f(x)=x3+3x2−2xx>0f(x) = x^3 + 3x^2 - 2\sqrt{x} \quad x > 0f(x)=x3+3x2−2xx>0
Show that f(x)=0f(x) = 0f(x)=0 has a root in the interval [0.6,0.7][0.6, 0.7][0.6,0.7]
Find f′(x)f'(x)f′(x)
Starting with x0=0.65x_0 = 0.65x0=0.65, apply the Newton-Raphson procedure once to find an approximate solution to the equation f(x)=0f(x) = 0f(x)=0 giving your answer to 3 decimal places.
