Question

Find the smallest positive root by using the bisection method

2sin(2x)+2x^3-5

with 3 decimal plates

Answer 1

Given,

$2x^3+2\sin (2x)-5=0$

$\Rightarrow f(x)=2x^3+2\sin(2x)-5$

Now, x=0, 1, 2...

1st Iteration:

$f(1)=-1.1814<0$ and $f(2)=9.4864>0$

Now, root lies between 1 and 2.

$x_o=\frac{1+2}{2}=1.5$

$f(x_o)=f(1.5)=2*1.5^3+ 2\sin(3)-5=2.0322>2$

2nd Iteration:

$f(1)=-1.1814<0$

and $f(1.5)=2.0322>0$

Root lies between 1 and 1.5

$x_1=\frac{1+1.5}{2}=1.25$

$f(x_1)=f(1.25)=2*1.253+2\sin(2.5)-5=0.1032>0$

3rd Iteration:

Here $f(1)=-1.1814<0$ and $f(1.25)=0.1032>0$

Now, Root lies between 1 and 1.25

$x_2=\frac{1+1.25}{2}=1.125$