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A hard question


Usman
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I've got no Idea how you do this. My older cousin gave it to me to have a go

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Usman
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Zeeshan Ahmed
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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(x)=3x2−2x+1, we have f(2)=9 and lim(x2)f(x)=0, therefore (by the definition of continuous at a number) f(x) is continuous at 2.

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Usman
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@zeeshan

I still dont understand how it works cos it's pretty much like subbing 2 in twice but thanks. Is this even in our spec cos I dont see anything on convergence and continuity

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Zeeshan Ahmed
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@usman

No its degree level maths, I dont think its in A-Level.

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