- 2072 Asoj · 8 marks
Bisection: 3x - √(1+sin x)
Short answer
Bisection with a=0,b=1 (f(0)<0,f(1)>0). After 11 iterations, interval width <0.001, giving root 0.392.
Using the Bisection method, find a real root of the equation f(x) = 3x - √(1+sin x) correct up to three decimal points.
Bisection method: if f(a)·f(b) < 0, the root lies in [a, b]. Take c = (a + b)/2; if f(a)·f(c) < 0 the root is in [a, c], otherwise in [c, b]. Repeat until the interval is shorter than the required accuracy.
Step 1: Bracket the root
f(x) = 3x − √(1 + sin x) (radians) f(0) = −1 < 0 and f(1) = 3 − √1.8415 = 1.643 > 0, so a root lies in (0, 1). Take a = 0, b = 1.
Step 2: Iterations
| n | a | b | c=(a+b)/2 | f(c) | root lies in |
|---|---|---|---|---|---|
| 1 | 0.0000 | 1.0000 | 0.5000 | 0.2837 | [a, c] |
| 2 | 0.0000 | 0.5000 | 0.2500 | -0.3669 | [c, b] |
| 3 | 0.2500 | 0.5000 | 0.3750 | -0.0439 | [c, b] |
| 4 | 0.3750 | 0.5000 | 0.4375 | 0.1193 | [a, c] |
| 5 | 0.3750 | 0.4375 | 0.4062 | 0.0376 | [a, c] |
| 6 | 0.3750 | 0.4062 | 0.3906 | -0.0032 | [c, b] |
| 7 | 0.3906 | 0.4062 | 0.3984 | 0.0172 | [a, c] |
| 8 | 0.3906 | 0.3984 | 0.3945 | 0.0070 | [a, c] |
| 9 | 0.3906 | 0.3945 | 0.3926 | 0.0019 | [a, c] |
| 10 | 0.3906 | 0.3926 | 0.3916 | -0.0006 | [c, b] |
| 11 | 0.3916 | 0.3926 | 0.3921 | 0.0006 | [a, c] |
After 11 iterations the interval width is 0.00049 (< 0.001), so the root is fixed to 3 decimals.
Answer: root ≈ 0.392
✓ Answer checked by Claude Opus 5.5 (high effort)
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