[ralph243]

(Original post by NoFunnyBusiness)
Does anyone have a really hard differentiation or integration question that they could post on here?

I can't do this differentiation question from the text book - it's 12(iii) on exercise 4D.

A function is defined by f(x)= (e^2x)/(x^2)
Show that for a certain value of k,

f(x-1) = k((1/(x-1))^2)f(x)

State the value of k.

[Alasda1r]

(Original post by SS*)
Is this always the case? What about reflections?

Always true, reflections can be considered S stretches of sf 1. So, a reflection in the y axis is given by f(-x) so this would be a stretch in the x direction. In the x axis it's -f(x) so a stretch in the y by s.f. -1

[NoFunnyBusiness]

(Original post by ralph243)
I can't do this differentiation question from the text book - it's 12(iii) on exercise 4D.

A function is defined by f(x)= (e^2x)/(x^2)
Show that for a certain value of k,

f(x-1) = k((x/(x-1))^2)f(x)

State the value of k.

k((x/(x-1))^2) is the same as (kx^2)/(x-1)^2

k((x/(x-1))^2)f(x) = ((kx^2)/(x-1)^2)) (e^2x/x^2)= (ke^2x)/((x-1)^2)

f(x-1) = e^2(x-1)/(x-1)^2 = (e^2x * e^-2)/(x-1)^2 ......because 2(x-1) = 2x -2

Therefore if f(x-1) = k((x/(x-1))^2)f(x)

Then ke^2x = e^-2 * e^2x

k= e^-2

[NoFunnyBusiness]

Can anyone do the example on page 100, the second part of the question?

[ralph243]

(Original post by NoFunnyBusiness)
k((x/(x-1))^2) is the same as (kx^2)/(x-1)^2

k((x/(x-1))^2)f(x) = ((kx^2)/(x-1)^2)) (e^2x/x^2)= (ke^2x)/((x-1)^2)

f(x-1) = e^2(x-1)/(x-1)^2 = (e^2x * e^-2)/(x-1)^2 ......because 2(x-1) = 2x -2

Therefore if f(x-1) = k((x/(x-1))^2)f(x)

Then ke^2x = e^-2 * e^2x

k= e^-2

I just realized that I made a copy error from the text book...

It should read:
f(x-1) = k((1/(x-1))^2)f(x)
not
f(x-1) = k((x/(x-1))^2)f(x)

When I use this I get down to: e^-2 = k/x^2

The answer is k = e^-2 (like you said) but i can't get that answer :/

[TrampingRampage]

(Original post by SS*)
Is this always the case? What about reflections?

Yes, most of the time it doesn't really matter, but to transform a curve f(x) onto a specified given curve f(ax+b), these are the order of steps you would take.

I would reflect first in the y-axis so f(-x) first.

You could take a look at it through treating the insides of the brackets first (so reflect in y-axis f(-x), translate in x-direction f(x-1) and stretch parallel to x-axis f(2x)), and then sorting out the outside (reflect in x-axis -f(x), stretch parallel to y-axis 2f(x) and translate in y-direction 2+f(x))

Hope this helps

[nasira372]

http://www.mei.org.uk/files/papers/c309ju_4lkp.pdf

Can anyone explain Q4 please?

[ralph243]

(Original post by nasira372)
http://www.mei.org.uk/files/papers/c309ju_4lkp.pdf

Can anyone explain Q4 please?

At point b, x=0 and the y co-ordinate is b. Input these coordinates into the equation and you get: b = 2*|0-1|
re-arrange to get b=2

At point a, y=0 and the x co-ordinate is a. You get: 0 = 2*|a-1|
Hence a must be 1

[chunderseverywhere]

good luck for the exam everyone! hope it's an alright one

[Amy-Rose]

yeah good luck everyone

[fayola1337]

Good luck everyone. I'm banking on this being less horrible than the January exam. I got a C in that, and I need at least a B for Uni. I hope it's fairly easy.

[Eamonn123]

Last exam, really need to smash this! Good luck everyone!!!


This was posted from The Student Room's iPhone/iPad App

[hhh123]

Wasn't too bad?!


This was posted from The Student Room's iPhone/iPad App

[Amy-Rose]

I think it went pretty well, a hell of a lot better than january! Just didn't enjoy the proof or the final part of the last q.

[shredfest]

thought that was pretty standard

[Alicelouise134]

Was fairly happy, except the last part of the last question just wouldn't work for me

[Jodieeee....]

Thought it went well! Wish it was my last exam though =[

[Pectodux]

What did everyone get for b and a was it 2 and 1? For the last question, the last part.

[steven hawking]

For the last question did anyone get something like:
4ln2 -0.5 units squared ?

[Alasda1r]

Last part of the last question you were using the integral of f(x-1), that is, a shift of the graph in the x direction by +1. So you needed to integrate between the bounds of 2 and 1, in order to get the area that would be between 0 and 1 on a normal graph. I got ln16 - 1/2.