Core 1 - Maxima and Minima

So I was strolling through some core 1 revision until I hit a brick wall; I know that I should have sufficient knowledge to complete the following question, but it has seemed to prove quite challenging (For me atleast). I'll outline the question and try to show you what I've done so far:

(For anyone on OCR, this is from the Core 1 and 2 textbook, revision exercise 2)

Q) Deduce the coordinates of the stationary point on the following curve:

y = \frac{2}{x - 1} - \frac{2}{(x-1)^2}

So what did I foolishly do?

I thought right:

y = \frac{2}{x - 1} - \frac{2}{(x-1)^2}

Therefore:

y = \frac{4}{(x - 1)^2} - \frac{2}{(x-1)^2}

\rightarrow \frac{2}{(x - 1)^2}

I think I've already gone wrong :s-smilie:

(edited 13 years ago)

Reply 1

little_wizard123

19

Original post by Femto

x

The amount of mistakes in that is crazy :P:

The line after therefore: you just squared the first term?

Reply 2

boromir9111

17

You're supposed to multiply by (x-1)....

Reply 3

Femto

OP

8

Aah right sorry my mistake

Reply 4

Dekota-XS

2

When it says 'deduce...' it means, you're not meant to do any more arithmetic, you're meant to just write down the coordinates. I presume, the previous parts of the question required to draw the curve?

Reply 5

lubus

12

common denominator dudee

Reply 6

Eloades11

17

looks like you had a good attempt at it, you were supposed to multiply top and bottom terms by (x+1) to get the same denominator for the 2 fractions so you could simplify. Think of it as this, multiplying by (x+1)/(x+1) which is just 1 isnt it? So you dont actually change it, just place it in terms of the next fraction. Squaring top and bottom doesnt quite have the same effect, which is what you done.

This is what the other guys said though, hope this helps

Reply 7

Femto

OP

8

Original post by Dekota-XS

When it says 'deduce...' it means, you're not meant to do any more arithmetic, you're meant to just write down the coordinates. I presume, the previous parts of the question required to draw the curve?

Well the previous part was to find the stationary points on the curve:

y = \frac{1}{x} - \frac{1}{x^2}

Which I could do. But I just couldn't find them for this curve.

(edited 13 years ago)

Reply 8

Dekota-XS

2

If Curve y = 1/x - 1/x^2 has stationary points at (x1, y1) and (x2, y2) then Curve y = 1/(x-1) - 1/(x-1)^2 has stationary points at (x1+1, y1+1) and (x2+1, y2+1). i.e. simply use the coordinates of the previous curve and add 1 to each coordinate to deduce the stationary points of the second curve.

(edited 13 years ago)

Reply 9

Femto

OP

8

Aah right yeah and the y value is stretched by 2.

I just want to apologise to everyone for wasting their time on such a simple question - I'm an idiot.

(edited 13 years ago)

Core 1 - Maxima and Minima

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