Question

If f is the function, f(x) = (1/2)x^2 +2x - (5/8) a) use algebra to find the fixed points of f and classify them as attracting or repelling?

 

for a) i have found the 2 fixed points of (1/2) or 0.5 and (-5/2) or -2.5
which i have classified as both attracting as they are <1

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

c) find the exact values of the second and third terms of the sequence xn obtained by iterating f with x0= -(7/2) express as fractions in their lowest terms, hence state the long term behavior of this sequence and explain

for c) i get x1 = -3/2
x2 = -5/2

presuming that x0 is first term and x1 is second term and x2 is third term

i am really stuck on part b) if anyone can help and if any one can point out if part a) and c) are correct i would really appreciate it

 

for b) ive now found that for fixed point 1/2 using the gradient criterion we get

-1<(1/2x)+(2)<1

which gives us -1<(1/2x)+2=-6<x
(1/2x)+2<1=-2<x

thus -6<x<-2

so interval of attraction for fixed point 1/2 is (-6,-2)

Answer 1

you have part (a) incorrect

1/2x^2 + 2x - 5/8 --> 4x^2 + 16x - 5

the fixed points are -2 + ((-sqrt21)/2) or -2 - ((-sqrt21)/2)

part (b) not sure what you have done either

but you should be looking at -1 < x < 1

which would -1 < x+2 < 1 ---> -3 < x < -1

and bit more for you to do

Yes part (c) looks like you are on the right tracks