Question

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(c) (2 pts) Let f(x) = 4xe-*. Find the fixed points and their stability. (d) (2 pts) Let f(x) = 2.x2 – 3. Find the fixed points along with their stability and determine the 2-cycle along with its stability. (e) (2 pts) Let c > 0 and fc(x) = c (2 - V3x + 4). Find the interval of stability of 0.

Answer 1

O fire) = 2x°-3 To find fixed point in → = 2600*)2 -32** 2600g 2 -x*-320 A ! ,4 - - - are the fixed point of fire), f'(x) = 4x 2. fx is a smooth function. f'(-1) = -4<, -" is a stable point p! () = 4.3 = 6 >1,2"Ź " is a unstable point, 3rd part . x92-1, x 3rd pant - = 8 p?(n) = f(pers) = f( 2x2-3) 2. 2.(2033)?-3 = $ * ,19*118-3 2864_2449715 f? GC) = 0 zy 88934_240*23715225 ż 8(x *)4 _ 2400*)2 _ x +15=0. => 8273 (Q*+1) – 872 (*+1) - 168*(x*+i) + 1563*+1)=0 27 (am +) (8X93 - 80*2 - 1605 +15) = 0 => (atel) (2**-3) (47*2 +2** -5): 0

=> 3 4 = H, 3, 1 I 127 2. Fixed point of $2(X) are, 293 01, , -14. But, 1,.& are the fixed point off. Again, f(-185) = -tar f(-+ + ) = -1 and f? ( -4*) - -* * ofen F 2 (-1-red) - Lowon This shows that set {-12 , teras forms the period 2- eyele of the map f. © - fe al = e (2-13x74). eso To find fixed point in fecax) a za : o e (2 - 53 28 44 )2X* 27 - 13** Ty = I -2 => [?(3x*+4) = 2402 * x*2_4ex* 2% a*?- (3 22+4)** + 4e2-4e2.0 zy am cam - 34274c) = 0 z4 g * 20 i 30? + 4C

. (1) 2 -20 231+Ч эс Р! (0) = -30 fece) is a stable if 1*, C+2) 2) - г) 1 - 1 - 1 21 2 -2 -2 21 - 2 - 2 1514ч 2-1 => - 2 2 1 зугу 21 27 -3 2 2 Гэмеч гэс by be a sontu <3 2; 9 - 2 ч. 9 2) Фе“ - 2 3 4 *-ч эе”-6 23x 2 эе. 4 9 А е'- 1, 4 2 е--144 Зе 2 x, e”- 6 12

© fea) = 40 er to find the fixed point of fex) a sem => 40% endt = ** 3 x*( 4 ex* _1) 20 ext ex* = 4. na awgy 1 o, logy are the fixed point of fex). f!() = 4e-x - Home-x since f Sie is a smooth Rretion then fiches Li, xo is attracting f! (a ) >1, * is repelling of ICO) = 4 >1 , 'O' is a unstable point el clony) - 48 -1684). _ yigie (494 - - q comey 21-6gey 1 2072 A . -013862 21 In logot is a stable point.