Seven at One Stroke wrote:Hmm...I'm not sure if my answer is correct so check with the answer book (haven't done math in a long time).

To have a point discontinuity, then the limit of the function approaching the discontinuity needs to be the same, i.e. it cannot be asymptotic (I had to look this up on the web, sad). Since the denominator is x-a, then towards the discontinuity, there's going to be a sign shift of the value of the function as x->a. So we have to make sure that the discontinuity the function approaches to from the left and the right is a point even though the two values next to the discontinuity has opposite signs, then it makes sense that the discontinuity should be at lim(a->0)=0, which is at the x-intercepts. Then you have that at a=1,2 or 3, the funciton has a point discontinuity. (Honestly, I got the graph of the function first, and formulated my reasoning second, so there's a huge chance that this is wrong, so do think it through yourself).

I had come to the same conclusion myself, but then i actually got stuck actually determining the values of a. How did you get that?

Ranbir wrote:Yay, maths. You're using unfamiliar language to me, but no matter, america in winning in the math language department, so I guess I should get to know it. We've already accepted the american billion as being 'the billion'

I agree with Seven.

I also plotted the graph, and set a=x. It automatically set it as 0. Therefore a must be equal or greater than 1.

God I hope thats right...my credibility as a mathematician is on the line...

If x=a, wouldn't the function be undefined?