Using interpolants to solve a polynomial.

[riddle]

Homework Statement

Show that a root of the equation x³ - 3x - 5 =0 lies in the interval [2,3], and then find the root using linear interpolation correct to one decimal place.

Homework Equations

n/a

The Attempt at a Solution

This is my first ever time using interpolants ( well at least in the sense of solving a polynomial. I think I've used them before... but instinctively.)

This is what I did:
3 \div 13 = (x₁ - 2) \div (3-x₁)
to end up with x₁ = 2.1875.
I went on to get the third approximation only to find out that it was outside the range (less than 2).
I checked what the solution said, and it said:
3 \div 13 = (3-x₁) \div (x₁ - 2)

I'm utterly confused. Please help.

 

[LCKurtz]

The first thing your problem asked was to show that there was a root of your function between x = 2 and x = 3. Did you do that? And how did you do it?

Your x₁ looks OK, and it is between 2 and 3. Did you check whether you got lucky and x₁ is your root? If it isn't, then how do you know whether the root is in [2, x₁] or in [x₁, 3]? You have to know that in order to know which interval to do the next step with.

 

[riddle]

Oh. I forgot to mention that I'd got the first part. Yes. The root is in between 2 and 3.
I just substituted "x" with 3 and 2 and saw that f(3) / f(2) = -s. i.e., the sign changed, so there had to have had been a number which had f(x) = 0, i.e., the root of the equation.
My first approximation is different from what the book says, the same with all my further approximations. I thought that it might have been a typo, but then I couldn't do the other questions either, and in the end I checked to see if the answers that the book gave worked, and they did. So there's got to be something wrong in what I'm doing.

 

[LCKurtz]

Oh. I forgot to mention that I'd got the first part. Yes. The root is in between 2 and 3.
I just substituted "x" with 3 and 2 and saw that f(3) / f(2) = -s. i.e., the sign changed, so there had to have had been a number which had f(x) = 0, i.e., the root of the equation.
My first approximation is different from what the book says, the same with all my further approximations. I thought that it might have been a typo, but then I couldn't do the other questions either, and in the end I checked to see if the answers that the book gave worked, and they did. So there's got to be something wrong in what I'm doing.

Your first approximation for x₁ of 35/16=2.1875 is correct. The value of the function at that point is negative so you want to use x = 2.1875 and x = 3 for your calculation of x₂. I get 2.250619230 for it.

 

[riddle]

Ok. I'm getting the same thing.
I understand how this works now. But I'm still having trouble believing that the publishers messed up so bad. Can you just have a look at what they did. I'm really tired right now, and I won't be able to use the book for a week or two so I won't be able to look at it later.