Originally Posted by

**Kiwigirl** **Question One.**

Use an appropriate method to find the root of *x*³ + *x * + 1 = 0 correct to 2 decimal places. Start with initial interval -0.8 and -0.6. Show all working.

Let:

$\displaystyle f(x)=x^3+x+1$.

We wish to find a zero of $\displaystyle f$ in the interval $\displaystyle [-0.8,-0.6]$. We know it has a zero

in this interval as it changes sign between $\displaystyle -0.8$ and $\displaystyle -0.6$.

Here we are going to use binary chop, In the following tableau we have

in each row a $\displaystyle x_1$ lower limit of the interval containing the zeros, $\displaystyle f_1$ the

value of the function at the lower limit, $\displaystyle x_2$ the upper limit of the interval

containing the zeros, $\displaystyle f_2$ the value of the function at the upper limit, $\displaystyle x_{mid}$

the mid-point of the interval and $\displaystyle f_{mid}$ the value of the function at the mid=point.

To obtain the following row from a row the value of the upper or lower limit

of the interval is replaced by the mid-point according to the sign of the

function at the mid-point. This process is continued until the upper and

lower limits of the interval are equal to two significant digits.

$\displaystyle \begin{array}{|c|c|c|c|c|c|}

\hline x_1 & f_1 & x_2 & f_2 & x_{mid} & f_{mid}\\ \hline

-0.8 & -0.312 & -0.6 & 0.184 & -0.7 & -0.043\\

-0.7 & -0.043 & -0.6 & 0.184 & -0.65 & 0.074\\

-0.7 & -0.043 & -0.65 & 0.074 & -0.675 & 0.175\\

-0.7 & -0.043 & -0.675 & 0.175 & -0.6875 & -0.0124\\

-0.6875 & -0.0124 &-0.675 & 0.175&-0.681&0.0026\\

-0.6875 & -0.0124 &-0.681&0.0026&-0.684&-0.0046\\

-0.684&-0.0046&-0.681&0.0026&\ & \ \\ \hline

\end{array}$

So the zero we seek is $\displaystyle x\approx -0.68$.

RonL