For which values of $a,\ b,$ and $c$ function $f(x)=ax^3+bx^2+cx$ is a Diffeomorphism. I need your help to determine all $a,b,c \in \mathbb{R}$ for which mapping $f:\mathbb{R} \to \mathbb{R}$, $$f(x)=ax^3+bx^2+cx$$ is a Diffeomorphism.
 A: $f'(x)=3ax^2+2bx+c$, $\Delta=4b^2-12ac$. $\Delta<0$ is i.e $b^2<3ac$, $f$ is strictly monotone, since $lim_{x\rightarrow -\infty}f(x)=-\infty$ if $a>0, +\infty$ if $a<0$ and $lim_{x\rightarrow +\infty}f(x)=+\infty$ if $a>0, -\infty$ if $a>0$, $f$ is bijective and locally invertible thus is a diffeomorphim.
$\Delta\geq 0$ implies there exists $x$ such that $f'(x)=0$, so $f$ is not a diffeomorphism since the differential of a diffeomorphism must be invertible.
A: A differentiable function $f: \Bbb R \to \Bbb R$ is a diffeomorphism if and only if:
$(1)$ $f'(x) > 0$ for all $x$, or $f'(x) < 0$ for all $x$.
$(2)$  $\lim_{x \to - \infty} f(x) = -\infty$ and $\lim_{x \to + \infty }f(x) = + \infty$, or $\lim_{x \to - \infty} f(x) = +\infty$ and $\lim_{x \to + \infty }f(x) = - \infty$
I leave the proof of this as an exercise for you.
Suppose that $a = 0$. If $b \neq 0$, then clearly $f$ is not a diffeomorphism. If $b =0$, then $f$ is a diffeomorphism iff $c \neq 0$.
Suppose that $a \neq 0$. We have $f'(x) = 3ax^2 + 2bx + c$. So condition $(1)$ is satisfied iff $b^2 - 3ac < 0$. Now, supposing that, we have that $f$ is a cubic polynomial, so it clearly satisfies $(2)$.
Therefore, $f$ is a diffeomorphism iff:


*

*$a = 0$, $b = 0$ and $c \neq 0$


or


*

*$a \neq 0$ and $b^2 < 3ac$.

