f(x)=x3+2x2−5xx>0f(x) = x^3 + 2x^2 - 5\sqrt{x} \quad x > 0f(x)=x3+2x2−5xx>0
Show that f(x)=0f(x) = 0f(x)=0 has a root in the interval [1.3,1.4][1.3, 1.4][1.3,1.4]
Find f′(x)f'(x)f′(x)
Starting with x0=1.35x_0 = 1.35x0=1.35, 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.
