h(x)=ln(x+2)+0.5x2−4x>−2h(x) = \ln(x + 2) + 0.5x^2 - 4 \quad x > -2h(x)=ln(x+2)+0.5x2−4x>−2
Show that h(x)=0h(x) = 0h(x)=0 has a root in the interval [2.2,2.3][2.2, 2.3][2.2,2.3]
Find h′(x)h'(x)h′(x)
Using x0=2.2x_0 = 2.2x0=2.2 as a first approximation, apply the Newton-Raphson procedure to find a second approximation, giving your answer to 3 decimal places.
