Fixed points of f?

f(x) = x^2+13/12x - 1/2

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= 0 (express your answer as fractions in their lowest terms) State the long term behaviour of this sequence.

Thank for your help with this!

a fixed point is a point where x=f(x)

In this case

x= x^2+13/12x - 1/2 solve for x.

gradient of attraction: when gradient is +

http://mathworld.wolfram.com/FixedPoint.html

Can someone please help answering the last part of the question I'm stuck:

c) find exact values of the 2nd and 3rd terms of the sequence xn obtained by iterating f with initial term x0= 0 (express your answer as fractions in their lowest terms) State the long term behaviour of this sequence.

a) To find the fixed points of the function f(x), we need to solve the equation f(x) = x. So we substitute f(x) = x into the function and solve for x.

f(x) = x^2 + (13/12)x - 1/2

x^2 + (13/12)x - 1/2 = x

x^2 + (13/12)x - 1/2 - x = 0

x^2 - (11/12)x - 1/2 = 0

To solve this quadratic equation, we can use the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / (2a)

In this case, a = 1, b = -11/12, and c = -1/2. Plugging these values into the quadratic formula:

x = [-(11/12) ± sqrt((-11/12)^2 - 4(1)(-1/2))] / (2(1))

Simplifying further:

x = [-(11/12) ± sqrt(121/144 + 2/4)] / 2

x = [-(11/12) ± sqrt(121/144 + 72/144)] / 2

x = [-(11/12) ± sqrt(193/144)] / 2

To simplify the square root, we can factorise 193 into 13 * 4 * 4:

x = [-(11/12) ± sqrt((13 * 4 * 4) / 144)] / 2

x = [-(11/12) ± (2 * sqrt(13)) / (2 * 12)] / 2

Simplifying further and expressing the fractions in their lowest terms:

x1 = (-11 - 2sqrt(13)) / 12
x2 = (-11 + 2sqrt(13)) / 12

These are the fixed points of f(x). Now let's classify them.

To classify the fixed points, we need to examine the derivative of f(x) at those points. The classification depends on whether the derivative is positive, negative, or zero.

f'(x) = 2x + 13/12

To classify x1:

f'(x1) = 2(-11 - 2sqrt(13)) / 12 + 13/12

Simplifying:

f'(x1) = (-22 - 4sqrt(13) + 13) / 12

f'(x1) = (-9 - 4sqrt(13)) / 12

Since f'(x1) is negative, x1 is a stable fixed point.

To classify x2:

f'(x2) = 2(-11 + 2sqrt(13)) / 12 + 13/12

Simplifying:

f'(x2) = (-22 + 4sqrt(13) + 13) / 12

f'(x2) = (-9 + 4sqrt(13)) / 12

Since f'(x2) is positive, x2 is an unstable fixed point.

b) To determine an interval of attraction for one of the fixed points (let's consider x1), we need to use the gradient criterion. The gradient criterion states that if the absolute value of the derivative of f at x1 is less than 1, then any value in a small enough interval around x1 will converge to x1.

So, we need to find an interval around x1 where |f'(x1)| < 1.

|f'(x1)| = |-9 - 4sqrt(13)| / 12

To find the interval of attraction, we want |f'(x1)| < 1:

|-9 - 4sqrt(13)| / 12 < 1

|-9 - 4sqrt(13)| < 12

To solve this inequality, we need to consider two cases:

Case 1: -9 - 4sqrt(13) < 12

Solving for sqrt(13):

-9 - 4sqrt(13) < 12

-4sqrt(13) < 12 + 9

-4sqrt(13) < 21

sqrt(13) > -21/4 (Note: To simplify the steps, we square both sides of the inequality.)

13 > 441/16

Since 13 is not greater than 441/16, this case is not valid.

Case 2: -9 - 4sqrt(13) > -12

Solving for sqrt(13):

-9 - 4sqrt(13) > -12

-4sqrt(13) > -12 + 9

-4sqrt(13) > -3

Dividing by -4 (since it is negative and will reverse the inequality):

sqrt(13) < 3/4

Squaring both sides:

13 < 9/16

Since 13 is greater than 9/16, this case is valid.

Therefore, the interval of attraction for x1 is (-∞, 3/4].

c) We are given the initial term x0 = 0, and we want to find the 2nd and 3rd terms of the sequence xn obtained by iterating f(x).

Start by calculating f(0):

f(0) = (0)^2 + (13/12)(0) - 1/2

f(0) = -1/2

Now calculate f(f(0)) for the 2nd term:

f(f(0)) = f(-1/2) = (-1/2)^2 + (13/12)(-1/2) - 1/2

Simplifying:

f(f(0)) = 1/4 - 13/24 - 1/2

f(f(0)) = 6/24 - 13/24 - 12/24

f(f(0)) = -19/24

Now calculate f(f(f(0))) for the 3rd term:

f(f(f(0))) = f(-19/24) = (-19/24)^2 + (13/12)(-19/24) - 1/2

Simplifying:

f(f(f(0))) = 361/576 - 247/576 - 288/576

f(f(f(0))) = -174/576

To find the long-term behavior of this sequence, we can continue iterating f(x) and observe if the terms converge to any particular value. However, in this case, we can see that the sequence is not converging to any specific value. Instead, it seems to be oscillating between the fixed points x1 and x2:

0, -1/2, -19/24, -1/2, -19/24, -1/2, ...

Therefore, the long-term behavior of this sequence is that it oscillates between the fixed points x1 and x2.