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 $1,1.5$ iteration interval midpoint sign 1 $1,1.5$ ? 2 ? 3 ? 4 ? 5 ? 6 ? 7 ? 8 ?

Explanation:

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:

$1.25$