Polynomial Function

Quote:

Originally Posted by impreza02

not sure how to do this question:

Approximate the zeros of each function(only gonna show one, the other i'll attempt)
f(x) = 1/4x3-2x2-2x+12

the smaller numbers are the powers

First sketch the function, see attachment. The function is a cubic and
so will have no more than three zeros. From the sketch we can see that
these will be close to $\pm 3$ and $8$ .

Now we will refine these estimates of the zeros using Newton Raphson
iteration. Which is the iteration:

$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $

Using our function gives:

$ x_{n+1}=x_n-\frac{0.25x_n^3-2x_n^2-2x_n+12}{0.75x_n^2-4x_n-2} $

For $x_0=-3$ , this gives:

$ \begin{array}{ccc}n&x_n&x_{n+1}\\0&-3&-2.597\\1&-2.597&-2.546\\2&-2.546&-2.546\end{array} $

So the root near $x=-3$ is $\approx -2.546$ .

For $x_0=3$ , this gives:

$ \begin{array}{ccc}n&x_n&x_{n+1}\\0&3&2.275\\1&2.27 5&2.281\\2&2.281&2.281\end{array} $

So the root near $x=3$ is $\approx 2.281$ .

For $x_0=8$ , this gives:

$ \begin{array}{ccc}n&x_n&x_{n+1}\\0&8&8.286\\1&8.28 6&8.265\\2&8.265&8.265\end{array} $

So the root near $x=8$ is $\approx 8.265$ .

RonL