Analyze a quadratic

mark

Let
[dmath]f(x) = x^2 - \frac{54}{5}x + 34.[/dmath]
Show that [imath]f[/imath] has two fixed points - one attractive and one repulsive.

joezayatz

Answer:

[dmath]
x = x^2 - (54/5)x + 34
[/dmath]
[dmath]
0 = x^2 - (59/5)x + 34
[/dmath]
[dmath]
x = ((59/5) ± \sqrt{(59/5)^2 - 4*34)}/2
[/dmath]
[dmath]
x = 5, 6.8
[/dmath]

Solve the derivative of f(x) at each point:

[dmath]
f'(x) = 2*x - 54/5
[/dmath]
[dmath]
f'(5) = 10 - 54/5 = -4/5
[/dmath]
[dmath]
|f'(5)| = 4/5
[/dmath]

Since 0 < |f'(5)| < 1, this fixed point is attractive.
[dmath]
f'(6.8) = 13.6 - 54/5 = 2.8
[/dmath]
Since |f'(6.8)| > 1, this fixed point is repulsive