g(x)=ln(3x−1)+x2−6x>13\displaystyle g(x) = \ln(3x - 1) + x^2 - 6 \quad x > \frac{1}{3}g(x)=ln(3x−1)+x2−6x>31
Show that g(x)=0g(x) = 0g(x)=0 has a root in the interval [2.0,2.1][2.0, 2.1][2.0,2.1]
Find g′(x)g'(x)g′(x)
Using x0=2.0x_0 = 2.0x0=2.0 as a first approximation, apply the Newton-Raphson procedure to find a second approximation, giving your answer to 3 decimal places.
