f(x)=ln(2x+1)+x2−5x>−0.5f(x) = \ln(2x + 1) + x^2 - 5 \quad x > -0.5f(x)=ln(2x+1)+x2−5x>−0.5
Show that f(x)=0f(x) = 0f(x)=0 has a root in the interval [1.8,1.9][1.8, 1.9][1.8,1.9]
Find f′(x)f'(x)f′(x)
Using x0=1.8x_0 = 1.8x0=1.8 as a first approximation, apply the Newton-Raphson procedure to find a second approximation, giving your answer to 3 decimal places.
