Question
find the smallest positive root of the equation f(x)=2sinx+x-2=0 using the false positive method.
Answers
Steve
well, you know that
f(0) = -2
f(pi/2) = pi/2
So, start with those two values
a1 = 0, b1=pi/2
so, c1 = pi/2-(pi/2)(pi/2-0)/(pi/2-(-2)) = 0.88
f(c1)=f(0.88) = 0.421
Since f(c1) > 0,
a2 = a1, b2 = c1 = 0.421
I assume you can follow the method, as explained at
https://en.wikipedia.org/wiki/False_position_method
Come back if you get stuck, and show your calculations.
f(0) = -2
f(pi/2) = pi/2
So, start with those two values
a1 = 0, b1=pi/2
so, c1 = pi/2-(pi/2)(pi/2-0)/(pi/2-(-2)) = 0.88
f(c1)=f(0.88) = 0.421
Since f(c1) > 0,
a2 = a1, b2 = c1 = 0.421
I assume you can follow the method, as explained at
https://en.wikipedia.org/wiki/False_position_method
Come back if you get stuck, and show your calculations.