Newton-Raphson process FP1 question..

I have some hwk and do not know how to solve this question:

The equation f(x) = 0 has root b(beta) in the interval [-2,-1]. Taking -1.5 as a first approximation to b(beta), apply the Newton Raphson process once to f(x) to obtain a second approximation to b(beta).

Up the page there is the function f(x) = (x^4)/2 - x^3 + x - 3

I know how the process works, but I get an approximation of around 2.2 (2 s.f.).... Plus the fact that this is a 5 mark question leads me to wander what I am doing wrong.

Thanks for help

Reply 1

BabyMaths

Original post by Henry3669

I have some hwk and do not know how to solve this question:

The equation f(x) = 0 has root b(beta) in the interval [-2,-1]. Taking -1.5 as a first approximation to b(beta), apply the Newton Raphson process once to f(x) to obtain a second approximation to b(beta).

Up the page there is the function f(x) = (x^4)/2 - x^3 + x - 3

I know how the process works, but I get an approximation of around 2.2 (2 s.f.).... Plus the fact that this is a 5 mark question leads me to wander what I am doing wrong.

Thanks for help

You should be getting -1.3875. Post working?

Reply 2

Henry3669

OP

-1.5 - [f(x)]/[f'(x)]

Then I substituted -1.5 as x in the function and it's derivative to get;

-1.5 - [(-117)/(32)]/1
= 2.15...

My notation my be a little convoluted... I plugged -1.5 into the function and its derivative and that got me -1.5 - minus (117/32)/1

Reply 3

BabyMaths

Original post by Henry3669

-1.5 - [f(x)]/[f'(x)]

Then I substituted -1.5 as x in the function and it's derivative to get;

-1.5 - [(-117)/(32)]/1
= 2.15...

My notation my be a little convoluted... I plugged -1.5 into the function and its derivative and that got me -1.5 - minus (117/32)/1

I would not expect you to see any intermediate results if you're doing this efficiently.

Try this.

-1.5 =

ANS-\frac{ANS^4/2-ANS^3+ANS-3}{2 ANS ^3 - 3 ANS^2+1}

=
=
=

Reply 4

Henry3669

OP

I'm not getting anything near where I should. Perhaps you could show me your working? Thanks for your help btw.

Reply 5

Henry3669

OP

I posted this yesterday and got some help but still could not obtain a correct answers;

The equation f(x) = 0 has root b(beta) in the interval [-2,-1]. Taking -1.5 as a first approximation to b(beta), apply the Newton Raphson process once to f(x) to obtain a second approximation to b(beta).

Up the page there is the function f(x) = (x^4)/2 - x^3 + x - 3

I know how the process works, but I get an approximation of around 2.2 (2 s.f.).... Plus the fact that this is a 5 mark question leads me to wander what I am doing wrong.

I ould be really grateful if you could show some working. Thanks.

Original post by Henry3669

I posted this yesterday and got some help but still could not obtain a correct answers;

The equation f(x) = 0 has root b(beta) in the interval [-2,-1]. Taking -1.5 as a first approximation to b(beta), apply the Newton Raphson process once to f(x) to obtain a second approximation to b(beta).

Up the page there is the function f(x) = (x^4)/2 - x^3 + x - 3

I know how the process works, but I get an approximation of around 2.2 (2 s.f.).... Plus the fact that this is a 5 mark question leads me to wander what I am doing wrong.

I ould be really grateful if you could show some working. Thanks.

Looks like you are finding the wrong root (2.16...). Show YOUR working please.

Reply 7

Henry3669

OP

Ok, I used the Newton Raphson process (I forgot to mention that this question asks for it). I differentiated the above function and done;

-1.5 - [f(x)]/[f'(x)]

When I substituted x as -1.5;

-1.5 - (-117/32)/1

And I got around 2.15 (2.2)

Reply 8

BabyMaths

Original post by Henry3669

I'm not getting anything near where I should. Perhaps you could show me your working? Thanks for your help btw.

My working is in post 4.

What did you get for f'(x)?

Although your f(x) seems to be clearly written, can I get you to confirm that

f(x)=\frac{x^4}{2}-x^3+x-3

?

Reply 9

BabyMaths

Original post by Henry3669

I posted this yesterday and got some help but still could not obtain a correct answers;

The equation f(x) = 0 has root b(beta) in the interval [-2,-1]. Taking -1.5 as a first approximation to b(beta), apply the Newton Raphson process once to f(x) to obtain a second approximation to b(beta).

Up the page there is the function f(x) = (x^4)/2 - x^3 + x - 3

I know how the process works, but I get an approximation of around 2.2 (2 s.f.).... Plus the fact that this is a 5 mark question leads me to wander what I am doing wrong.

I ould be really grateful if you could show some working. Thanks.

There's no need to start a new thread. You could have bumped the other one. I'm surprised that Mr M didn't tell you off. He's usually quite strict. :K:

Reply 10

sophieshoesmith

I get f(-1.5)=2.531+3.375-1.5-3 = about 1.4 ish (not what you make it)

I'm only at GCSE maths level but hope that helps?

Reply 11

brianeverit

9

Original post by Henry3669

Ok, I used the Newton Raphson process (I forgot to mention that this question asks for it). I differentiated the above function and done;

-1.5 - [f(x)]/[f'(x)]

When I substituted x as -1.5;

-1.5 - (-117/32)/1

And I got around 2.15 (2.2)

Show your working for f(-1.5) and f'(-1.5). I don't see where you got (-117/32)/1 from.

Reply 12

Henry3669

OP

f(-1.5) = 0.5(-1.5)^4 - (-1.5)^3 - 1.5 - 3

f'(-1.5) = 2(-1.5)^3 - 3(-1.5)^2 + 1

= [(-117/32)]/1

Reply 13

BabyMaths

Original post by Henry3669

f(-1.5) = 0.5(-1.5)^4 - (-1.5)^3 - 1.5 - 3

f'(-1.5) = 2(-1.5)^3 - 3(-1.5)^2 + 1

= [(-117/32)]/1

Your first two lines there make sense. You could work them out to get 45/32 and -25/2.

The result would then be -1.5 - (45/32)/(-25/2) = -3/2 + 9/80 = -1.3875.