Question

find by Newton's method the real root of x^3+3x-1=0 correct to 2 decimal places​

Answer 1

Answer:

write it as x=1/3+x^3/3. Since the derivative of the rhs is <1 for all |x|<1, it follows that the repetative method converges. So peak any x in (0,1)-{0}. Btw, the three roots are -1.87939, 0.347296 and 1.53209. So guess what happens for these three initial values of yours.

Answer 2

Answer:

hope it helps you properly mate