Solutions to Assignment 1
Question 1 (4 marks), p.14, 1(b)(d)
(b), Let f (x) = (x 2)2 ln(x), obviously f (x) continuous on [1, 2] and [e, 4],
f (1)
f (2)
f (e)
f (4)
=
=
=
=
1,
0.6931,
0.4841,
2.6137,
since f (1) f (2) < 0 and f (e) f (4) < 0, by I.V.T. t
Solutions to Assignment 2
Question 1 (4 marks)
p54, No. 6(d)
This is the Bisection Method.
Input the function F(x) in terms of x
For example: cos(x)
x+1-2*sin(pi*x)
Input endpoints A < B on separate lines
0
0.5
Input tolerance
1e-5
Input maximum number of
Suggested Solutions to Assignment 1
Question 1 (4 marks), p.14, 1(a)(c)
(a)
f (x) = x cos x 2x2 + 3x 1 is continuous on [0.2, 0.3] with f (0.2) = 0.2840
and f (0.3) = 0.0066, the Intermediate Value Theorem implies that a number
x must exist in (0.2, 0.3)
Suggested Solutions to Assignment 2
Note: Question 5 should be answered by computer, others can be
answered manually or by computer.
Question 1, p.54 6(d) (a)
(1) For the interval [0, 0.5], using the Bisection method gives a1 = 0, and
1
b1 = 0.5, so f (a1