1.09a Sign change methods: locate roots

$$f(x) = 2x^{-\frac{2}{3}} + \frac{1}{2}x - \frac{1}{3x - 5} - \frac{5}{2} \quad x \neq \frac{5}{3}$$ The table below shows values of \(f(x)\) for some values of \(x\), with values of \(f(x)\) given to 4 decimal places where appropriate.

1. Complete the table giving the values to 4 decimal places. [2]

The equation \(f(x) = 0\) has exactly one positive root, \(\alpha\). Using the values in the completed table and explaining your reasoning,

1. determine an interval of width one that contains \(\alpha\). [2]
2. Hence use interval bisection twice to obtain an interval of width 0.25 that contains \(\alpha\). [3]

Given also that the equation \(f(x) = 0\) has a negative root, \(\beta\), in the interval \([-1, -0.5]\)

1. use linear interpolation once on this interval to find an approximation for \(\beta\). Give your answer to 3 significant figures. [3]

$$\text{f}(x) = 3\sqrt{x} + \frac{18}{\sqrt{x}} - 20.$$

1. Show that the equation f\((x) = 0\) has a root \(a\) in the interval \([1.1, 1.2]\). [2]
2. Find \(f'(x)\). [3]
3. Using \(x_0 = 1.1\) as a first approximation to \(a\), apply the Newton-Raphson procedure once to f\((x)\) to find a second approximation to \(a\), giving your answer to 3 significant figures. [4]

$$f(x) = \frac{1}{2}x^4 - x^3 + x - 3$$

1. Show that the equation \(f(x) = 0\) has a root \(\alpha\) between \(x = 2\) and \(x = 2.5\) [2]
2. Starting with the interval \([2, 2.5]\) use interval bisection twice to find an interval of width \(0.125\) which contains \(\alpha\). [3]

The equation \(f(x) = 0\) has a root \(\beta\) in the interval \([-2, -1]\).

1. Taking \(-1.5\) as a first approximation to \(\beta\), apply the Newton-Raphson process once to \(f(x)\) to obtain a second approximation to \(\beta\). Give your answer to 2 decimal places. [5]

$$f(x) = 3^x - x - 6.$$

1. Show that \(f(x) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\). [2]
2. Starting with the interval \((1, 2)\), use interval bisection three times to find an interval of width 0.125 which contains \(\alpha\). [2]

$$f(x) = 1 - e^x + 3 \sin 2x$$ The equation \(f(x) = 0\) has a root \(\alpha\) in the interval \(1.0 < x < 1.4\). Starting with the interval \((1.0, 1.4)\), use interval bisection three times to find the value of \(\alpha\) to one decimal place. [3]

$$f(x) = 0.25x - 2 + 4 \sin \sqrt{x}.$$

1. Show that the equation \(f(x) = 0\) has a root \(\alpha\) between \(x = 0.24\) and \(x = 0.28\). [2]
2. Starting with the interval \([0.24, 0.28]\), use interval bisection three times to find an interval of width 0.005 which contains \(\alpha\). [3]

$$f (x) = x^3 + 8x - 19.$$

1. Show that the equation \(f(x) = 0\) has only one real root. [3]
2. Show that the real root of \(f(x) = 0\) lies between 1 and 2. [2]
3. Obtain an approximation to the real root of \(f(x) = 0\) by performing two applications of the Newton-Raphson procedure to \(f(x)\) , using \(x = 2\) as the first approximation. Give your answer to 3 decimal places. [4]
4. By considering the change of sign of \(f(x)\) over an appropriate interval, show that your answer to part (c) is accurate to 3 decimal places. [2]

The root of the equation f(x) = 0, where $$f(x) = x + \ln 2x - 4$$ is to be estimated using the iterative formula \(x_{n+1} = 4 - \ln 2x_n\), with \(x_0 = 2.4\).

1. Showing your values of \(x_1, x_2, x_3, \ldots\), obtain the value, to 3 decimal places, of the root. [4]
2. By considering the change of sign of f(x) in a suitable interval, justify the accuracy of your answer to part (a). [2]

\(\text{f}(x) = x^3 + x^2 - 4x - 1\). The equation f(x) = 0 has only one positive root, \(\alpha\).

1. Show that f(x) = 0 can be rearranged as $$x = \sqrt{\frac{4x+1}{x+1}}, \quad x \neq -1.$$ [2]

The iterative formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) is used to find an approximation to \(\alpha\).

1. Taking \(x_1 = 1\), find, to 2 decimal places, the values of \(x_2\), \(x_3\) and \(x_4\). [3]
2. By choosing values of \(x\) in a suitable interval, prove that \(\alpha = 1.70\), correct to 2 decimal places. [3]
3. Write down a value of \(x_1\) for which the iteration formula \(x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}\) does not produce a valid value for \(x_2\). Justify your answer. [2]

The curve \(C\) has equation \(y = \text{f}(x)\), where $$\text{f}(x) = 3 \ln x + \frac{1}{x}, \quad x > 0.$$ The point \(P\) is a stationary point on \(C\).

1. Calculate the \(x\)-coordinate of \(P\). [4]
2. Show that the \(y\)-coordinate of \(P\) may be expressed in the form \(k - k \ln k\), where \(k\) is a constant to be found. [2]

The point \(Q\) on \(C\) has \(x\)-coordinate 1.

1. Find an equation for the normal to \(C\) at \(Q\). [4]

The normal to \(C\) at \(Q\) meets \(C\) again at the point \(R\).

1. Show that the \(x\)-coordinate of \(R\)
   1. satisfies the equation \(6 \ln x + x + \frac{2}{x} - 3 = 0\),
   2. lies between 0.13 and 0.14. [4]

1. Sketch, on the same set of axes, the graphs of $$y = 2 - e^{-x} \text{ and } y = \sqrt{x}.$$ [3] [It is not necessary to find the coordinates of any points of intersection with the axes.] Given that f(x) = \(e^{-x} + \sqrt{x} - 2\), \(x \geq 0\),
2. explain how your graphs show that the equation f(x) = 0 has only one solution, [1]
3. show that the solution of f(x) = 0 lies between \(x = 3\) and \(x = 4\). [2]

The iterative formula \(x_{n+1} = (2 - e^{-x_n})^2\) is used to solve the equation f(x) = 0.

1. Taking \(x_0 = 4\), write down the values of \(x_1\), \(x_2\), \(x_3\) and \(x_4\), and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. [4]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]

The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).

1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
2. Show that \(-1 < \alpha < 0\). [2]

The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).

1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]

The equation \(2x^3 + 4x - 35 = 0\) has one real root.

1. Show by calculation that this real root lies between 2 and 3. [3]
2. Use the iterative formula $$x_{n+1} = \sqrt[3]{17.5 - 2x_n},$$ with a suitable starting value, to find the real root of the equation \(2x^3 + 4x - 35 = 0\) correct to 2 decimal places. You should show the result of each iteration. [3]

$$\text{f}(x) = x^2 + 5x - 2 \sec x, \quad x \in \mathbb{R}, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}.$$

1. Show that the equation \(\text{f}(x) = 0\) has a root, \(\alpha\), such that \(1 < \alpha < 1.5\) [2]
2. Show that a suitable rearrangement of the equation \(\text{f}(x) = 0\) leads to the iterative formula $$x_{n+1} = \cos^{-1} \left( \frac{2}{x_n^2 + 5x_n} \right).$$ [3]
3. Use the iterative formula in part (ii) with a starting value of 1.25 to find \(\alpha\) correct to 3 decimal places. You should show the result of each iteration. [3]

A curve has equation \(y = 2\ln(k - 3x) + x^2 - 3x\), where \(k\) is a positive constant.

1. Given that the curve has a point of inflection where \(x = 1\), show that \(k = 6\). [5] It is also given that the curve intersects the \(x\)-axis at exactly one point.
2. Show by calculation that the \(x\)-coordinate of this point lies between 0.5 and 1.5. [2]
3. Use the Newton-Raphson method, with initial value \(x_0 = 1\), to find the \(x\)-coordinate of the point where the curve intersects the \(x\)-axis, giving your answer correct to 5 decimal places. Show the result of each iteration to 6 decimal places. [3]
4. By choosing suitable bounds, verify that your answer to part (c) is correct to 5 decimal places. [1]

The equation \(6\arcsin(2x - 1) - x^2 = 0\) has exactly one real root.

1. Show by calculation that the root lies between 0.5 and 0.6. [2]

In order to find the root, the iterative formula \(x_{n+1} = p + q\sin(rx_n^2)\), with initial value \(x_0 = 0.5\), is to be used.

1. Determine the values of the constants \(p\), \(q\) and \(r\). [2]
2. Hence find the root correct to 4 significant figures. Show the result of each step of the iteration process. [2]

In this question you must show detailed reasoning. \includegraphics{figure_5} The diagram shows the curve with equation \(y = \frac{2x - 3}{4x^2 + 1}\). The tangent to the curve at the point \(P\) has gradient 2.

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$4x^3 + 3x - 3 = 0.$$ [5]
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.5 and 1. [2]
3. Show that the iteration $$x_{n+1} = \frac{3 - 4x_n^3}{3}$$ cannot converge to the \(x\)-coordinate of \(P\) whatever starting value is used. [2]
4. Use the Newton-Raphson method, with initial value 0.5, to determine the coordinates of \(P\) correct to 5 decimal places. [5]

The equation \(x^3 - 3x + 1 = 0\) has three real roots.

1. Show that one of the roots lies between \(-2\) and \(-1\) [2 marks]
2. Taking \(x_1 = -2\) as the first approximation to one of the roots, use the Newton-Raphson method to find \(x_2\), the second approximation. [3 marks]
3. Explain why the Newton-Raphson method fails in the case when the first approximation is \(x_1 = -1\) [1 mark]

A function \(f\) with domain \((-\infty,\infty)\) is defined by \(f(x) = 6x^3 + 35x^2 - 7x - 6\).

1. Determine the number of roots of the equation \(f(x) = 0\) in the interval \([-1, 1]\). [2]
2. Use the Newton-Raphson method to find a root of the equation \(f(x) = 0\). Starting with \(x_0 = 1\),
   1. write down the value of \(x_1\),
   2. determine the value of the root correct to one decimal place. [4]
3. It is suggested that another iterative sequence $$x_{n+1} = \sqrt{\frac{7x_n + 6 - 6x_n^3}{35}},$$ starting with \(x_0 = -3\), could be used to find a root of the equation \(f(x) = 0\). Explain why this method fails. [2]

The function \(f\) is given by $$f(x) = \frac{25x + 32}{(2x - 5)(x + 1)(x + 2)}.$$

1. Express \(f(x)\) in terms of partial fractions. [4]
2. Show that \(\int_1^2 f(x) dx = -\ln P\), where \(P\) is an integer whose value is to be found. [5]
3. Show that the sign of \(f(x)\) changes in the interval \(x = 2\) to \(x = 3\). Explain why the change of sign method fails to locate a root of the equation \(f(x) = 0\) in this case. [2]

The function \(f\) is defined by $$f(x) = x^3 + 4x^2 - 3x - 1.$$

1. Show that the equation \(f(x) = 0\) has a root in the interval \([0, 1]\). [1]
2. Using the Newton-Raphson method with \(x_0 = 0 \cdot 8\),
   1. write down in full the decimal value of \(x_1\) as given in your calculator,
   2. determine the value of this root correct to six decimal places. [4]
3. Explain why the Newton-Raphson method does not work if \(x_0 = \frac{1}{3}\). [2]

Find a small positive value of \(x\) which is an approximate solution of the equation. $$\cos x - 4\sin x = x^2.$$ [4]