f(x) for x e^x - cos x is zero between:

0 and 1, f(0) = -1, f(1) = 2.178

The secant formula uses:

The last two iterates and their function values

Short answer

Use secant method with x0=0, x1=1 on f(x)=x e^x - cos x. Iterates converge to 0.5178, so the root is approximately 0.5178.

Find the root of the equation x e^x - cos x = 0 using the secant method correct to four decimal places.

Secant formula: x(n+1) = x(n) − f(x(n))·[x(n) − x(n−1)] / [f(x(n)) − f(x(n−1))].

f(x) = x·eˣ − cos x f(0) = −1 < 0 and f(1) = 2.7183 − 0.5403 = 2.1780 > 0, so a root lies in (0, 1). Take x0 = 0 and x1 = 1.

n	x(n-1)	x(n)	f(x(n-1))	f(x(n))	x(n+1)
1	0.0000	1.0000	-1.0000	2.1780	0.3147
2	1.0000	0.3147	2.1780	-0.5199	0.4467
3	0.3147	0.4467	-0.5199	-0.2035	0.5317
4	0.4467	0.5317	-0.2035	0.0429	0.5169
5	0.5317	0.5169	0.0429	-0.0026	0.5177
6	0.5169	0.5177	-0.0026	-0.0000	0.5178
7	0.5177	0.5178	-0.0000	0.0000	0.5178

The last two iterates agree to four decimals (both are 0.517757).

Answer: root ≈ 0.5178

✓ Answer checked by Claude Opus 5.5 (high effort)

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