1. ## help!

show that the cubic y=x^3+2x^2+5 has zero somewhere between x=-2.5 and x=-3. pick a suitable starting estimate of the root, x0, and perfrom 3 iterations of newtons method. test your final estimate to see how good it is.

thanks
jimmy

2. The first part is the intermediate value theorem.

The second bit is newton Rhapson method.

Newton's Method -- from Wolfram MathWorld

3. i dont understand the intermediate value therom, do you know any good references for this?

4. y=x^3+2x^2+5
$f(x)=x^3+2x^2+5$

$f(-3)=(-3)^3+2(-3)^2+5=-27+18+5=-4<0$

$f(-2.5)>0$ (I can't work this out in my head!)

f(x) is continous since it's a polynomial (and all polynomials are continuous).

By the Intermediate Value Theorem f(x) must assume every value between f(-3) and f(-2.5) since it's continuous. Therefore $\exists \ x_0 \in \mathbb{R} \ s.t \ f(x_0)=0$. $x_0$ is a solution to f(x).