Pure Interval Bisection Only

1. (a) Show that the equation \(4 x - 2 \sin x - 1 = 0\), where \(x\) is in radians, has a root \(\alpha\) in the interval [0.2, 0.6]
   [0pt] (b) Starting with the interval [0.2, 0.6], use interval bisection twice to find an interval of width 0.1 in which \(\alpha\) lies.
   (3)
   

1. $$f ( x ) = 3 ^ { x } + 3 x - 7$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 1\) and \(x = 2\).
2. Starting with the interval \([ 1,2 ]\), use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).

2

1. Show that the equation $$x ^ { 3 } + x - 7 = 0$$ has a root between 1.6 and 1.8.
2. Use interval bisection twice, starting with the interval in part (a), to give this root to one decimal place.

$$f(x) = 2x^{-\frac{2}{3}} + \frac{1}{2}x - \frac{1}{3x - 5} - \frac{5}{2} \quad x \neq \frac{5}{3}$$ 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]

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

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]