If function f(x) = begin{cases} x, & text{if | vedclass

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(B) For $f(x)$ to be continuous at a point $x = a$,the limit $\lim_{x 	o a} f(x)$ must exist and be equal to $f(a)$.Consider the point $x = a$. For a

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$1$

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(B) For $f(x)$ to be continuous at a point $x = a$,the limit $\lim_{x 	o a} f(x)$ must exist and be equal to $f(a)$.Consider the point $x = a$. For any neighborhood of $a$,there exist both rational and irrational numbers.If $x$ is rational,$f(x) = x$,and if $x$ is irrational,$f(x) = 1 - x$.For the limit to exist at $x = a$,we must have $a = 1 - a$,which implies $2a = 1$,or $a = 1/2$.Now,check continuity at $a = 1/2$:$f(1/2) = 1/2$ (since $1/2$ is rational).However,in any neighborhood of $1/2$,there are irrational numbers $x$ such that $f(x) = 1 - x$,which approaches $1 - 1/2 = 1/2$.Wait,let us re-evaluate: $\lim_{x 	o 1/2} f(x)$ exists if the values of $f(x)$ for rational and irrational numbers approach the same value.For rational $x$,$f(x) = x 	o 1/2$.For irrational $x$,$f(x) = 1 - x 	o 1 - 1/2 = 1/2$.Since both limits are equal to $1/2$,the limit exists at $x = 1/2$.Thus,$f(x)$ is continuous only at $x = 1/2$.Therefore,the number of points of continuity is $1$.

If function $f(x) = \begin{cases} x, & \text{if } x \text{ is rational} \\ 1 - x, & \text{if } x \text{ is irrational} \end{cases}$,then $f(x)$ is continuous at how many points?

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