f(x)=2x3−6x+3xx>0f(x) = 2x^3 - 6x + 3\sqrt{x} \quad x > 0f(x)=2x3−6x+3xx>0
Show that f(x)=0f(x) = 0f(x)=0 has a root in the interval [1.2,1.3][1.2, 1.3][1.2,1.3]
Find f′(x)f'(x)f′(x)
Starting with x0=1.25x_0 = 1.25x0=1.25, 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.
