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 andOriginally Posted by impreza02
so will have no more than three zeros. From the sketch we can see that
these will be close to $\displaystyle \pm 3$ and $\displaystyle 8$.
Now we will refine these estimates of the zeros using Newton Raphson
iteration. Which is the iteration:
$\displaystyle
x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}
$
Using our function gives:
$\displaystyle
x_{n+1}=x_n-\frac{0.25x_n^3-2x_n^2-2x_n+12}{0.75x_n^2-4x_n-2}
$
For $\displaystyle x_0=-3$, this gives:
$\displaystyle
\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 $\displaystyle x=-3$ is $\displaystyle \approx -2.546$.
For $\displaystyle x_0=3$, this gives:
$\displaystyle
\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 $\displaystyle x=3$ is $\displaystyle \approx 2.281$.
For $\displaystyle x_0=8$, this gives:
$\displaystyle
\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 $\displaystyle x=8$ is $\displaystyle \approx 8.265$.
RonL