A hard question

@zeeshan @zubair 

I've seen these types of questions before in a module called convergence and continuity. There are few ways to answer that. I would personally use limits.

i.e. if you want to prove that f(x) is continuous at x=2, that basically means we have to prove that:

lim(x->2) f(x) = f(2)

where f(2) = 3(2)^2 - 2(2) + 1  = 9

To prove  that you can split up the lim of f(x) as the following.

f(x) = 3x^2 - 2x + 1.

so lim(x->2) [3x^2 -2x + 1] = lim(x->2)[3x^2] - lim(x->2)[2x] + lim(x->2)[1]   (sum and difference property)

= 3(lim(x->2)[x])^2 - 2lim(x->2)[x] + lim(x->2)[1]. (power and constant multiple property)

= 3(2)^2 - 2(2) + 1 

So we have shown that for $f\left(x\right)=3x^2-2x+1$, we have $f\left(2\right)=9$ and $\underset{x\to \mathrm{2)}}{lim(}f\left(x\right)=0$, therefore (by the definition of continuous at a number) $f\left(x\right)$ is continuous at $2$.