Transforming Cubics into Limited Range Cubics

Question

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.

It's not clear what your question is. Are you asking why they are equal for $x>0$ or $x<0$? Are you asking why the transformed function is twice differentiable? — kccu, May 09 '17 at 02:55

score 1 · Accepted Answer · answered May 09 '17 at 15:47

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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 $a\neq 0$.

answered May 09 '17 at 15:47

kccu

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if $a=0$, it's still twice differentiable. – Jacob Claassen May 10 '17 at 02:31
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@JacobClaassen I should have clarified - when $a=0$ it is infinitely differentiable, but when $a \neq 0$ it is only twice differentiable. – kccu May 10 '17 at 10:05

Transforming Cubics into Limited Range Cubics

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