k(x)=2ln(x+1)+x2−8x>−1k(x) = 2\ln(x + 1) + x^2 - 8 \quad x > -1k(x)=2ln(x+1)+x2−8x>−1
Show that k(x)=0k(x) = 0k(x)=0 has a root in the interval [2.3,2.4][2.3, 2.4][2.3,2.4]
Find k′(x)k'(x)k′(x)
Using x0=2.3x_0 = 2.3x0=2.3 as a first approximation, apply the Newton-Raphson procedure to find a second approximation, giving your answer to 3 decimal places.
