Answer:
The given function is
f(x)=cos 3x-7 x²+ 4x
f'(x)=-3 sin 3 x-14 x+4
When you will draw the graph of the function , you will find that root of the function lie between (-1,0).
Consider initial root as,
[tex]x_{0}=0[/tex]
Using Newton method to find the roots of the equation
[tex]x_{n+1}=x_{n}-\frac{f{x_n}}{f'{x_{n}}}\\\\x_{1}=x_{0} - \frac{cos 3x_{0}-7 x_{0}^2+ 4x_{0}}{-3 sin 3 x_{0}-14 x_{0}+4}\\\\x_{1}=-\frac{\cos 0^{\circ}-0+0}{-3 \times 0-0+4}\\\\x_{1}=\frac{-1}{4}\\\\x_{1}= -0.25\\\\x_{2}=x_{1} - \frac{cos 3x_{1}-7 x_{1}^2+ 4x_{1}}{-3 sin 3 x_{1}-14 x_{1}+4}\\\\x_{2}=-0.25 -\frac{cos (-0.75)-7\times (0.0625)- 1}{-3 sin (-0.75)+3.50+4}\\\\x_{2}= -0.176054[/tex]
[tex]x_{3}=x_{2} - \frac{cos 3x_{2}-7 x_{2}^2+ 4x_{2}}{-3 sin 3 x_{2}-14 x_{2}+4}\\\\x_{3}=-0.176054 -\frac{cos (3\times -0.176054)-7\times (-0.176054)^2+4 \times -0.176054}{-3 sin (-0.176054)-14 \times (-0.176054)+4}\\\\x_{3}= -0.1689[/tex]
So, root of the equation is
=0.1688878
=0.1689(approx)
