f(x)=x3−2x2+4x−5x>0f(x) = x^3 - 2x^2 + 4\sqrt{x} - 5 \quad x > 0f(x)=x3−2x2+4x−5x>0
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)
Starting with x0=1.85x_0 = 1.85x0=1.85, 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.
