- #1
kellymh
- 2
- 0
Investing function fc(x) = (6/x)+(x/2)-c where 0<= c <=3
a) Use alegbra to find the positive fixed point (in terms of c) and identify its exact interval of existence
b) Use algebra and calculus to find the exact interval of stability of the fixed point
c) Use algebra to find the points of the 2 cycle (in terms of c) and its exact interval of existence
d) Use algebra and calculus to find the exact interval of stability of the 2 cycle
I have found so far a positive fixed point of p(c) =(c^2 +12)^(1/2) -c
I then think the interval of existence is 0 <=c <=3
----
Yet, if this is correct I am having trouble finding the interval of stability.
I have taken the derivative of the function
so f'(x) = (1/2)-(6/x^2)
then I plugged in the fixed point, so f'(p(c))=(1/2)(6/p(c)^2)
I know a function is stable if the absolute value is less than 1, but do not know how to solve this to get an answer.Then the two cycle would be
p1(c) =(c^2 +12)^(1/2) -c
p2(c) =-(c^2 +12)^(1/2) -c
I am unsure what the interval of existence is and then again, unsure how to find the interval of stability
a) Use alegbra to find the positive fixed point (in terms of c) and identify its exact interval of existence
b) Use algebra and calculus to find the exact interval of stability of the fixed point
c) Use algebra to find the points of the 2 cycle (in terms of c) and its exact interval of existence
d) Use algebra and calculus to find the exact interval of stability of the 2 cycle
I have found so far a positive fixed point of p(c) =(c^2 +12)^(1/2) -c
I then think the interval of existence is 0 <=c <=3
----
Yet, if this is correct I am having trouble finding the interval of stability.
I have taken the derivative of the function
so f'(x) = (1/2)-(6/x^2)
then I plugged in the fixed point, so f'(p(c))=(1/2)(6/p(c)^2)
I know a function is stable if the absolute value is less than 1, but do not know how to solve this to get an answer.Then the two cycle would be
p1(c) =(c^2 +12)^(1/2) -c
p2(c) =-(c^2 +12)^(1/2) -c
I am unsure what the interval of existence is and then again, unsure how to find the interval of stability