Question

use the bisection method to approximate, accurate to two decimal places, the value of the root of the given function in the given interval.
$f(x)=x^{2}+2x - 4$ on the interval $1,1.5$

Explanation:

Step1: Recall bisection - method principle

The bisection method finds the root of a continuous function $y = f(x)$ in an interval $[a,b]$ by repeatedly bisecting the interval and narrowing down the sub - interval that contains the root.

Step2: Initial function and interval

We have $f(x)=x^{2}+2x - 4$ and the interval $[a_1,b_1]=[1,1.5]$. First, calculate $f(1)=1^{2}+2\times1 - 4=1 + 2-4=-1$ and $f(1.5)=1.5^{2}+2\times1.5 - 4=2.25 + 3-4 = 1.25$.

Step3: First iteration

The mid - point $c_1=\frac{1 + 1.5}{2}=1.25$. Calculate $f(1.25)=1.25^{2}+2\times1.25 - 4=1.5625+2.5 - 4=-0.9375$. Since $f(1.25)$ and $f(1.5)$ have opposite signs, the new interval is $[1.25,1.5]$.

Step4: Second iteration

The mid - point $c_2=\frac{1.25 + 1.5}{2}=1.375$. Calculate $f(1.375)=1.375^{2}+2\times1.375 - 4=1.890625+2.75 - 4=0.640625$. Since $f(1.25)$ and $f(1.375)$ have opposite signs, the new interval is $[1.25,1.375]$.

Step5: Third iteration

The mid - point $c_3=\frac{1.25 + 1.375}{2}=1.3125$. Calculate $f(1.3125)=1.3125^{2}+2\times1.3125 - 4=1.72265625+2.625 - 4=0.34765625$. Since $f(1.25)$ and $f(1.3125)$ have opposite signs, the new interval is $[1.25,1.3125]$.

Step6: Fourth iteration

The mid - point $c_4=\frac{1.25 + 1.3125}{2}=1.28125$. Calculate $f(1.28125)=1.28125^{2}+2\times1.28125 - 4=1.6412109375+2.5625 - 4=0.2037109375$. Since $f(1.25)$ and $f(1.28125)$ have opposite signs, the new interval is $[1.25,1.28125]$.

Step7: Fifth iteration

The mid - point $c_5=\frac{1.25 + 1.28125}{2}=1.265625$. Calculate $f(1.265625)=1.265625^{2}+2\times1.265625 - 4=1.6017089844+2.53125 - 4=0.1329589844$. Since $f(1.25)$ and $f(1.265625)$ have opposite signs, the new interval is $[1.25,1.265625]$.

Step8: Sixth iteration

The mid - point $c_6=\frac{1.25 + 1.265625}{2}=1.2578125$. Calculate $f(1.2578125)=1.2578125^{2}+2\times1.2578125 - 4=1.58203125+2.515625 - 4=0.09765625$. Since $f(1.25)$ and $f(1.2578125)$ have opposite signs, the new interval is $[1.25,1.2578125]$.

Step9: Seventh iteration

The mid - point $c_7=\frac{1.25 + 1.2578125}{2}=1.25390625$. Calculate $f(1.25390625)=1.25390625^{2}+2\times1.25390625 - 4=1.57234375+2.5078125 - 4=0.07015625$. Since $f(1.25)$ and $f(1.25390625)$ have opposite signs, the new interval is $[1.25,1.25390625]$.

Step10: Eighth iteration

The mid - point $c_8=\frac{1.25 + 1.25390625}{2}=1.251953125$. Calculate $f(1.251953125)=1.251953125^{2}+2\times1.251953125 - 4=1.5673828125+2.50390625 - 4=0.0712890625$. Since $f(1.25)$ and $f(1.251953125)$ have opposite signs, the new interval is $[1.25,1.251953125]$.

Step11: Approximation to two decimal places

Rounding the mid - point of the final interval to two decimal places, we get $1.25$.

Answer:

$1.25$