use the bisection method to approximate, accurate to two decimal places, the value of the root of the given…

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 $1,1.5$ iteration interval midpoint sign 1 $1,1.5$ ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 8 ?
Answer
planation:
Step1: Calculate midpoint of first interval
The midpoint $m_1$ of the interval $[a_1,b_1]=[1,1.5]$ is $m_1=\frac{1 + 1.5}{2}=1.25$.
Step2: Evaluate function at midpoint
$f(1.25)=(1.25)^2+2\times1.25 - 4=1.5625 + 2.5-4=-0.9375<0$. Since $f(1)=1^2+2\times1 - 4=-1<0$ and $f(1.5)=(1.5)^2+2\times1.5 - 4=2.25 + 3 - 4 = 1.25>0$, the new interval is $[1.25,1.5]$.
Step3: Calculate midpoint of new interval
The midpoint $m_2$ of the interval $[a_2,b_2]=[1.25,1.5]$ is $m_2=\frac{1.25 + 1.5}{2}=1.375$.
Step4: Evaluate function at new midpoint
$f(1.375)=(1.375)^2+2\times1.375 - 4=1.890625+2.75 - 4=0.640625>0$. So the new interval is $[1.25,1.375]$.
Step5: Calculate midpoint of new - new interval
The midpoint $m_3$ of the interval $[a_3,b_3]=[1.25,1.375]$ is $m_3=\frac{1.25 + 1.375}{2}=1.3125$.
Step6: Evaluate function at new - new midpoint
$f(1.3125)=(1.3125)^2+2\times1.3125 - 4=1.72265625+2.625 - 4=0.34765625>0$. So the new interval is $[1.25,1.3125]$.
Step7: Calculate midpoint of new - new - new interval
The midpoint $m_4$ of the interval $[a_4,b_4]=[1.25,1.3125]$ is $m_4=\frac{1.25 + 1.3125}{2}=1.28125$.
Step8: Evaluate function at new - new - new midpoint
$f(1.28125)=(1.28125)^2+2\times1.28125 - 4=1.641015625+2.5625 - 4=0.203515625>0$. So the new interval is $[1.25,1.28125]$.
Step9: Calculate midpoint of new - new - new - new interval
The midpoint $m_5$ of the interval $[a_5,b_5]=[1.25,1.28125]$ is $m_5=\frac{1.25 + 1.28125}{2}=1.265625$.
Step10: Evaluate function at new - new - new - new midpoint
$f(1.265625)=(1.265625)^2+2\times1.265625 - 4=1.601806640625+2.53125 - 4=0.133056640625>0$. So the new interval is $[1.25,1.265625]$.
Step11: Calculate midpoint of new - new - new - new - new interval
The midpoint $m_6$ of the interval $[a_6,b_6]=[1.25,1.265625]$ is $m_6=\frac{1.25 + 1.265625}{2}=1.2578125$.
Step12: Evaluate function at new - new - new - new - new midpoint
$f(1.2578125)=(1.2578125)^2+2\times1.2578125 - 4=1.58203125+2.515625 - 4=0.09765625>0$. So the new interval is $[1.25,1.2578125]$.
Step13: Calculate midpoint of new - new - new - new - new - new interval
The midpoint $m_7$ of the interval $[a_7,b_7]=[1.25,1.2578125]$ is $m_7=\frac{1.25 + 1.2578125}{2}=1.25390625$.
Step14: Evaluate function at new - new - new - new - new - new midpoint
$f(1.25390625)=(1.25390625)^2+2\times1.25390625 - 4=1.572265625+2.5078125 - 4=0.079978125>0$. So the new interval is $[1.25,1.25390625]$.
Step15: Calculate midpoint of new - new - new - new - new - new - new interval
The midpoint $m_8$ of the interval $[a_8,b_8]=[1.25,1.25390625]$ is $m_8=\frac{1.25 + 1.25390625}{2}=1.251953125$.
Accurate to two decimal places, the root is $1.25$.
Answer:
$1.25$