Irrational Exuberance

Part 1: Let us prove first that $f(x)$ defined in that way is continuous at 0 and 1, and discontinuous at any other number.

Continuity at 0:

Let us consider any number $x$ such that $-1<x<1$ then $x^{3}\leq f(x) \leq x^{2}$ . Then using the Sandwich theorem we get that $\lim_{x\to 0}f(x)=0=f(0).$ so the function $f(x)$ is continuous at 0.

Continuity at 1:

We need to consider two cases: the case when $x>1$ and the case when $0<x<1$ . If $x>1$ then $x^{2}\leq f(x) \leq x^{3}.$ If $0<x<1$ then $x^{3}\leq f(x) \leq x^{2}.$ Therefore, using the Sandwich theorem again, the one-side limits $\lim_{x\to 1+}f(x)=1=f(1)$ and $\lim_{x\to 1-}f(x)=1=f(1)$ , so the function $f(x)$ is continuous at 1.

Discontinuity at any number $c$ different from 0 and 1:

If $c$ is any number different from 0 and 1, then we can consider two sequences ${\alpha_{n}}$ and ${\beta_{n}}$ , where any number $\alpha_{n}$ is rational and any $\beta_{n}$ is irrational such that $\lim_{n\to \infty} \alpha_{n}=c$ and $\lim_{n\to \infty} \beta_{n}=c$ . Then $\lim_{n\to \infty}f(\alpha_{n})=\lim_{n\to \infty}(\alpha_{n})^{2}=c^{2}$ and $\lim_{n\to \infty}f(\beta_{n})=\lim_{n\to \infty}(\beta_{n})^{3}=c^{3}$ . Since $c \neq 1$ and $c\neq 0$ then $c^2\neq c^3,$ and, therefore, the $\lim_{x \to c}f(x)$ does not exist. Therefore, $f(x)$ is discontinuous at $c.$

Part 2: Now, we will prove that the function $f(x)$ is differentiable only at $0$ :

Of course, $f(x)$ can not be differentiable at any point where it is discontinuous. So $f(x)$ might be differentiable only at 0 and 1. The next thing I will do is to prove that $f(x)$ is not differentiable at 1 and it is differentiable at 0.

Not differentiable at x=1:

In a similar way to what we did before, we are going to consider two sequences tending to 1, one is going to be ${\alpha_{n}}$ formed by rational numbers different from 1, and the other is going to be ${\beta_{n}}$ formed by irrational numbers different from 1. Then $\lim_{n\to \infty}\frac{f(\alpha_{n})-f(1)}{\alpha_{n}-1}=\lim_{n\to \infty}\frac{(\alpha_{n})^{2}-1}{\alpha_{n}-1}=2$ and $\lim_{n\to \infty}\frac{f(\beta_{n})-f(1)}{\beta_{n}-1}=\lim_{n\to \infty}\frac{(\beta_{n})^{3}-1}{\beta_{n}-1}=3.$ This proves that the limit of the difference quotient of $f(x)$ does not exist.

Differentiability at x=0:

Since $f(0)=0$ , $\frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}$ . For $-1<x<1$ , as we have stated before, $x^3\leq f(x) \leq x^2.$ Considering this inequality for a $x$ in the given interval , and $x\neq 0$ and dividing by $x,$ we get $x^{2}\leq \frac{f(x)}{x} \leq x.$ Now using the Sandwich theorem we get that $\lim_{x\to 0} \frac{f(x)}{x}=0$ , so the function $f(x)$ is differentiable at x=0.

Summarizing:

The function $f(x)$ is continuous at 0 and 1 only, and differentiable only at 0. So the answer to the problem is 3.