Question: 3x We consider the DTDS xt+1 = f(xt) for t = 0, 1, 2, 3, ..., where the updating function is f(x): 2 +0.4x

3x We consider the DTDS xt+1 = f(xt) for t = 0,1, 2, 3, ..., where the updating function is f(x): 2 +0.4x

3x We consider the DTDS xt+1 = f(xt) for t = 0, 1, 2, 3, ..., where the updating function is f(x): 2 +0.4x a) Find the equilibrium point(s). Separate each value by a semi-colon. Answer: b) Compute f'(x). Answer: c) If p1 80 < P2 are the two equilibrium points that you have found in (a), compute ' (p) and ' (p2). Give exact values in your answers. Answer: f'(p) = ' (p) = e) Starting from x = 5, calculate x1, x2 and 3. Give your answers with an accuracy of two decimal places. Answer: 1 Number x2 x3 = Number Number

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