A math odyssey
Today I spent two hours on calculus exercises. Among other things, this involved drawing thirteen graphs.1 That part was a bit tiresome, but I did learn from it. My favorite was the graph of $$f(x) = \frac{3x+|x|}{x}\text{,}$$ which didn’t look at all the way I had imagined before thinking it through. Click here to see the graph.
Cute function!
At the risk of assigning homework in someone else’s blog:
For some reason, this function reminds me of
$$
f(x)=\begin{cases}
+x^2,&\text{\ if $x\ge0$}\\
-x^2,&\text{\ if $x<0$.}
\end{cases}
$$
This function looks qualitatively like $x^3$, but for those of you who know a little calculus, what happens when you take $f'(x)$? $f''(x)$?
And for those of you who don't know calculus, is there a way to write $f(x)$ using absolute values instead of cases? Are there any other functions like $f(x)$ that are worth playing with?
Finally, for those of you being particularly careful, don't worry about the point $x=0$ unless you feel like being particularly particularly careful.
And finally finally, for those of you who are Olly, it won't bother me a bit if you delete this comment if I'm overstepping my bounds as a reader.
Nah, this is a forum for discussion, too. You are allowed to pose problems. I will think about this when I get an opportunity.
Is the use of the circles at the undefined zero value of x a standard graphing convention?
Yes.
Tim, I don’t remember enough about differentiation to combine your suggestions to the two groups, and to differentiate f(x) = x âĸ |x|, though it seems it should look like f'(x) = 2x
Oh, no it should not, it should look like f'(x) = 2âĸ|x|, since at x<0, the slope is positive
then, I can’t differentiate f'(x), but it looks as if it would bring us back (sorta) to the Graph of the Day
But then you can differentiate it, right?
$$
f'(x)=\begin{cases}
+2x,&\text{if $x\geq0$}\\
-2x,&\text{if $x<0$}\\
\end{cases}
$$$$
f^{”}(x)=\begin{cases}
+2,&\text{if $x>0$}\\
-2,&\text{if $x<0$}\\
\text{DNE},&\text{if $x=0$}
\end{cases}
$$
which is indeed almost like Olly's function.
There's no reason the derivative should have a single simple formula. If you wanted, though, you could use the idea in Olly's function to try to write $f'$ and $f''$ using absolute values instead of cases.
What I love about this is that this $f(x)$ looks a lot like $c(x)=x^3$, but it gets funky when you start differentiating. The function $c(x)$ has derivatives of all orders, but $f'(x)$ doesn't have a derivative at $x=0$, and $f''(x)$ isn't even continuous at $x=0$.
So $|x|$ is continuous everywhere, but doesn't have a derivative at $x=0$. Our $f(x)$ is everywhere continuous and differentiable, but it doesn't have second derivative at $x=0$. Could there be a function that has 0th – 4th derivatives everywhere, but not a 5th derivative at $x=0$? Or 0th – 16th and 17th?
(I only picked 4th because, for those who don't know, today is Star Wars Day. May the Fourth be with you!)
OK, the cases got messed up. It should have said that $f'(x)$ is either $2x$ or $-2x$, and that therefore $f”(x)$ is either $2$ or $-2$.
I’ve tried to fix them. Let me know if they still aren’t right.
đđ
I just love breathing in the energy of these great minds.
Leave a Reply