Originally Posted by

**transgalactic** A.

show that $\displaystyle \frac{x}{2}=\sin x$ has a root on [1,2]

use the two steps of the crossing method that contains the root.

B.

find the root of the equation using newton referson methos.

solution of the book:

f(0)=0

f(pi/3)>0

so it rises and the is no root in [0,pi/3] interval

then they say that in [pi/3,pi] f(x) going down because f'(x)<0

so there is a solution in this interval

but how does f'(x)<0 shows that there is a soltion??

in order to see that there is a solution we need to have two point one positive one negative and that f(x) is going down

why they check from 0 to pi

and not from [1,2] like asked in the question