0

For my calculus exam, I need to be able to identify if a function is indifferentiable at any point without a graph. I thought this would be rather simple, but I messed up on the question x^(2/3) because I did not realize it had a "cusp" at x = 0.

How would I identify, or look for cusps based on the formula of a function, without graphing it?

I plugged in other similar formulas into a graphing calculator, like x^(1/3), x^(1/4), x^(3/4) but couldn't identify a simple pattern I could quickly use on my test

Kartik Chugh
  • 236
  • Look for points where the derivative has a limit of $\infty$ (or a limit of $-\infty$). – Ben Grossmann Jul 21 '16 at 02:14
  • 1
    Also if it's left and right derivatives at a point don't match then it doesn't have a derivative there. Obvious example: $y=|x|$. –  Jul 21 '16 at 02:17
  • @Bye_World by the definition of "cusp" that I'm used to, y=|x| wouldn't qualify. There might be different uses, though – Ben Grossmann Jul 21 '16 at 02:35
  • 1
    You know, I don't know that I've ever seen a definition of "cusp". I remember my professors just explaining that they are the "pointy" parts of the graphs. And $y=|x|$ is fairly pointy (less so than $x^{2/3}$ though). –  Jul 21 '16 at 02:39
  • I have never seen a clear definition either; my teacher just taught them to us as "corners" whereas some books divide them into "corners" (like absolute value) and "cusps" (x^2/3) – Kartik Chugh Jul 21 '16 at 21:43

2 Answers2

1

Probably your problem is that you are not checking the conditions for theorems to hold, and that is why you do not see where they fail.

$\frac{d(x^n)}{dx} = n x^{n-1}$ for every positive integer $n$.

$\frac{d(x^n)}{dx} = n x^{n-1}$ for every integer $n$ and variable $x \ne 0$.

$\frac{d(x^n)}{dx} = n x^{n-1}$ for every rational $n$ and real variable $x \ne 0$ such that $x^n$ is defined.

$\frac{d(x^n)}{dx} = n x^{n-1}$ for every real $n$ and real variable $x > 0$.

Clearly if you apply the appropriate theorem for $n = \frac23$, you will see that $( x \mapsto x^{2/3} )$ is guaranteed to have a derivative everywhere except $x = 0$. Thus you can focus on that single point and check whether it has a derivative there or not. In this case it does not, but in general it might or might not; you've to check.

Community
  • 1
user21820
  • 57,693
  • 9
  • 98
  • 256
  • So are these four fundamental theorems of applying the chain rule? – Kartik Chugh Jul 21 '16 at 04:52
  • @KartikChughヅ: No. The first is trivial by product rule and induction. The second also, after first proving that $\frac{d(x^{-1})}{dx} = -x^{-2}$ for $x \ne 0$. The third is most easily proven by first proving that $\frac{d(x^{1/k})}{dx} = \frac1k x^{1/k-1}$ for positive integer $k$ by implicit differentiation (equivalently chain rule) and then using induction and product rule again. The last is highly non-trivial to define (see http://math.stackexchange.com/a/1790356) not to say prove, so for now it suffices that you know how to use it correctly. – user21820 Jul 21 '16 at 06:43
  • @KartikChughヅ: And by the way $\frac{d(x^{1/3})}{dx}$ also does not exist; just because the graph is smooth does not mean that it is differentiable, since the linear approximation could be vertical. – user21820 Jul 21 '16 at 06:46
  • Sorry I didn't mean to say chain rule, that was the incorrect term. And thanks. – Kartik Chugh Jul 21 '16 at 21:41
  • 1
    @KartikChughヅ: Sure. And you're welcome! – user21820 Jul 22 '16 at 06:10
1

I need to be able to identify if a function is indifferentiable at any point

The common way to do that is to actually determine the derivative and inspect it for singularities. This is generally easy with elementary functions.

In your example:

$$ f(x) = x ^ \frac{2}{3} $$ $$ f'(x) = \frac{2}{3} x ^ \frac{-1}{3} = \frac{2}{3 \sqrt[3] x} \;\; for \;\; x \ne 0 $$

On cursory inspection (or by applying the definition), it's obvious that $f'(0)$ doesn't exist, so $f(x)$ is not differentiable at $0$.

How would I identify, or look for cusps based on the formula of a function, without graphing it?

How would you if you could graph it?

You can't draw an infinite graph at infinite resolution. Maybe the "cusp" is at $x = 10^{10^{10}}$, or maybe it's for $f(x) = 0.00001$.

Or, try graphing $f(x) = x \sin \frac{1}{x}$ and finding the "cusp" there.

dxiv
  • 76,497
  • So I should find the derivative and see where it doesn't exist? – Kartik Chugh Jul 21 '16 at 04:52
  • @KartikChughヅ Yes, since your stated goal is to identify if a function is indifferentiable at any point. – dxiv Jul 21 '16 at 05:02
  • @KartikChughヅ: No that's incorrect. If the derivative does not exist you cannot possibly find it. If you manage to find something that does not exist, it means your reasoning is flawed. Don't think that you can just get away with this kind of approach; otherwise you'll keep having the same problem in all areas of mathematics and not just calculus. – user21820 Jul 21 '16 at 06:35
  • @user21820 ..I meant find the general derivative of the function, and see at which points it cannot exist.. – Kartik Chugh Jul 21 '16 at 21:42
  • Since one cannot find something nonexistent, I'm not sure how I would go about doing that alleged "approach" anyways – Kartik Chugh Jul 22 '16 at 01:16
  • @KartikChughヅ: Oh that's fine. The alleged approach (which I've seen almost all high-school students and teachers do) is that you differentiate $x^{2/3}$ and get $\frac23 x^{-1/3}$ and then conclude that when $x = 0$ you have division by zero and so it does not exist. The problem is that if you have division by zero means you already made an illegal step in reasoning! If you don't do this then it's fine. – user21820 Jul 22 '16 at 06:07