Explain why

f(x) = ln x - 1/x - 1 , x>0

have one and only one zero point

Use Newton's method to find approximately value for the zero-point. The answer must state with 2 correct decimals.

How do I solve this?

Why?

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- Nov 17th 2005, 06:57 AMmr_pj[SOLVED] Finding zeropoint with Newton's method
Explain why

f(x) = ln x - 1/x - 1 , x>0

have one and only one zero point

Use Newton's method to find approximately value for the zero-point. The answer must state with 2 correct decimals.

How do I solve this?

Why? - Nov 17th 2005, 09:26 AMCaptainBlackQuote:

Originally Posted by**mr_pj**

you tell us what course/level you are studying.

Look at the definition of $\displaystyle f(x)$, it is the

difference of two positive functions of $\displaystyle x$:

$\displaystyle ln(x), \mbox{ and }\ 1/x+1$,

$\displaystyle f(x)$ is zero when these two functions are

equal. Now $\displaystyle ln(x)$ is an increasing function and

$\displaystyle 1/x+1$ is a decreasing function of $\displaystyle x$.

So it they are equal at some point, this is the only point at

which they are ever equal. This is clear if you sketch the

two functions (see attachment).

Look at your notes to find what the Newton's Method

is. You will find it gives you a new estimate of the

root of a function in terms of the old estimate.

You start with a guess, get a new estimate and feed

that back, repeating the procedure until it has converged

sufficiently for your purposes.

In this case starting with a guess of 2 I get:

>itert(10,2)

$\displaystyle n\ \ \ \ \ \ \ x_n$

$\displaystyle 0\ \ \ \ \ \ 2.000000$

$\displaystyle 1\ \ \ \ \ \ 3.075804 $

$\displaystyle 2\ \ \ \ \ \ 3.543637 $

$\displaystyle 3\ \ \ \ \ \ 3.590737 $

$\displaystyle 4\ \ \ \ \ \ 3.591121 $

$\displaystyle 5\ \ \ \ \ \ 3.591121 $

$\displaystyle 6\ \ \ \ \ \ 3.591121 $

$\displaystyle 7\ \ \ \ \ \ 3.591121 $

$\displaystyle 8\ \ \ \ \ \ 3.591121 $

$\displaystyle 9\ \ \ \ \ \ 3.591121 $

$\displaystyle 10\ \ \ \ 3.591121 $

>