c3 - iterations

Hi,
I am not very good at iterations (no wonder I'm retaking C3) but anyways, i tried to do the following question: -

The function f is defined by :

f(x) = \frac{1}{4}ln(3x +6)

; x >-2

and then a sketch of the graph is shown.

a) Explain how the graph shows that f is one to one and find the inverse

b) The equation

x= \frac {1}{4}ln(3x+6)

has two roots, alpha and beta, where alpha <0 and beta >0.

i) Show that aplha <-1.9995 and hence state the value of alpha correct to 3 d.p

ii) Use an iteration process based on the equation

x= \frac {1}{4}ln(3x+6)

to find the value of beta to 3 d.p

iii) Explain why the curves with equations y = f(x) and

Unparseable latex formula:

y = f^-^1(x)

meet at the points (alpha , aplha) and (beta, beta)

Ooo k

I've done a) and dont have a clue from b) onwards.

THanks a lot.

Reply 1

Kyalimers

21

Are you using Newton-Raphson method?

Reply 2

wizz_kid

OP

16

sohanshah

Are you using Newton-Raphson method?

Nope. That is uni stuff init?

Reply 3

JohnnySPal

2

Are ya sure (b) is right? I don't think x=0.25ln(3x+6) has two roots. Could be wrong, but 'tis my instinct.

My advice would be to draw the line y=x on the graph at first, and look at where the two graphs intersect.

Reply 4

Kyalimers

21

wizz_kid

Nope. That is uni stuff init?

No. It is both C3 and FP1 stuff.

Reply 5

wizz_kid

OP

16

sohanshah

No. It is both C3 and FP1 stuff.

I'll take ur word for it. However, our maths teacher introduced the newton raphson method to us last year in a cover lesson and it semed quite complex.

JohnnySPal

Are ya sure (b) is right? I don't think x=0.25ln(3x+6) has two roots. Could be wrong, but 'tis my instinct.

That's the question in my book. Could be wrong.

Reply 6

tazarooni89

20

wizz_kid

a) Explain how the graph shows that f is one to one and find the inverse

b) The equation

x= \frac {1}{4}ln(3x+6)

has two roots, alpha and beta, where alpha <0 and beta >0.

How can a function have two roots and also be one-to-one?

There are two values for x for which f(x) = 0 so doesn't that make it many-to-one?

Reply 7

caaakeeey

11

sohanshah

No. It is both C3 and FP1 stuff.

Not in Edexcel...

Reply 8

tazarooni89

20

caaakeeey

Not in Edexcel...

There's definitely Newton-Raphson in FP1... I don't remember it being in C3 though :s-smilie:

Reply 9

wizz_kid

OP

16

tazarooni89

How can a function have two roots and also be one-to-one?

I dunno.

That's wht the book says. It could be wrong.

Reply 10

caaakeeey

11

tazarooni89

There's definitely Newton-Raphson in FP1... I don't remember it being in C3 though :s-smilie:

That's what I meant.

Reply 11

Edenr

14

tazarooni89

How can a function have two roots and also be one-to-one?

There are two values for x for which f(x) = 0 so doesn't that make it many-to-one?

The first part is defined for x>-2, but the second part is not.

Reply 12

wizz_kid

OP

16

Edenr

The first part is defined for x>-2, but the second part is not.

Does that mean that the book has worded the question incorrenctly?

Reply 13

Edenr

14

wizz_kid

Does that mean that the book has worded the question incorrenctly?

No, the function f is only defined on x>-2. The first part of b) doesn't mention f(x).

Reply 14

ClaireHogben

13

sohanshah

No. It is both C3 and FP1 stuff.

not C3, its in FP1 though

Reply 15

wizz_kid

OP

16

Edenr

No, the function f is only defined on x>-2. The first part of b) doesn't mention f(x).

Rite. So does that mean that there is no solution for b?

Reply 16

Edenr

14

It does not mean that. If you want my advice, I would do b) i) and ii) as I would if I hadn't seen a) and the pre-amble at the top.

Reply 17

wizz_kid

OP

16

Edenr

It does not mean that. If you want my advice, I would do b) i) and ii) as I would if I hadn't seen a) and the pre-amble at the top.

Kool.

Tbh, i havnt got a clue in doing b i or ii

Reply 18

JohnnySPal

2

Edenr

No, the function f is only defined on x>-2. The first part of b) doesn't mention f(x).

You do still require that x>-2 though. That's indisputable, given that part of the equation involves log(3x+6) and that log is only defined on positive numbers.

Besides, I think the last part of b(i) needs the fact that x>-2. Knowing that x>-2 and x<-1.995 gives you a nice little bound on what alpha can be to 3sf.

Part bii wants you to apply a very basic principle that you can do with your calculator. Start off with some value of x, any you like so long as it's not too close to alpha. Type that into your calculator and press the equals button. Now that will be stored in the memory of the calculator that you can recall as "ANS". Are you familiar with this concept? Now, type into your calculator 0.25log(3ANS + 6). This will give you the first "iterate" and that number will now be stored as ANS. Again, type in 0.25log(3ANS + 6) to get the second iterate. As you keep applying this process you'll home in on the true root of the equation x=0.25log(3x+6).

Just so you have something tangible (I suppose I use the word "tangible" loosely :p:

) to aim for, I started off with x=1, and I very rapidly homed in on the root 0.504139...

Reply 19

wizz_kid

OP

16

JohnnySPal

You do still require that x>-2 though. That's indisputable, given that part of the equation involves log(3x+6) and that log is only defined on positive numbers.

Besides, I think the last part of b(i) needs the fact that x>-2. Knowing that x>-2 and x<-1.995 gives you a nice little bound on what alpha can be to 3sf.

Personally I'm not at all convinced that the question is correct. I'd be tempted to skip it and try a similar example, to be honest.

I see what u mean. I have worked out alpha as 2.000

Thanks

c3 - iterations

Related discussions

Latest

Trending

Trending

Articles for you