In any neighbourhood of 1 there are rational and irrational points, so there

are points arbitrarily close to 1 where f(x) is arbitrarily close to 5, and other

points arbitrarily close where f(x) is close to 7, so f must be discontinuous

at x=1. So if you want you can find a sequence x_n of rational points

converging to 1 and for this sequence f(x_n), similarly you can find a

sequence of irrational points y_n which converges to 1 and for this sequence

f(y_n) converges to 7, thus proving that f is not continuous at x=1.

All sequences x_n points all converging to 2 have f(x_n) converging to 10, so

f(x) is continuous at x=2.

Similarly the function is continuous whenever 5x = x^2 + 6, so it is

continuous at x=2 and x=3.

The above should be convertible into whatever form you use to demonstrate

continuity without too much difficulty.

RonL