1. **State the problem:** We have a discrete-time dynamical system (DTDS) defined by the updating rule $$N_{t+1} = \frac{2}{4 + N_t^3}$$ for $t=0,1,2,\ldots$. We want to analyze the fixed points and approximate an equilibrium. 2. **Part (a): Find the updating function $f(x)$:** Replace $N_t$ by $x$ to write the function as $$f(x) = \frac{2}{4 + x^3}.$$ This function gives the next population value from the current value $x$. 3. **Part (b): Write the fixed point equation $f(x) = x$ as $g(x) = 0$ where $g(x)$ is a polynomial:** The fixed points satisfy $$x = \frac{2}{4 + x^3}.$$ Multiply both sides by $4 + x^3$: $$x(4 + x^3) = 2,$$ which expands to $$4x + x^4 = 2.$$ Rearranging to zero, $$x^4 + 4x - 2 = 0.$$ Thus, $$g(x) = x^4 + 4x - 2.$$ This is the polynomial form for the fixed points. 4. **Part (c): Use the Intermediate Value Theorem (IVT) to find a closed interval $[a,b]$ containing a biologically relevant equilibrium:** We want to find $a,b$ such that $g(a)g(b) < 0$, indicating a root in $(a,b)$. Evaluate $g(0)$: $$g(0) = 0^4 + 4\cdot0 - 2 = -2 < 0.$$ Evaluate $g(1)$: $$g(1) = 1 + 4 - 2 = 3 > 0.$$ Since $g(0) < 0$ and $g(1) > 0$, by IVT there is at least one root in $[0,1]$. Because population values are nonnegative and $[0,1]$ contains a root, this interval is biologically relevant. 5. **Part (d): Use Newton's method to approximate the equilibrium starting from $x_0$ in $[0,1]$:** Newton's method formula: $$x_{n+1} = x_n - \frac{g(x_n)}{g'(x_n)}.$$ Calculate derivative: $$g'(x) = 4x^3 + 4.$$ Choose initial guess $x_0 = 0.5$ (midpoint of $[0,1]$). Calculate iterations: - $x_0 = 0.5$ - $g(0.5) = (0.5)^4 + 4(0.5) - 2 = 0.0625 + 2 - 2 = 0.0625$ - $g'(0.5) = 4(0.5)^3 + 4 = 4(0.125) + 4 = 0.5 + 4 = 4.5$ - $x_1 = 0.5 - \frac{0.0625}{4.5} = 0.5 - 0.0139 = 0.4861$ - $g(0.4861) \approx (0.4861)^4 + 4(0.4861) - 2 = 0.0558 + 1.9444 - 2 = 0.0002$ - $g'(0.4861) = 4(0.4861)^3 + 4 = 4(0.1149) + 4 = 0.4596 + 4 = 4.4596$ - $x_2 = 0.4861 - \frac{0.0002}{4.4596} = 0.4861 - 0.000045 = 0.4861$ - $g(0.4861) \approx 0$ (very close), so further iterations will not change much. - $x_3 \approx 0.4861$ - $x_4 \approx 0.4861$ Final approximation after 4 iterations is about $0.486$. **Answer for (d):** $[0.5, 0.486]$. --- **Summary:** - (a) $f(x) = \frac{2}{4 + x^3}$ - (b) $g(x) = x^4 + 4x - 2$ - (c) $[0,1]$ - (d) $[0.5, 0.486]$