1. ## Continuous Functions Proof

Define f : R R by f(x) = 5x if x is rational and f(x) = x² + 6 if x is irrational. Prove that f is discontinuous at 1 and continuous at 2. Are there any other point besides 2 at which f is continuous?

2. at x=1 any open neighborhood about it will contain both rational and irrational numbers, but the irrational part is converging to 7 while the rational part is converging to 5 hear.

however at 2 both parts converge to 10.
to find the other point consider where
$x^2+6=5x \Rightarrow x^2-5x + 6 =0 \Rightarrow (x-3)(x-2)=0$

3 might be another good place to look.

3. Originally Posted by bearej50
Define f : R R by f(x) = 5x if x is rational and f(x) = x² + 6 if x is irrational. Prove that f is discontinuous at 1 and continuous at 2. Are there any other point besides 2 at which f is continuous?
Hi bearej50.

To prove that $f$ is discontinuous at 1, let $\epsilon=1.$ Then for any $\delta>0,$ the interval $(1-\delta,\,1+\delta)$ contains irrational numbers. If $y$ is one of them, then $|f(y)-f(1)|=|y^2+6-5|=|y^2+1|\geq1=\epsilon.$

To prove that $f$ is continuous at 2, let $\epsilon>0$ be given. Then if $\delta=\min\left\{1,\frac\epsilon5\right\},$ consider $y$ in the interval $(2-\delta,\,2+\delta).$ If $y$ is rational, we have $|f(y)-f(2)|=|5y-10|=5|y-2|<5\delta\leq\epsilon.$ If $y$ is irrational, then $|f(y)-f(2)|=|y^2-4|=|y+2||y-2|<3\delta<\epsilon.$

There is one other point at which $f$ is continuous, and that is found by solving the equation $5x=x^2+6$ for $x.$

4. Or consider $(a_n) = 1+ \frac{\sqrt{2}}{n}$. Then $a_n \to 1$. But $f(a_n) \not \to f(1)$. Thus $f$ is discontinuous at $x = 1$. It seems that if both rational parts and the irrational parts are equal at $x$ then $f$ is continuous at $x$ (e.g. in the case of $x = 2$).