Iteration

The starting value is the input for the first use of the formula.

1. x_1=\frac{2+6}{2}.
2. Therefore x_1=4.

Use the formula once to get the first new value, then apply it again.

1. x_1=2\times3-1=5.
2. x_2=2\times5-1=9.

Evaluate inside the root using the starting value.

1. x_1=\sqrt{2+7}=\sqrt{9}.
2. Therefore x_1=3.

Use the previous result as the next input and round only the final answer.

1. Substitute x_0=4: x_1=\frac{10}{4+1}=2.
2. Substitute x_1=2: x_2=\frac{10}{2+1}=3.3333\ldots
3. To 3 decimal places, x_2=3.333.

The subscript increases by one after each use of the formula.

1. The first use gives x_1, the second gives x_2, and the third gives x_3.

Use the previous result as the next input and round only the final answer.

1. Substitute x_0=2: x_1=\sqrt{2\times2+6}=\sqrt{10}=3.16227\ldots
2. Use the full value of x_1 (the ANS key keeps it): x_2=\sqrt{2\times3.16227\ldots+6}=3.51063\ldots
3. x_3=\sqrt{2\times3.51063\ldots+6}=3.60849\ldots
4. Round only at the end: x_3=3.608 to 3 decimal places.

Work backwards through one application of the formula.

1. The first application gives 3=\frac{x_0+5}{3}.
2. So 9=x_0+5, giving x_0=4.

Use the previous result as the next input and round only the final answer.

1. Substitute x_0=3: x_1=\frac12\left(3+\frac83\right)=2.83333\ldots
2. Use the full value of x_1: x_2=\frac12\left(2.83333\ldots+\frac{8}{2.83333\ldots}\right)=2.828431\ldots
3. To 4 decimal places, x_2=2.8284.

Isolate the cubic term before undoing the cube.

1. Subtract x and add 9: x^3=9-x.
2. Take the cube root of both sides: x=\sqrt[3]{9-x}.
3. x=\sqrt{9-x} is wrong: a square root undoes a square, not a cube.

Keep applying the same rule to the latest result.

1. x_1=\frac{0+4}{2}=2.
2. x_2=\frac{2+4}{2}=3.
3. x_3=\frac{3+4}{2}=\frac72.

At the limiting value, the input and output of the iteration are equal.

1. When the values settle at L, putting L into the formula gives L back, so L=\sqrt{3L+4}.
2. Square both sides: L^2=3L+4, so L^2-3L-4=0.
3. Factorise: (L-4)(L+1)=0, so L=4 or L=-1.
4. L is positive, so L=4.

Get both terms in x on one side, then take out x as a common factor.

1. Add 12 to both sides: x^3+3x=12.
2. Take out the common factor x: x(x^2+3)=12.
3. Divide both sides by x^2+3: x=\frac{12}{x^2+3}.
4. x=\frac{12}{x^2}-3 is wrong: dividing x^3=12-3x by x^2 gives x=\frac{12}{x^2}-\frac{3}{x}, not \frac{12}{x^2}-3.

Evaluate the polynomial separately at both ends of the interval.

1. Substitute x=2: f(2)=2^3-2-7=-1.
2. Substitute x=3: f(3)=3^3-3-7=17.
3. f(2) is negative and f(3) is positive. The graph of y=f(x) is an unbroken curve, so it crosses y=0 between x=2 and x=3: there is a solution there.

Use the previous result as the next input and round only the final answer.

1. Substitute x_0=2: x_1=\frac{11}{2+2}=2.75.
2. x_2=\frac{11}{2.75+2}=2.315789\ldots
3. Use the full value of x_2: x_3=\frac{11}{2.315789\ldots+2}=2.548780\ldots
4. To 3 decimal places, x_3=2.549.

Set the next value equal to the current value, square, then clear the fraction.

1. L=\sqrt{5+\frac{2}{L}} gives L^2=5+\frac{2}{L}.
2. Multiplying by L gives L^3=5L+2.
3. Therefore L^3-5L-2=0.

For the equation, the values settle where x=\sqrt[3]{2x+9}: undo the cube root.

Part (a)

1. Substitute x_0=2: x_1=\sqrt[3]{2\times2+9}=\sqrt[3]{13}=2.35133\ldots
2. Use the full value each time: x_2=\sqrt[3]{2\times2.35133\ldots+9}=2.39295\ldots
3. x_3=\sqrt[3]{2\times2.39295\ldots+9}=2.39779\ldots
4. To 4 decimal places, x_3=2.3978.

Part (b)

1. The values settle at a solution of x=\sqrt[3]{2x+9}.
2. Cube both sides: x^3=2x+9, so x^3-2x-9=0.
3. Compare with x^3+ax+b=0: a=-2.

Part (c)

1. From part (b), the equation is x^3-2x-9=0.
2. Compare with x^3+ax+b=0: b=-9.

Use the previous result as the next input and round only the final answer.

1. Substitute x_0=2: x_1=\sqrt{6+\frac32}=\sqrt{7.5}=2.73861\ldots
2. Use the full value each time: x_2=\sqrt{6+\frac{3}{2.73861\ldots}}=2.66372\ldots
3. x_3=\sqrt{6+\frac{3}{2.66372\ldots}}=2.66950\ldots
4. To 4 decimal places, x_3=2.6695.

Which numbers round to 1.92, correct to 2 decimal places?

1. A number rounds to 1.92 (to 2 decimal places) when it lies between the bounds 1.915 and 1.925.
2. f(1.915)=-0.062\ldots, which is negative, and f(1.925)=0.058\ldots, which is positive.
3. The sign changes, so the solution lies between 1.915 and 1.925. It rounds to 1.92, so Sam is right.
4. f(1.91)<0 and f(1.93)>0 is not enough: it only puts the solution between 1.91 and 1.93, so it could round to 1.91, 1.92 or 1.93.

Compare the rounded values, but keep full precision when calculating each next value.

1. Substitute x_0=4: x_1=\frac12\left(4+\frac{10}{4}\right)=3.25.
2. Keep the full values: x_2=3.16346\ldots, x_3=3.162277\ldots and x_4=3.162277\ldots
3. Rounded to 3 decimal places: x_1=3.250, x_2=3.163, x_3=3.162, x_4=3.162.
4. x_2 and x_3 are different, but x_3 and x_4 are the same, so n=3.

Get x^3 on its own, then take the cube root. Keep full calculator values between steps.

Part (a)

1. Add 4x and 11 to both sides: x^3=4x+11.
2. Take the cube root of both sides: x=\sqrt[3]{4x+11}, which gives x_{n+1}=\sqrt[3]{4x_n+11}.
3. \sqrt[3]{4x_n-11} has the wrong sign, and a square root would undo a square, not a cube.

Part (b)

1. Substitute x_0=2.5: x_1=\sqrt[3]{4\times2.5+11}=\sqrt[3]{21}=2.75892\ldots
2. x_2=\sqrt[3]{4\times2.75892\ldots+11}=2.80355\ldots
3. x_3=\sqrt[3]{4\times2.80355\ldots+11}=2.81110\ldots
4. x_4=\sqrt[3]{4\times2.81110\ldots+11}=2.81237\ldots
5. To 4 decimal places, x_4=2.8124.