wink

    steps1.add(new StepsHelper("Step 1","This equation has a solution in [0,1) if and only if the polynomial<br/>"
                + "\\(f(x) = x^3 - tx + t\\)<br/>"
                + "has a zero in that interval"));
        steps1.add(new StepsHelper("Step 2","Take the derivative of \\(f(x)\\) to determine the intervals on which \\(f(x)\\) is " +
                "strictly increasing and decreasing;<br/> on each interval on which \\(f(x)\\) is monotone,"
                + " there can be at most one zero.<br/>"
                + "\\(f'(x) = 3x^2-t\\)"));
        steps1.add(new StepsHelper("Step 3","if \\(t > 0\\) then \\(f'(x) > 0\\) for all \\(x\\). Since \\(f(0) = t<0\\) and \\(f(1)=1>0\\), there is exactly one solution of the equation in \\((0,1)\\)."));
        steps1.add(new StepsHelper("Step 4", "if \\(t = 0\\) then the equation has a solution at \\(x = 0\\), but no solutions at \\((0,1)\\)"));
        steps1.add(new StepsHelper("Step 5", "if \\(0<t<3)\\) then \\(f'(x) = 0\\) at the point \\(x = \\sqrt{t/3}\\), which is on the interval \\(0,1)\\)."
                + "At this point, <br/>"
                + "\\(f(\\sqrt{t/3}) = \\frac{t}{3}*\\sqrt{t/3}-t*\\sqrt{t/3}+t\\)<br/>"
                + "\\(= -\\frac{2t}{3}\\sqrt{t/3}+t\\)<br/>"
                + "\\(= t(-\\frac{2}{3}\\sqrt{t/3}+1)\\)<br/>"
                + "Since \\(\\sqrt{t/3} < 1 \\) and \\(t > 0\\), this quantity is positive. Since \\(f(t)\\) is decreasing to on one side of the critical point"
                + "and increasing on the other,<br/> there can be no solutions on the interval \\((0,\\sqrt{t/3})\\) or on the interval \\(\\sqrt{t/3},1)\\)"));
        steps1.add(new StepsHelper("Step 6", "if \\(t \\ge 3\\) then \\(f'(x) < 0\\) on \\((0,1)\\);"
                + "hence \\(f(x)\\) is decreasing from one positive value to another on \\((0,1)\\). <br/>"
                + "Hence "+lastStep));