Consider the curve y = 2x^3 - 2x - 12.1) y-intercept. When x=0, y= -12 3) when x tends to infinity...y tends to infinity and when x tends to negative infinity...y tends to negative infinity 4) stationary points (i.e. where gradient of curve=0)). dy/dx=6x^2 - 2. At a given stationary point, dy/dx=0. Solve quadratic equation to conclude that stationary points exist at x values +/- sqrt (1/3). Determine that the y values of both stationary points are negative. 5) sketch the curve with what you know from above. Because the curve is a cubic and its two stationary points lie below the xAxis, it is evident that the curve crosses the xAxis at only one point - there is only one root to f(x)=0. 6) you know from your sketch that the root is greater than + sqrt (1/3). Calculate f(1): f(1) is negative, so the root is greater than f(1) (refer back to the sketch). So calculate f(2): it turns out that f(2)=0 so x=2 is the root of f(x)=0.