A function that is not?

Calculus Level 5

$f(x) = \begin{cases} \sin x & \text{for } x \in \mathbb{Q} \\ (\ln x)^2 & \text{for } x \notin \mathbb{Q} \end{cases}$

Consider the function $f$ defined over the set of positive real numbers.

Which of the following statements about the function $f$ is true?

Notations :

$\mathbb Q$ denotes the set of rational numbers .
$\mathbb R$ denotes the set of real numbers .

f

is continuous at a finite number of points, nowhere differentiable

f

is nowhere continuous, nowhere differentiable

f

is everywhere continuous, nowhere differentiable

f

is continuous between a given interval

[a, b]

, nowhere differentiable

f

is nowhere continuous, everywhere differentiable

f

is everywhere continuous, differentiable at a finite number of points None of the other choices

1 solution

Efren Medallo
Oct 6, 2016

Both rational and irrational numbers are dense in $\mathbb{R}$ . That means, between any two rational numbers, there will exist an irrational number. Similarly, between any two irrational numbers, there will exist a rational number.

Basically, the graph of this function will look like this

which, by the way, are not continuous graphs (again check the first paragraph).

Now, as you can see, there are two intersections for $\sin {x}$ and $\ln^2 x$ . Now, these intersections may be verified to be the only continuous points in the graph, because, as shown, they are the only points where $\lim {f(x)}$ exists, and that this limit at these points $x=a$ and $x=b$ are $f(a)$ and $f(b)$ , respectively.

Now, given that these are the only points where $\text{f}$ is continuous, we can now investigate for its differentiability.

Notice that the graph of f does not give us an assurance whether the number next to the continuous points are rational or not. That is, we can never tell where the graph leads once it leaves the points of continuity. This leaves us with the premise that $\text{f}$ is indeed nowhere differentiable .

The domain of the function $f$ is the positive reals $(0,\infty)$ , not $\mathbb{R}$ . A better way of proving non-differentiability at the two points of continuity is to note that the derivatives of $\sin x$ and $\ln^2x$ are different at these points.

Mark Hennings - 4 years, 8 months ago

Thanks. I see that this problem has been edited.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the menu. This will notify the problem creator (and eventually staff) who can fix the issues.

Brilliant Mathematics Staff - 4 years, 8 months ago

Good problem

Indraneel Mukhopadhyaya - 4 years, 8 months ago

A function that is not?

1 solution

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