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Evaluate whether a function $f$ is: 1. continuous at 0; 2. differentiable at 0

a) $$ f(x) = \left\{ \begin{array}{ll} 0 & \quad x \space rational \\ 1+x & \quad x \space irrational \end{array} \right. $$ b) $$ f(x) = \left\{ \begin{array}{ll} 0 & \quad x \space rational \\ x(1+x) & \quad x \space irrational \end{array} \right. $$ c) $$ f(x) = \left\{ \begin{array}{ll} 0 & \quad x \space rational \\ x^2(1+x) & \quad x \space irrational \end{array} \right. $$

Apparently, one can use differentiation rules to solve this problem. But I have got no clue what x rational and x irrational means

Thanks in advance

Hopeful Student
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  • $x$ rational means it can be written as $a/b$, where $a,b$ are integers. So for example, $3/5$ is rational. $x$ irrational means it is a real number but cannot be written in this form. $\pi, \sqrt{2}$ are examples of irrational numbers. – Ethan Alwaise Nov 28 '15 at 09:38

1 Answers1

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A number is called rational iff it is the fraction $p/q$ of some integers $p,q$ with $q \neq 0$; otherwise it is called irrational. For instance, the number $4$ is rational; for we have $4 = 8/2$. But the number $\sqrt{2}$ is not rational (can you prove this?).

I will do (a) and (b) for your reference; you may thus try to do (c) yourself, for the reason that the questions look like exercises.

For (a) observe that $|f(x) - f(0)| > 1$ for all irrational $x > 0$; hence $f$ is not continuous from the right at $0$, implying that $f$ is not continuous at $0$. Hence $f$ is not differentiable at $0$ (why?).

For (b) observe that, if $x \in \Bbb{R}$, then $|f(x) - f(0)| \leq |x||1+x|$; if in addition $|x| < 1$, then $|x+1| \leq |x|+1 < 2$, implying $|x||1+x| < 2|x|$; given any $\varepsilon > 0$, we have $2|x| < \varepsilon$ if in addition $|x| < \varepsilon/2$. Hence, for every $\varepsilon > 0$, if $|x| < \min \{1, \varepsilon/2 \}$ then $|f(x) - f(0)| < \varepsilon$; this shows that $f$ is continuous at $0$. To investigate differentiability of $f$ at $0$, note that, if $x > 0$ is irrational, then $$ \frac{|f(x) - f(0)|}{|x|} = |1+x| > 1; $$ hence $f$ is not differentiable at $0$.

Yes
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