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Chapter 1: Problem 83

Show that the function \(f(x)=\left\\{\begin{array}{ll}0, & \text { if } x \text { is rational } \\ k x, & \text { if } x \text { is irrational }\end{array}\right.\) is continuous only at \(x=0\). (Assume that \(k\) is any nonzero real number.)

Short Answer

Expert verified
The given function \(f(x)\) is continuous only at \(x=0\) and discontinuous for all other \(x\).

Step by step solution

01

Definition of Continuity

A function \(f(x)\) is said to be continuous at a point \(x = a\) if \(\lim_{{x \to a^-}} f(x) = \lim_{{x \to a^+}} f(x) = f(a)\). We will apply this definition to our function to ascertain where it is continuous.
02

Analysis at \(x = 0\)

For \(x = 0\), we see that \(f(x) = 0\) for all rational \(x\) and \(f(x) = 0\) for all irrational \(x\). Therefore, at \(x = 0\), \(f(0) = \lim_{{x \to 0^-}} f(x) = \lim_{{x \to 0^+}} f(x) = 0\). Hence, the function is continuous at \(x = 0\).
03

Analysis at \(x 鈮� 0\)

For \(x 鈮� 0\), let's assume \(a\) is any non-zero number. When \(x\) approaches \(a\), for rational \(x\), \(f(x) = 0\), but for irrational \(x\), \(f(x) = kx\). This means that as \(x\) approaches \(a\), the limits from the positive and negative side do not coincide, implying \(\lim_{{x \to a^-}} f(x) 鈮� \lim_{{x \to a^+}} f(x)\). Therefore, this function is not continuous for \(x 鈮� 0\).

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Most popular questions from this chapter

Sketch the graph of the function. Use a graphing utility to verify your graph. $$ f(x)=\arctan x+\frac{\pi}{2} $$

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