C3 Quotient Rule help?

Original post by hash007

Did you get this for the second part? $(5,6e^{-5}) and (1,-2e^{-1})$

I found the second bit where you had to differentiate far simpler than the first part!

The first derivitive is

-e^{-x} (x^2 -6x +5)

and the second derivitive is

e^{-x} (x^2 - 8x +11)

If you have done all of the other stuff correctly then your answer is fine. (:

Reply 22

Original post by omgilyx

That's what I said. -.-

No, you suggested that he divided throughout by e^-x; which I pointed out was a pointless step.

Reply 23

Original post by H.C. Chinaski

No, you suggested that he divided throughout by e^-x; which I pointed out was a pointless step.

Umm so how else do you get rid of it? Numbers don't randomly disappear, you know?

Reply 24

Original post by omgilyx

Umm so how else do you get rid of it?

Let me explain ...

If A x B = 0 then either A=0 and/ or B=0.
No division required.
I have taught this stuff for long enough to know precisely how it works.

Original post by omgilyx

Numbers don't randomly disappear, you know?

well firstly, e^-x is not a number, depending on context it may be a term in an expression, it may be a function etc ..., e^-x will of course take a numeric value when you assign a value for the independent variable.

Your suggested approach of "dividing throughout by e^-x" is often used by weaker students and unfortunately they frequently do exactly the same thing in cases like
sin x. (x^2-9) = 0; subsequently losing a whole load of solutions.

Reply 25

Original post by H.C. Chinaski

Let me explain ...

If A x B = 0 then either A=0 and/ or B=0.
No division required.
I have taught this stuff for long enough to know precisely how it works.

well firstly, e^-x is not a number, depending on context it may be a term in an expression, it may be a function etc ..., e^-x will of course take a numeric value when you assign a value for the independent variable.

Your suggested approach of "dividing throughout by e^-x" is often used by weaker students and unfortunately they frequently do exactly the same thing in cases like
sin x. (x^2-9) = 0; subsequently losing a whole load of solutions.

Because 0/A = 0 or 0/B = 0.

Reply 26

e^{-x}

can never equal zero, chaps.

Reply 27

Original post by Perpetuallity

e^{-x}

can never equal zero, chaps.

Which is why it's perfectly viable to divide through by it.

Reply 28

soup

Original post by Perpetuallity

e^{-x}

can never equal zero, chaps.

Yes but the point is that you shouldn't tell people to divide through by

e^{-x}

because people may get in to the habit of dividing through by terms which do in fact equal zero and they can miss a possible solution. It is much better to write

e^{-x} = 0

and then say that there are no solutions to that equation.

Reply 29

Original post by omgilyx

Which is why it's perfectly viable to divide through by it.

Original post by soup

Yes but the point is that you shouldn't tell people to divide through by

e^{-x}

because people may get in to the habit of dividing through by terms which do in fact equal zero and they can miss a possible solution. It is much better to write

e^{-x} = 0

and then say that there are no solutions to that equation.

My point is that

e^{-x}

cannot be zero, therefore

x^2 -4x +1

must be equal to zero. I'm not condoning dividing through by it at all.

Reply 30

hash007

Original post by Perpetuallity

e^{-x}

can never equal zero, chaps.

Do I need to know why it doesn't = 0 for C3 maths?

Reply 31

nuodai

Original post by hash007

Do I need to know why it doesn't = 0 for C3 maths?

Yes; in fact, you should know from C1 Maths why powers of non-zero numbers are always non-zero.

Reply 32

Original post by hash007

Do I need to know why it doesn't = 0 for C3 maths?

Yes, you should know. You'll find it quite intuitive if you try to find a value of x for which

e^{-x} = 0

. (You won't be able to find a value :tongue:

)

On the graph of

e^{-x}

, the function can approch zero, but it never get's there, i.e. there is an asymptote at y=0.

And so the limit as x tends to infinity of

e^{-x}

is 0

(edited 12 years ago)

Reply 33

Original post by Perpetuallity

My point is that

e^{-x}

cannot be zero, therefore

x^2 -4x +1

must be equal to zero. I'm not condoning dividing through by it at all.

Lol, you don't have to condone it; either way works.

Reply 34

Original post by omgilyx

Because 0/A = 0 or 0/B = 0.

I really am not sure why you want to continue with this ... however if you really are struggling to understand the implications of A x B x C x .... x Z=0 ... and can only proceed by performing a pointless division then carry on by all means.

The fact of the matter is that there are many bright students on here who do not need the waters muddying by what is essentially bad advice and indeed something that only weaker students would do. Sorry, but that really is how it is.

Reply 35

Original post by omgilyx

Which is why it's perfectly viable to divide through by it.

Being viable and being useful are not the same thing.

Reply 36

Oh my Ms. Coffey

Original post by hash007

Well I didn't really use it much in C1 + C2? I don't recall doing many questions involving it, and I got an A at AS Maths.

Lol, this is a howler worse than Torres's.

Reply 37

IrrationalNumber

Original post by omgilyx

Because 0/A = 0 or 0/B = 0.

That is not correct. The reason AxB = 0 implies A=0 or B=0 is because the real numbers form an euclidean domain. It's certainly not because 0/A = 0 or 0/B = 0 (where does this even come from?)

Original post by H.C. Chinaski

Being viable and being useful are not the same thing.

To be fair, dividing by e^(-x) does yield the correct answer, and can be justified by saying that e^(-x) is non zero.

Reply 38