# maths - fixed points

posted by
**Anonymous**
.

Fixed points of f?

f(x) = 1/4x^2 - 1/8x - 5/8

a) use algebra to find the fixed points and classify them.

b) use the gradient criterion to determine an interval of attraction for one of the fixed points of f

c) find exact values of the 2nd and 3rd terms of the sequence xn obtained by iterating f with initial term x0= -3.

State the long term behaviour of this sequence

My answers;

a) fixed points 5 and -1/2

f '(5) = 1/2 x 5 - 1/8 = 19/8 >1 therefore it is repelling

f '(-1/2) = 1/2 x -1/2 - 1/8 = -3/8 <1 and is therefore attracting

b) -1 < 1/2x - 1/8 < 1

So, 1/2x - 1/8 < 1 = x < 9/4

And, -1 < 1/2x - 1/8 = -7/4 < x

thus, -7/4 < x < 9/4

so, interval of attracting for point -1/2 is (-7/4 , 3/4)

c) x0 = -3

x1 = (1/4 x -3^2) - (1/8 x -3) - 5/8

= 2

x2 = (1/4 x 2^2) - (1/8 x 2) - 5/8

= 1/8

long term behaviour is;

xn tends to -1/2 as n tends to infinity.

If someone could check my answers I would be very grateful. Thanks