X 3 0 5x July 8, 2019 16 32 64 128 F x )= 3 0 5x X 3 2 5x 0 X 3 )( 4 5x )= 0 X 3 4x 2 5x 0 X 3 6x 2 5x 0 X 2x 3 2 0 5x X 3 3x 2 5x 0 X 2y 3 y =- 0 5x X 5 6x 3 5x 0 X 3 5x 3 0 by newton raphson method X 3 5x 6 0 X 5 5x 3 0 X 3 5x 1 0 regula falsi method Y x 3 0 5x y 3 (- 5x 7 )(- x 3 )= 0
