Differentiation

Selphie

This is probably a stupid question, so bear in mind my absolute uslessness at maths while answering.
The book says: we will be interested in asymptotes of the form
y= f(x)/g(x) where both f(x) and g(x) are polynomials... i.e. 2x+1/x-3

So I have a question where the top line is not a polynomial. It's 2/x+3
Do the same rules apply?? I presume not... so can anyone tell me what to do?

ZorbaTehZ

Yup they do.

Vertical Asymptote, let the bit below the line equal zero (X+3=0)
Therefore the Vertical Asymptote is at x = -3.

Horizontal Asymptote, find the limit of the whole thing as x tends to infinity
lim(x->inf) 2/x+3
lim(x->inf) (2/x)/1+(3/x) . . .Divide everything by x
lim(x->inf) 0/1+0 . . .(1/x tends to zero as x tends to infinity)
= 0/1 = 0
Therefore Horizontal Asymptote is at y = 0.

Selphie

Thanks a lot Zorba

Darren1o1

Selphie wrote:

This is probably a stupid question, so bear in mind my absolute uslessness at maths while answering.
The book says: we will be interested in asymptotes of the form
y= f(x)/g(x) where both f(x) and g(x) are polynomials... i.e. 2x+1/x-3

So I have a question where the top line is not a polynomial. It's 2/x+3
Do the same rules apply?? I presume not... so can anyone tell me what to do?

You could use the quotient rule from the tables. Where

y=F(x)/G(x)

dy/dx (F'(x)g(x)-G'(x)F(x))/G(x)^2

Where G'(x) is G(x)dx and F'(x) is F(x)dx

You will probably need to use the quotient rule again to find the dfferentials of G(x) and F(x)

ZorbaTehZ

When above the line is a constant, not a function then the Horizontal Asymptote is always y = 0. You don't really need to remember this or anything, since doing to limits will give you the answer anyways.

Naikon

I presume you leave it in the format 0/x+3 then solve it using the quotient rule.
Dont take my word as 100% as I do differentiation in Ordinary maths so it might differ somewhat.

Darren1o1

Darren1o1 wrote:

You could use the quotient rule from the tables. Where

y=F(x)/G(x)

dy/dx (F'(x)g(x)-G'(x)F(x))/G(x)^2

Where G'(x) is G(x)dx and F'(x) is F(x)dx

You will probably need to use the quotient rule again to find the dfferentials of G(x) and F(x)

Sorry my bad read it wrong, thought you were looking for the differential

Naikon

ZorbaTehZ wrote:

Yup they do.

Vertical Asymptote, let the bit below the line equal zero (X+3=0)
Therefore the Vertical Asymptote is at x = -3.

Horizontal Asymptote, find the limit of the whole thing as x tends to infinity
lim(x->inf) 2/x+3
lim(x->inf) (2/x)/1+(3/x) . . .Divide everything by x
lim(x->inf) 0/1+0 . . .(1/x tends to zero as x tends to infinity)
= 0/1 = 0
Therefore Horizontal Asymptote is at y = 0.

got 0 too there a few minutes ago using the quotient rule:/

ninebeanrows

I really just don't like Differentiation and the questions are usually made much harder seen as everyone does the questions.

My advice do question 3/4/5 and Integration..

Integration is my favourite question as it is almost always standard and the part c's can actually be enjoyable as i know what i'm doing for a change.

JC 2K3

Q6&7 are easier than 1&2 IMO.

There's usually an absolute bitch of a part c in either 1 or 2.

4&5 are very easy and 3 is usually grand.

Selphie

JC 2K3 wrote:

Q6&7 are easier than 1&2 IMO.

Really?? I always lose my way in differentiation. Integration is easy enough, I much prefer it, and I count on 1 - 5 then to help me build up marks.

JC 2K3

Well put it this way. we've been doing algebra since first year, it gives them more of a "right" to give an absolute bitch of a question here than anywhere else.