# If $f$ has graph satisfying $|y|=|x^2-x^3|$, at how many points must it be differentiable?

I'm trying to do problem 2 here.

Let $f(x)$ be a function defined for all real $x$ such that the coordinates of each point of its graph satisfy $|y|=|x^2-x^3|$. The total number of points at which $f(x)$ must be differentiable is

(A) none

(B) $1$

(C) $2$

(D) $3$

(E) infinite

The correct answer is B, but I'm completely stumped as to why that is. Is there an explanation for this answer?

-
I see this relation presents no function in $\mathbb R^2$. – S. Snape Sep 16 '12 at 7:13
I guess the question should be "the minimal number of points at which $f$ must be differentiable is". – Philippe Malot Sep 16 '12 at 7:20
@Babak: It describes an infinite family of functions, one for each possible choice of algebraic signs. – Brian M. Scott Sep 16 '12 at 7:39
@girianshiido: No rewording is necessary: that’s the normal interpretation of the question as it is actually written. – Brian M. Scott Sep 16 '12 at 7:40
@BrianM.Scott: Really? I guess I just learn something today. Thank you! – Philippe Malot Sep 16 '12 at 15:13

You have a function $f:\Bbb R\to\Bbb R$ such that $|f(y)|=|x^2-x^3|=x^2|1-x|$ for all $x\in\Bbb R$. There are many such functions: for each $x$ except $0$ and $1$ you can choose either $x^2|1-x|$ or $-x^2|1-x|$ to be the value of $f(x)$.

Suppose that $a\ne 0$ and $a\ne 1$, so that $f(a)\ne 0$. Then $f$ need not even be continuous at $a$: $\lim\limits_{x\to a}f(x)$ exists if and only if there is an $\epsilon>0$ such that $f(x)$ has the same algebraic sign for all $x\in(a-\epsilon,a)\cup(a,a+\epsilon)$, and that limit is $f(a)$ if and only if $f(a)$ also has that algebraic sign. To see what can go wrong, imagine that you set

$$f(x)=\begin{cases} x^2|1-x|,&\text{if }x\text{ is rational}\\\\ -x^2|1-x|,&\text{if }x\text{ is irrational}\;; \end{cases}$$

This function clearly can’t be continuous at any irrational.

Now what happens at $x=1$? It’s not hard to see that since $\lim\limits_{x\to 1}x^2|1-x|=0$, $f(x)$ is at least continuous at $x=1$, but must it have a derivative there? Must the limit

$$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1}=\lim_{x\to 1}\frac{f(x)}{x-1}$$

exist? $\dfrac{f(x)}{x-1}=\dfrac{x^2|1-x|}{x-1}=\pm x^2$, depending on whether $f(x)$ is $x^2|1-x|$ or $-x^2|1-x|$. What if we chose to set $f(x)=x^2|1-x|$ for all $x$? Then $$\lim_{x\to 1^-}\frac{f(x)}{x-1}=\lim_{x\to 1^-}\frac{x^2|1-x|}{x-1}=\lim_{x\to 1^-}\frac{x^2(1-x)}{x-1}=-1\;,$$ but $$\lim_{x\to 1^+}\frac{f(x)}{x-1}=\lim_{x\to 1^+}\frac{x^2|1-x|}{x-1}=\lim_{x\to 1^+}\frac{x^2(x-1)}{x-1}=1\;,$$ and $f$ is not differentiable at $x=1$.

The only point that remains to be considered is $x=0$. As with $x=1$, it’s not hard to check that $f$ must at least be continuous at $x=0$. What about the derivative? That would be

$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to 0}\frac{f(x)}x=\lim_{x\to 0}\frac{x^2|1-x|}x=\lim_{x\to 0}x|1-x|=0\;.$$

In other words, such a function $f$ must be differentiable at $x=0$, where its derivative must be $0$.

(This is a cute problem; I like it.)

-
I got where my wrong idea was. Yes there are many infinite families of functions satisfying the relation. Thanks Brian. I like your complete answer too (+1). :) – S. Snape Sep 16 '12 at 7:50
Thanks Brian, this is a very nice answer. – Nastassja Sep 16 '12 at 7:56
Nice answer indeed. – Philippe Malot Sep 16 '12 at 15:16

Let $A$ be the set of points $a$ for which $f(a) = a^2 - a^3$ and $B$ the set of points for which $f(b) = b^3 - b^2$. If we choose a point $x$, then either $x$ has a complete neighborhood within one of the two sets, and the function will be differentiable at $x$, or it doesn't, in which case we have to take a closer look.

If you choose $A$ to be the rational numbers and $B$ to be the irrational, for instance, then no point will have such a neighborhood, so that is no comfort.

If there is no point completely contained in one of the sets $A$ or $B$ (which is the same as saying neither $A$ nor $B$ contains any intervals), let's see what happens to the derivative at some $x$, defined as $$\lim_{x'\to x} \frac{f(x') - f(x)}{x' - x}$$ Now, let's say $x\in A$ (swap for $B$ for the other side of the argument). Then the limit will intuitively have two values. One for $x'\in A$, and one for $x' \in B$. If $f(x) \neq 0$ then the limit will approach $\frac{\pm \infty}{0}$ for $x'\in B$ (swap for $A$), so no differentiating in that case.

Thus we are left with two points of interest. $x = 1$ and $x=0$. For none of these points does the limit above go toward infinity, but depending on whether $x'$ is in $A$ or $B$, the fraction above is positive or negative. For $x=1$ this means the limit approaches both $+1$ and $-1$, so it does not exist.

For $x=0$, it approaches 0 no matter which way you put it, so the limit does exist, and we conclude that it is differentiable with $f'(0) = 0$

-

First, note that $|f(h)-f(0) - 0.h| = |f(h)| = h^2|h-1|$. If we choose $|h| < \min (\frac{\epsilon}{2}, 1)$, then $|f(h)-f(0) - 0.h| \leq \epsilon |h|$, hence $f$ is differentiable at $x=0$ with derivative $0$.

Now let $f(x) = x^2(x-1) (2 \cdot1_{\mathbb{Q}}(x)-1)$. Then $f$ satisfies the hypothesis, but is only continuous at $x=0$ and $x=1$. Hence $x=1$ is the only other possible location at which $f$ must be differentiable. Suppose $f$ is differentiable at $x=1$ with derivative $\alpha$. Then for each $\epsilon>0$, there must be a $\delta>0$ such that if $|h| < \delta$, then $|f(1+h)-f(1)-\alpha h| \leq \epsilon |h|$, or equivalently, $|(1+h)^2 (2 \cdot 1_{\mathbb{Q}}(1+h)-1)-\alpha| \leq \epsilon$. However, this cannot be true, as choosing $h=\frac{1}{n}$ (for $n$ sufficiently large) yields $\alpha = 1$, whereas choosing $h=\frac{\pi}{n}$ (for $n$ sufficiently large) yields $\alpha = -1$. hence $f$ is not differentiable at $x=1$.

It follows that $x=0$ is the only point at which $f$ must be differentiable.

-