As we know, through the IVT, all cubic functions of the form $y=ax^3+bx^2+cx+d$ will have a range of $\mathbb{R}$. I've noticed how $y=ax^3+bx^2+cx+d$>>>$y=a*abs(x)x^2+bx^2+cx+d$ will turn a cubic into a twice differentiable function of range $r<n$ or $r>n$ that is the piecewise function of two cubics with the split at $x=0$. Here's a desmos page with all of these things and more. https://www.desmos.com/calculator/hun3zgzxrp
If anybody could explain why this is, I'd love to hear the answer, thank you.
When $x>0$, $|x|x^2=x^3$, which is differentiable with derivative $3x^2=3|x|x$. When $x<0$, $|x|x^2=-x^3$, which is differentiable with derivative $-3x^2=3|x|x$. You can check that $|x|x^2$ is also differentiable at $0$ with derivative equal to $0$, since $$\lim_{h \to 0} \frac{3|h|h^2-0}{h} = \lim_{h \to 0} 3|h|h =0.$$ This is also the value of $3|x|x$ at $x=0$. So $\frac{d}{d}(|x|x^2)=3|x|x$ everywhere. Similarly, you can check that $|x|x$ is differentiable with derivative $2|x|$. But $|x|$ is not differentiable (at $0$). So the function $a|x|x^2+bx^2+cx+d$ is twice differentiable when $a\neq 0$.