $ and also $\lim _{x\to -\infty \:}f\left(x\right)\:=\:-\infty $ and from the defenition of the limits, we can choose $x_0>0$ and $x_1<0$ that for every $x>x_0$ and every $x<x_1$ (I write it as two conditions bcause of my laziness, but of course I mean them as two separate cases), there are such $M>0$ and $m<0$ for which $f(x)>M$ and $f(x)<m$. And thats why we can get any number from $R$ in the image. Is this explanation good enough or you would add to it something else, especially in the end? 
p2: To find the equation of the tangent $f^{-1}\left(x\right)$ at $x_0 = 1$: 
$$$$ So, I need to find the slope at  $x_0 = 1$ and the image of $f^{-1}\left(1\right)$ I do remember that there is a formula for derivative for inverse functions, that I want to find for the slop of the tangent at  $x_0 = 1$: $$\left(f^{-1}\left(x\right)\right)'=\frac{1}{f\:'\left(f^{-1}\left(x\right)\right)}$$ but I forgot how to use it and what it means, so I tried to find the derivative by those steps: $y(1) = 5$
$$y\left(x\right)=x^3+3x+1$$
$$y'\left(x\right)=3x^2+3$$
$$\frac{y'\left(x\right)-3}{3}=\left(x\left(y\right)\right)^2$$
$$+-\sqrt{\frac{y'\left(x\right)-3}{3}}=x\left(y\right)$$
Then because I need $y(1) = 5$ I take the positive equation and calculate $x(5)$ by that calculation: $y'(1)=6$ so $y'(5) = 1$ and after that calculate the slop by: $$x'\left(5\right)=\frac{1}{2\sqrt{\frac{y'\left(1\right)-3}{3}}}\cdot y''\left(1\right)=\frac{1}{2}\cdot 6=3$$
So eventually I get the equation: $$x\left(y\right)=1+3\left(y-1\right)$$
$$x\left(y\right)=3y-2$$ or $$y\left(x\right)=3x-2$$
Did I cheat in the whole thing and I had to use he formula that I didnt remember how to use it or this solution is also correct ?
Can somebody write the right solution for the second part please? 
Thank you.
 A: Question 1: Because the derivative is strictly monotonic and is defined and continous on $\mathbb{R}$ the function $f$ is injective. 
Question 2: According to the inverse function theorem:
"If $J \subseteq \mathbb{R}$ and $f$ is a function $f: J \rightarrow \mathbb{R}$ and $f$ is continous and $I=f(J)$ is an interval, than:
$f: J \rightarrow I$ is invertible if and only if $f$ is strictly monotonic on $J$ and in that case $f$ has a continous inverse $g: I \rightarrow J$.
Furthermore: if $f$ is differentiable on $J$ and $f'(x)\neq0 \ \forall x \in J$ than: $$g'(f(x))=\frac{1}{f'(x)}$$ for all $x\in J$. 
In this case we can take $J=\mathbb{R}$ and we see that $g'(f(x))=\frac{1}{3x^2+3}$. Taking $x_0=1$: $$g'(f(x_0))=\frac{1}{3x_0^2+3}=\frac{1}{6}$$.
A: To calculate the tangent of $f^{-1}\left(x\right)
 $ at $x=1
 $ we have to calculate $$y=\left(f^{-1}\left(1\right)\right)'\left(x-1\right)+f^{-1}\left(1\right)
 $$ and in this case we can use the formula you wrote $$\left(f^{-1}\left(1\right)\right)'=\frac{1}{f'\left(f^{-1}\left(1\right)\right)}.
 $$ Now note that trivially $$f\left(0\right)=1
 $$ then $$\frac{1}{f'\left(f^{-1}\left(1\right)\right)}=\frac{1}{f'\left(0\right)}=\frac{1}{3}
 $$ then the tangent is $$y=\frac{x-1}{3}.
 $$
