1. a. f(0) = -2 and f(1) = 2. As signs are different there must be a root between x = 0 and x = 1
b.Root is 0.60
2. a.
b. When x = -1, x2 − x − 1 = 1, which is positive
When x = 0, x2 − x − 1 = -1, which is negative, showing the root is between x = -1 and x = 0.
c. The positive root lies between x = 1 and x = 2.
d. The Bisection method give a root of 1.6, to one decimal place.
3. The root is 1.71 (to 2 decimal places.)
4. x = 5.45
5. a. f(1.05) = − 0.073 which is negative and f(1.15) = 0.409 which is positive. So the graph crosses the x-axis between x = 1.05 and x = 1.15.
b.
| Interval | mid-point | f(mid-point) |
| [1.05, 1.15] | 1.1 | 0.044 |
| [10.05, 1.1] | 1.075 |
6. a. f(0) = -1 which is negative and f(2) = 7.8 which is positive. So the graph crosses the x-axis between x = 0 and x = 2.
b. The interval at the end of the second iteration is [0.5, 1]
