Let f(x) = (cos(x))^2 - 3*x

Show that abs(f''(x)/f'(y))<=B, for x,y exist in [0,1], B is constant>0

Hence or otherwise deduce that Newton's method will converge.

I found that

abs(f''(x)/f'(y)) = abs( (2(sin(x)^2) - 2*(cos(x)^2)) / (-2cosy*siny - 3))

I can deduce that abs(2(sin(x)^2) - 2*(cos(x)^2)) is <=2

and that abs(-2cosy*siny - 3)>=3

so abs( (2(sin(x)^2) - 2*(cos(x)^2)) / (-2cosy*siny - 3))<=2/3

and so B=1? Am I correct??

Thank you very much!!!