Quick graphs question (1 Viewer)

[fine ambiguity]

For example if you're given a graph with a vertical asymptote and asked to sketch y = 1/f(x), is this graph defined for the asymptote? I was marked down in my trial for putting an open circle on this kind of question and my teacher told me 1/(undefined) was equal to zero. But the past HSC answers have it as being undefined. What is correct?

(I hope that made sense)
Thanks for any help

 

[100percent]

For example if you're given a graph with a vertical asymptote and asked to sketch y = 1/f(x), is this graph defined for the asymptote? I was marked down in my trial for putting an open circle on this kind of question and my teacher told me 1/(undefined) was equal to zero. But the past HSC answers have it as being undefined. What is correct?

(I hope that made sense)
Thanks for any help

i think it's undefined (open circle), since there is no x value on the asymptote, so how can you have 1/(nothing)??

 

[icycloud]

Consider tan(x),
1/tan(x) = cot(x)

tan(pi/2) is undefined,
however cot(pi/2) = 0.

Do you put an open circle at x = pi/2 on the graph of y = cot(x)? I think not...

Just some food for thought .

 

[KFunk]

I remember debating this some time ago and the conclusion we ended up with was that you can safely treat undefined x values of f(x) as zeros of 1/f(x). Examples exist where this is not the case but I don't think we have to worry about it for the level we're working at and given the kind of functions we ussually deal with.

 

[100percent]

hmm, i find it weird, coz they don't give you a defined x,
like if you square 5 or -5 both answers are 25
but if you square root 25, it's only 5.

but i'll stick with what kfunk said and put it as zero.
i guess you can assume that those points are at ±infinity and 1/infinity = zero?

 

[robbo_145]

looking at the suggested answers to Q3)a)i) from 2003 HSC 1/undefined remains undefined. However this could be because they do not explicitly give you the equation of the graph.

 

[underthebridge]

when you have a square root of something, it is ALWAYS positive. Hence when you say x^2=5 u put x=+ or - root 5, implying that the root 5 itself is positive.