Bisection Method

Question

Suppose you need to find an approximate root of the equation (x*sin(x)) - 1 = 0 by using Bisection method.

Show that there is a root in the interval [0, 2]

Answer 1

f(x) = xsinx - 1

f(0) = -1 <0

f(2) = 2sin2-1 = -0.93 <0 if x is measured in degrees and x=2 means 2 degrees

when the sign of f(x) changes, there's a zero or root between the two value of x. There's no sign change, with degrees

unless x was in measured in radians, with 1 radian = 180/pi = about 180/3.14159 = about 57.3 degrees, then

f(2) = 2sin(57.3) -1 = about 1.7-1 = +0.7 >0

then there's a sign change from f(0) to f(2), and at least one zero or root in the interval (0,2)

0<x<2