2) Estimate 10 to within 1 decimal place by using Bisection to solve the equation x³ = 10 on the interval [2, 3]. Organize the results of all the calculations in the table below. You don't need to write down more than 5 decimal places.

Here $x^3-10=0$

Let $f\left(x\right)=x^3-10$

1st iteration :

$$\begin{gathered} a_0=2 \\ {c}_0=3 \end{gathered}$$

Here  and f(3)=17>0

∴ Now, Root lies between $2$ and $3$

Half length $b_0=\frac{2+3}{2}$

$f\left(b_0\right)=f(2.5)={\left(2.5\right)}^3-10=5.625>0$


2nd iteration :

$$\begin{gathered} a_1=2 \\ {c}_1=b_0=2.5 \end{gathered}$$

Here $f\left(2\right)=-2<0$ and $f(2.5)=5.625>0$

Half length $b_1=\frac{2+2.5}{2}=2.25$

∴ Now, Root lies between $2$ and $2.5$

$f\left(b_1\right)=f(2.25)={\left(2.25\right)}^3-10=1.39062>0$


3rd iteration :

$$\begin{gathered} a_2=2 \\ {c}_2=b_1=2.25 \end{gathered}$$

Here $f\left(2\right)=-2<0$and $f(2.25)=1.39062>0$

∴ Now, Root lies between $2$ and $2.25$

Half length $b_2=\frac{2+2.25}{2}=2.125$

$f\left(b_2\right)=f(2.125)={\left(2.125\right)}^3-10=-0.4043<0$

4th iteration :

Here$f(2.125)=-0.4043<0$and $f(2.25)=1.39062>0$

$$\begin{gathered} a_3=b_2=2.125 \\ {c}_3=2.25 \end{gathered}$$

∴ Now, Root lies between $2.125$ and $2.25$

Half length $b_3=\frac{2.125+2.25}{2}=2.1875$

$f\left(b_3\right)=f(2.1875)={\left(2.1875\right)}^3-10=0.46753>0$


5th iteration :

Here $f(2.125)=-0.4043<0$ and $f(2.1875)=0.46753>0$

$$\begin{gathered} a_4=2.125 \\ {c}_4=b_3=2.1875 \end{gathered}$$

∴ Now, Root lies between $2.125$ and $2.1875$

Half length $b_4=\frac{2.125+2.1875}{2}=2.15625$

$f\left(b_4\right)=f(2.15625)={\left(2.15625\right)}^3-10=0.0253>0$

6th iteration :

Here $f(2.125)=-0.4043<0$ and $f(2.15625)=0.0253>0$

$$\begin{gathered} a_5=2.125 \\ {c}_5=b_4=2.15625 \end{gathered}$$

∴ Now, Root lies between $2.125$ and $2.15625$

Half length $b_5=\frac{2.125+2.15626}{2}=2.14062$

$f\left(b_5\right)=f(2.14062)={\left(2.14062\right)}^3-10=-0.19107<0$


7th iteration :

Here $f(2.14062)=-0.19107<0$and $f(2.15625)=0.0253>0$

$$\begin{gathered} a_6=b_5=2.14062 \\ {c}_6=2.1562 \end{gathered}$$

∴ Now, Root lies between $2.14062$ and $2.15625$

Half length $b_6=\frac{2.14062+2.15625}{2}=2.14844$

$f\left(b_6\right)=f(2.14844)={\left(2.14844\right)}^3-10=-0.08328<0$


8th iteration :

Here $f(2.14844)=-0.08328<0$and $f(2.15625)=0.0253>0$

$$\begin{gathered} a_7=b_6=2.14844 \\ {c}_7=2.15625 \end{gathered}$$

∴ Now, Root lies between $2.14844$ and $2.15625$

$b_7=\frac{2.14844+2.15625}{2}=2.15234$

$f\left(b_7\right)=f(2.15234)={\left(2.15234\right)}^3-10=-0.02909<0$


9th iteration :

Here f$(2.15234)=-0.02909<0$ and $f(2.15625)=0.0253>0$

$$\begin{gathered} a_8=b_7=2.15234 \\ {c}_8=2.15625 \end{gathered}$$

∴ Now, Root lies between $2.15234$ and $2.15625$

$b_8=\frac{2.15234+2.15625}{2}=2.1543$

$f\left(b_8\right)=f(2.1543)={\left(2.1543\right)}^3-10=-0.00192<0$

10th iteration :

Here $f(2.1543)=-0.00192<0$ and $f(2.15625)=0.0253>0$

$$\begin{gathered} a_9=b_8=2.1542 \\ {c}_3=2.15625 \end{gathered}$$
∴ Now, Root lies between $2.1543$ and $2.15625$

Half length $b_9=\frac{2.1543+2.15625}{2}=2.15527$

$f\left(b_9\right)=f(2.15527)={\left(2.15527\right)}^3-10=0.01168>0$

11th iteration :

Here $f(2.1543)=-0.00192<0$and $f(2.15527)=0.01168>0$

$$\begin{gathered} a_{10}=2.1543 \\ {c}_{10}=b_9=2.15527 \end{gathered}$$

∴ Now, Root lies between $2.1543$and $2.15527$

$b_{10}=\frac{2.1543+2.15527}{2}=2.15479$

$f\left(b_{10}\right)=f(2.15479)={\left(2.15479\right)}^3-10=0.00488>0$


12th iteration :

Here $f(2.1543)=-0.00192<0$ and $f(2.15479)=0.00488>0$

$$\begin{gathered} a_{11}=2.1543 \\ {c}_{11}=b_{10}=2.15479 \end{gathered}$$

∴ Now, Root lies between $2.1543$and $2.15479$

Half length $b_{11}=\frac{2.1543+2.15479}{2}=2.15454$

$f\left(b_{11}\right)=f(2.15454)={\left(2.15454\right)}^3-10=0.00148>0$


13th iteration :

 Here $f(2.1543)=-0.00192<0$and $f(2.15454)=0.00148>0$

$$\begin{gathered} a_{12}=2.1543 \\ {c}_{12}=b_{11}=2.15454 \end{gathered}$$

∴ Now, Root lies between $2.1543$ and $2.15454$

Half length $b_{12}=\frac{2.1543+2.15454}{2}=2.15442$

$f\left(b_{12}\right)=f(2.15442)={\left(2.15442\right)}^3-10=-0.00022<0$


14th iteration :

Here $f(2.15442)=-0.00022<0$ and $f(2.15454)=0.00148>0$

$$\begin{gathered} a_{13}=b_{12}=2.15442 \\ {c}_{13}=2.15454 \end{gathered}$$
∴ Now, Root lies between $2.15442$ and $2.15454$

Half length $b_{13}=\frac{2.15442+2.15454}{2}=2.15448$

$f\left(b_{13}\right)=f(2.15448)={\left(2.15448\right)}^3-10=0.00063>0$


15th iteration :

Here $f(2.15442)=-0.00022<0$ and$f(2.15448)=0.00063>0$

$$\begin{gathered} a_{14}=2.15442 \\ {c}_{14}=c_{13}=2.15448 \end{gathered}$$

∴ Now, Root lies between $2.15442$ and $2.15448$

$b_{14}=\frac{2.15442+2.15448}{2}=2.15445$

$f\left(b_{14}\right)=f(2.15445)={\left(2.15445\right)}^3-10=0.00021>0$


16th iteration :

Here $f(2.15442)=-0.00022<0$ and $f(2.15445)=0.00021>0$

$$\begin{gathered} a_{15}=2.15442 \\ {c}_{15}=b_{15}=2.15445 \end{gathered}$$

∴ Now, Root lies between $2.15442$and $2.15445$

$b_{15}=\frac{2.15442+2.15445}{2}=2.15443$

$f\left(b_{15}\right)=f(2.15443)={\left(2.15443\right)}^3-10=0<0$