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- Start with the given value at index \(0\) and apply the rule once. - Each new row depends on the value in the immediately preceding row. - Interpret the subscript as the number of recurrence steps from the starting index.

1. Apply the recurrence to \(r_0\): \(r_1=3\cdot2-1=5\). 2. Use the newly found value: \(r_2=3\cdot5-1=14\). 3. Continue once more: \(r_3=3\cdot14-1=41\). 4. The value \(r_2=14\) is the result after applying the recurrence rule twice starting from \(r_0\). It is also the input used to produce \(r_3\).

\(r_1=5\), \(r_2=14\), and \(r_3=41\). The term \(r_2\) is the value after two recurrence steps from the starting term.

- A fixed point gives the same value before and after the recurrence update. - Represent that unchanged value with one variable on both sides of the update rule. - Check the result by putting it back into the recurrence.

At a fixed point, the current and next values are equal, so \(L=\frac{1}{4}L+9\). Thus \(\frac{3}{4}L=9\), giving \(L=12\). Substituting verifies the result: \(\frac{1}{4}(12)+9=12\).

The fixed point is \(12\).

- Determine what fraction of the concentration remains after one \(6\)-hour interval. - Find a value that would remain unchanged by one complete recurrence step. - Compare each term's distance from that stable value with the previous term's distance.

1. After a \(6\)-hour interval, \(60\%\) of the previous concentration remains. The next dose then adds \(80\,\text{mg/L}\), so \(c_{n+1}=0.6c_n+80\), with \(c_0=0\). 2. A stable value \(L\) satisfies \(L=0.6L+80\), giving \(L=200\). 3. Subtract \(200\) from both sides of the recurrence: \(c_{n+1}-200=0.6(c_n-200)\). Thus the distance from \(200\) is multiplied by \(0.6\) after every dose. Since \(0<0.6<1\), this geometric difference approaches \(0\). 4. Therefore \(c_n\) approaches \(200\,\text{mg/L}\).

a) \(c_0=0\) and \(c_{n+1}=0.6c_n+80\). b) \(200\,\text{mg/L}\); the difference from \(200\) is multiplied by \(0.6\) at each step, so it approaches \(0\).

- Move the non-\(x\) terms to the other side, then divide by \(2\). - Set the calculator to radians. - Repeatedly apply \(x_{n+1}=g(x_n)\).

1. Rearrange the equation: \(2x=\cos(x)+1\), so \(x=\frac{1}{2}(\cos(x)+1)\). 2. In radians, \(x_1=\frac{1}{2}(\cos(0.8)+1)\approx0.8484\). 3. Then \(x_2=\frac{1}{2}(\cos(x_1)+1)\approx0.8306\). 4. Finally, \(x_3=\frac{1}{2}(\cos(x_2)+1)\approx0.8372\).

1. \(x=\frac{1}{2}(\cos(x)+1)\) 2. \(x_1\approx0.8484\); \(x_2\approx0.8306\); \(x_3\approx0.8372\)

- Translate the percentage drained into the percentage that remains. - Preserve the stated order: drain first, then add the fixed amount. - Use each day's result as the input for the next recurrence step.

1. Draining \(15\%\) leaves \(85\%\) of the current amount, so after the drain the tank contains \(0.85V_n\). 2. The \(40\)-gallon addition occurs afterward. Therefore \(V_{n+1}=0.85V_n+40\), with \(V_0=300\). 3. Iterating gives \(V_1=0.85\cdot300+40=295\), \(V_2=0.85\cdot295+40=290.75\), and \(V_3=0.85\cdot290.75+40=287.1375\).

a) \(V_0=300\) and \(V_{n+1}=0.85V_n+40\). b) \(V_1=295\) gallons, \(V_2=290.75\) gallons, and \(V_3=287.1375\) gallons.

- Iterate the recurrence several times before trying to describe the long-term behavior. - Separate even and odd indices when looking for an explicit rule. - Boundedness only limits where the terms can lie; convergence requires the terms to approach one value.

a) The terms are \(2,4,2,4,2,4\). b) The parity pattern is \(x_n=3-(-1)^n\). c) The sequence is bounded because every term is either \(2\) or \(4\). It does not converge because the even-indexed terms are always \(2\) while the odd-indexed terms are always \(4\), so the sequence never approaches one common value.

a) \(2,4,2,4,2,4\) b) \(x_n=3-(-1)^n\) c) The sequence is bounded but does not converge; it alternates forever between \(2\) and \(4\).

- The weekly increase is based on the remaining gap to \(1200\). - Write a recursion using the current number of known words. - The initial gap to the goal is \(900\) words. - Use logarithms to find the boundary time, then check whole weeks.

1. The recursive model is \(V_{n+1}=V_n+0.10(1200-V_n)\), with \(V_0=300\). 2. The equivalent explicit model is \(V_n=1200-900(0.9)^n\). 3. a) \(V_8=1200-900(0.9)^8\approx812.58\), so the student knows about \(813\) words. 4. b) Find the boundary by solving \(1200-900(0.9)^n=1000\). Then \(0.9^n=\frac{2}{9}\), so \(n\approx14.28\). 5. After \(14\) weeks, the student knows fewer than \(1000\) words; after \(15\) weeks, the student knows more than \(1000\).

a) About \(813\) words b) After \(15\) weeks

- First find the number of households represented by \(75\%\). - Each month, the increase is based on the households that still lack service. - You may use either a recursion or an explicit formula for the remaining gap. - Use logarithms to find the boundary time, then check whole months.

1. The target is \(0.75\cdot8000=6000\) households. 2. The recursive model is \(H_{n+1}=H_n+0.12(8000-H_n)\), with \(H_0=400\). 3. The equivalent explicit model is \(H_n=8000-7600(0.88)^n\). 4. Solve the boundary equation \(8000-7600(0.88)^n=6000\). Then \(0.88^n=\frac{5}{19}\), so \(n\approx10.44\). 5. After \(10\) months, about \(5883\) households have service; after \(11\) months, about \(6137\) do. Therefore, the target is first reached after \(11\) months.

After \(11\) months

- Focus on how the difference between the coffee and room temperatures changes. - Identify the temperature the sequence approaches. - Determine the multiplier for the portion of the difference that remains. - Evaluate terms until the temperature first falls below the target.

1. Since \(88\%\) of the temperature difference remains each minute, \(T_{n+1}-21=0.88(T_n-21)\). Therefore, \(T_{n+1}=0.88T_n+2.52\), with \(T_0=80\). 2. The initial difference from room temperature is \(80-21=59\). Thus, \(T_n=21+59\cdot0.88^n\). 3. Evaluate consecutive terms near the threshold: \(T_7\approx45.112^\circ\text{C}\) and \(T_8\approx42.218^\circ\text{C}\). 4. The coffee is first cooler than \(45^\circ\text{C}\) after \(8\) minutes.

a) \(T_0=80\); \(T_{n+1}=0.88T_n+2.52\) b) \(T_n=21+59\cdot0.88^n\) c) \(8\) minutes

- Determine how much of the previous amount remains before the next dose. - Find the fixed point that the sequence approaches. - Remember that \(a_0\) represents the amount after the first dose. - Evaluate terms until the amount first exceeds \(133\,\text{mg}\).

1. One-fourth of the previous amount remains, and the next dose adds \(100\,\text{mg}\). Thus, \(a_{n+1}=0.25a_n+100\), with \(a_0=100\). 2. The fixed point satisfies \(S=0.25S+100\), so \(S=\frac{400}{3}\). 3. The explicit form is \(a_n=S-(S-a_0)0.25^n=\frac{400}{3}-\frac{100}{3}\cdot0.25^n\). 4. The relevant terms are \(a_3=132.8125\,\text{mg}\) and \(a_4=133.203125\,\text{mg}\). 5. Because \(a_0\) is the amount after the first dose, \(a_4\) corresponds to the fifth dose. The amount first exceeds \(133\,\text{mg}\) after dose \(5\).

a) \(a_0=100\); \(a_{n+1}=0.25a_n+100\) b) \(a_n=\frac{400}{3}-\frac{100}{3}\cdot0.25^n\) c) After the fifth dose

- Build each year's balance from the previous balance, interest, and end-of-year payment. - Use the actual loan principal, not only the amount received, as the initial balance. - Calculate one year at a time. - Compare the two balances after year \(5\).

1. For Offer A, use \(K_{n+1}=1.055K_n-20{,}000\), with \(K_0=150{,}000\). 2. The balances after years \(1\) through \(5\) are \(\$138{,}250.00\), \(\$125{,}853.75\), \(\$112{,}775.71\), \(\$98{,}978.37\), and \(\$84{,}422.18\). 3. For Offer B, use \(K_{n+1}=1.03K_n-20{,}000\), with \(K_0=165{,}000\). 4. The balances after years \(1\) through \(5\) are \(\$149{,}950.00\), \(\$134{,}448.50\), \(\$118{,}481.96\), \(\$102{,}036.41\), and \(\$85{,}097.51\). 5. After five years, Offer A has the smaller balance by about \(\$675.33\). Based only on the remaining balance after five years, the company should choose Offer A.

After five years, Offer A has a balance of approximately \(\$84{,}422.18\), and Offer B has a balance of approximately \(\$85{,}097.51\). Offer A has the smaller remaining balance.

- Make a year-by-year table for each plan. - In Plan 2, the growth factor changes each year. - Apply the withdrawal after adding that year's interest.

1. For Plan 1, use \(K_{n+1}=1.035K_n-4000\), with \(K_0=50{,}000\). 2. The balances after years \(1\) through \(4\) are \(\$47{,}750.00\), \(\$45{,}421.25\), \(\$43{,}010.99\), and \(\$40{,}516.38\). 3. For Plan 2, apply the yearly factors \(1.02\), \(1.025\), \(1.03\), and \(1.035\), subtracting \(\$4000\) after each year. 4. The balances after years \(1\) through \(4\) are \(\$47{,}000.00\), \(\$44{,}175.00\), \(\$41{,}500.25\), and \(\$38{,}952.76\). 5. Plan 1 leaves more money after four years.

After four years, Plan 1 has approximately \(\$40{,}516.38\), and Plan 2 has approximately \(\$38{,}952.76\). Plan 1 leaves more money.

- Identify the surviving fraction and the fixed number added each month. - Substitute the recurrence parameters into the supplied explicit formula without rounding early. - After finding a stable value, compare the next term's distance from it with the current term's distance.

1. Since \(80\%\) of the fish survive and \(500\) new fish are added, \(a_{n+1}=0.8a_n+500\), with \(a_1=500\). 2. Substitute \(q=0.8\), \(c=500\), and \(n=5\): \(a_5=0.8^4\cdot500+500\cdot\frac{0.8^4-1}{0.8-1}=1680.8\). The model gives approximately \(1681\) fish. 3. A stable value \(L\) satisfies \(L=0.8L+500\), so \(L=2500\). 4. Subtracting \(2500\) from the recurrence gives \(a_{n+1}-2500=0.8(a_n-2500)\). Therefore the distance from \(2500\) is multiplied by \(0.8\) each month. Since \(0<0.8<1\), that distance approaches \(0\), so \(a_n\) approaches \(2500\).

a) \(a_{n+1}=0.8a_n+500\), with \(a_1=500\). b) \(a_5=1680.8\), or approximately \(1681\) fish. c) The population approaches \(2500\) fish because its distance from \(2500\) is multiplied by \(0.8\) each month.

- Determine what fraction remains after one night and use it in the recurrence. - After finding a stable value, examine how the distance from that value changes from one day to the next. - Compare the justified long-run value with the medical threshold rather than relying only on the first few terms. - For the doubled dose, identify what changes in the recurrence and what stays the same.

1. Since \(80\%\) remains overnight, \(a_{n+1}=0.8a_n+12\), with \(a_0=0\). 2. The first five amounts are \(a_1=12\), \(a_2=21.6\), \(a_3=29.28\), \(a_4=35.424\), and \(a_5=40.3392\), all in milligrams. 3. A stable value \(L\) satisfies \(L=0.8L+12\), so \(L=60\,\text{mg}\). Moreover, \(a_{n+1}-60=0.8(a_n-60)\). Thus the distance from \(60\) is multiplied by \(0.8\) each day, so it approaches \(0\). Therefore \(a_n\) approaches \(60\,\text{mg}\). 4. Because the sequence approaches \(60\,\text{mg}\), which is above \(50\,\text{mg}\), the \(50\,\text{mg}\) limit will eventually be exceeded. 5. With a \(24\,\text{mg}\) daily dose, the stable value satisfies \(L=0.8L+24\), giving \(L=120\,\text{mg}\). The same difference-from-the-fixed-point argument shows convergence because the multiplier remains \(0.8\).

a) Day 1: \(12\,\text{mg}\); Day 2: \(21.6\,\text{mg}\); Day 3: \(29.28\,\text{mg}\); Day 4: \(35.424\,\text{mg}\); Day 5: \(40.3392\,\text{mg}\). b) \(60\,\text{mg}\). c) No. The amount eventually exceeds \(50\,\text{mg}\). d) The limiting amount doubles to \(120\,\text{mg}\).

- Relate each year's amount to the previous year's amount using the retained fraction and the inflow. - After finding a stable value, rewrite the recurrence in terms of the difference from that value. - For the reduced inflow, compare the new stable value with the stated initial amount. - With no inflow, identify the familiar type of sequence that remains.

1. Since \(90\%\) remains and \(200\,\text{kg}\) is added, \(s_{n+1}=0.9s_n+200\), with \(s_0=1500\). 2. The successive values are \(s_1=1550\), \(s_2=1595\), \(s_3=1635.5\), and \(s_4=1671.95\), in kilograms. 3. A stable value satisfies \(L=0.9L+200\), so \(L=2000\,\text{kg}\). Also, \(s_{n+1}-2000=0.9(s_n-2000)\). Because \(0<0.9<1\), the difference from \(2000\) approaches \(0\), so the salt amount approaches \(2000\,\text{kg}\). 4. With an inflow of \(150\,\text{kg}\), the stable value is \(1500\,\text{kg}\). Since the initial amount is already \(1500\,\text{kg}\), the recurrence keeps it constant. 5. With no inflow, \(s_{n+1}=0.9s_n\), so \(s_n=1500(0.9)^n\), which approaches \(0\).

a) \(s_{n+1}=0.9s_n+200\), with \(s_0=1500\); \(s_2=1595\,\text{kg}\); \(s_4=1671.95\,\text{kg}\). b) \(2000\,\text{kg}\). c) \(1500\,\text{kg}\). d) The amount decreases geometrically toward \(0\,\text{kg}\).

- Factor \(x\) from the two terms containing it before isolating \(x\). - Use each unrounded iterate as the input for the next recurrence step. - Compare successive iterates with an absolute difference when applying the stopping rule. - Distinguish numerical evidence from a theorem that proves convergence or root existence.

1. Rearrange \(x^3+3x=5\) as \(x(x^2+3)=5\). Since \(x^2+3>0\), this gives \(x=\frac{5}{x^2+3}\). 2. The first iterations are \(x_1=\frac{5}{1^2+3}=1.2500\), \(x_2=\frac{5}{1.25^2+3}\approx1.0959\), and, using the unrounded value of \(x_2\), \(x_3\approx1.1902\). 3. Continuing with unrounded values gives \(x_{12}\approx1.1537\) and \(x_{13}\approx1.1544\), with \(|x_{13}-x_{12}|\approx0.0007<0.001\). Thus \(1.1544\) is a numerical estimate suggested by this iteration. The close consecutive values are numerical evidence of stabilization, not by themselves a proof that the iteration converges or that a root exists.

1. \(x=\frac{5}{x^2+3}\). 2. \(x_1=1.2500\), \(x_2\approx1.0959\), \(x_3\approx1.1902\). 3. The stopping condition is first met at \(x_{13}\approx1.1544\). This is a numerical estimate, not a proof of convergence or existence.

- Determine the hourly net change separately when the pump is off and when it is on. - Find the beginning-of-hour volume above which adding the pumped-case net amount would exceed capacity. - Apply the appropriate case in order, beginning with \(V_0=400\). - Convert \(75\%\) of the capacity to a volume before comparing the iterates.

1. If \(0\le V_n<1200\), the pump is off and \(V_{n+1}=V_n+350\). This case cannot exceed capacity because \(V_n+350<1550\). 2. If \(1200\le V_n\le1770\), the pump is on and \(V_{n+1}=V_n+350-120=V_n+230\). 3. If \(1770<V_n\le2000\), the pump is on but \(V_n+230>2000\), so overflow leaves \(V_{n+1}=2000\). Therefore, \(V_{n+1}=\begin{cases}V_n+350,&0\le V_n<1200\\V_n+230,&1200\le V_n\le1770\\2000,&1770<V_n\le2000\end{cases}\), with \(V_0=400\). 4. The first values are \(V_1=750\), \(V_2=1100\), \(V_3=1450\), and \(V_4=1680\), all in cubic meters. 5. Seventy-five percent of capacity is \(0.75\cdot2000=1500\,\text{m}^3\). Since \(V_3=1450\) and \(V_4=1680\), the basin is first more than \(75\%\) full after \(4\) complete hours.

a) \(V_0=400\) and \(V_{n+1}=\begin{cases}V_n+350,&0\le V_n<1200\\V_n+230,&1200\le V_n\le1770\\2000,&1770<V_n\le2000\end{cases}\). b) \(V_1=750\,\text{m}^3\), \(V_2=1100\,\text{m}^3\), \(V_3=1450\,\text{m}^3\), \(V_4=1680\,\text{m}^3\). c) After \(4\) complete hours.

- Use the order of operations in each recurrence to interpret the daily timing. - Find a stable value for each model, then compare how the distance from that value changes after one recurrence step. - For part c, convert \(90\%\) of Model A's justified limit to a target amount. - Check the two consecutive days around the first crossing of that target.

1. In Model A, the existing amount is reduced first and then the new dose is added. In Model B, the dose is added first and then \(25\%\) of the total is eliminated. 2. For Model A, a stable value satisfies \(L=0.75L+40\), so \(L_A=160\,\text{mg}\). Subtracting \(160\) from the recurrence gives \(W_n-160=0.75(W_{n-1}-160)\). Since \(0<0.75<1\), the difference from \(160\) approaches \(0\). 3. Model B can be written \(W_n=0.75W_{n-1}+30\). Its stable value is \(L_B=120\,\text{mg}\), and \(W_n-120=0.75(W_{n-1}-120)\). Therefore Model B also converges to its stable value. 4. Ninety percent of \(160\,\text{mg}\) is \(144\,\text{mg}\). For Model A, \(W_8\approx143.982\,\text{mg}\) and \(W_9\approx147.986\,\text{mg}\). Therefore at least \(90\%\) of the limiting amount is first reached on day \(9\).

a) Model A applies elimination before the dose; Model B applies the dose before elimination. b) Model A approaches \(160\,\text{mg}\); Model B approaches \(120\,\text{mg}\). In each model, the difference from its fixed point is multiplied by \(0.75\) each day. c) Day \(9\).

- Follow the stated order of growth and filtering when forming one recurrence step. - After finding a stable value, rewrite the recurrence in terms of the difference from that value. - Express the difference after \(n\) days as a geometric factor times the initial difference. - Translate \(1\%\) of the limiting value into an allowable absolute error.

1. Because the filter acts after the daily growth, \(A_n=0.85(A_{n-1}+3)=0.85A_{n-1}+2.55\), with \(A_0=10\). 2. A stable value satisfies \(L=0.85L+2.55\), so \(L=17\,\text{kg}\). 3. Subtracting \(17\) from the recurrence gives \(A_n-17=0.85(A_{n-1}-17)\). Hence \(A_n-17=(10-17)(0.85)^n=-7(0.85)^n\), which approaches \(0\). Therefore \(A_n\) approaches \(17\,\text{kg}\). 4. One percent of the limit is \(0.17\,\text{kg}\). The error is \(|A_n-17|=7(0.85)^n\). At \(n=22\), the error is approximately \(0.1960\,\text{kg}\); at \(n=23\), it is approximately \(0.1666\,\text{kg}\). 5. Therefore the algae mass is first within \(1\%\) of the long-term value after \(23\) days.

a) \(A_n=0.85(A_{n-1}+3)\), with \(A_0=10\). b) \(17\,\text{kg}\). c) After \(23\) days.

- Classify every complete tiling by what appears at its right-hand end. - Removing a final unit square shortens the strip by one, while removing a final domino shortens it by two. - Account separately for the two possible colors of a final unit square.

1. A tiling can end with a red unit square or a blue unit square. Removing that last square leaves any tiling of length \(n-1\), giving \(2t_{n-1}\) possibilities. 2. A tiling can instead end with a domino. Removing it leaves any tiling of length \(n-2\), giving \(t_{n-2}\) possibilities. 3. Therefore \(t_n=2t_{n-1}+t_{n-2}\) for \(n\ge2\), with \(t_0=1\) and \(t_1=2\). 4. The recurrence gives \(t_2=5\), \(t_3=12\), \(t_4=29\), and \(t_5=70\).

\(t_0=1\), \(t_1=2\), and \(t_n=2t_{n-1}+t_{n-2}\) for \(n\ge2\). Also, \(t_5=70\).

- After each update, replace the result by its least nonnegative residue modulo \(8\). - When a state repeats in a deterministic recurrence, the subsequent states repeat as well. - Use the cycle length to replace the distant index by its remainder modulo the period.

a) Reducing modulo \(8\) at each step gives \(1,4,5,0,1,4,5,0\). b) Once \(1\) returns, the deterministic recurrence repeats the same sequence of states. The cycle has length \(4\): \(1,4,5,0\). c) Since \(2026\equiv2\pmod{4}\), \(a_{2026}\) has the same position in the cycle as \(a_2\). Therefore \(a_{2026}=5\).

a) \(1,4,5,0,1,4,5,0\) b) The cycle length is \(4\). c) \(a_{2026}=5\).

- Each year, find \(18\%\) of the difference between \(3000\) and the current population. - Add that increase to the current population to get the next term. - An explicit form can be written using the initial distance from the carrying capacity. - The answer to part b must be a whole year that satisfies the condition.

1. The recursive model is \(B_{n+1}=B_n+0.18(3000-B_n)=0.82B_n+540\), with \(B_0=600\). 2. Repeatedly applying the recursion gives \(B_1=1032\), \(B_2\approx1386\), \(B_3\approx1677\), \(B_4\approx1915\), and \(B_5\approx2110\). 3. The equivalent explicit model is \(B_n=3000-2400(0.82)^n\). 4. Find the boundary by solving \(3000-2400(0.82)^n=2500\). Then \(0.82^n=\frac{500}{2400}\), so \(n=\frac{\ln\left(\frac{500}{2400}\right)}{\ln(0.82)}\approx7.90\). 5. Therefore, the population first exceeds \(2500\) fish in year \(8\).

a) <table> <tr><td>Year</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td></tr> <tr><td>Population</td><td>600</td><td>1032</td><td>1386</td><td>1677</td><td>1915</td><td>2110</td></tr> </table> b) Year \(8\)

- Focus on the percentage that is still missing from full charge. - The missing amount is multiplied by the same factor every \(5\) minutes. - Convert \(30\) minutes to a number of charging intervals. - After finding a noninteger number of intervals, check the next complete interval.

1. The initial gap to full charge is \(100-12=88\) percentage points. Each interval leaves \(85\%\) of the previous gap. 2. Therefore, \(C(n)=100-88(0.85)^n\). 3. b) Thirty minutes is \(6\) intervals, so \(C(6)=100-88(0.85)^6\approx66.81\%\). 4. c) The boundary equation is \(100-88(0.85)^n=90\), so \(0.85^n=\frac{10}{88}\). 5. This gives \(n\approx13.38\). Since the model changes only at complete \(5\)-minute intervals, check \(n=13\) and \(n=14\). The levels are about \(89.36\%\) and \(90.96\%\), respectively. 6. Therefore, the battery first reaches at least \(90\%\) after \(14\) intervals, or \(70\) minutes.

a) \(C(n)=100-88(0.85)^n\) b) About \(66.81\%\) c) \(70\) minutes

- Simplify division by \(0.5\). - A constant sequence cannot depend on \(n\). - Consider the behavior of a power with base greater than \(1\). - Track the sign of the coefficient multiplying the power.

1. Substitute into the explicit formula: \(a_n=1.5^{n-1}a_1-100\frac{1.5^{n-1}-1}{0.5}\). 2. Since dividing by \(0.5\) multiplies by \(2\), \(a_n=a_1\cdot1.5^{n-1}-200(1.5^{n-1}-1)\). 3. Therefore, \(a_n=200+(a_1-200)1.5^{n-1}\). 4. The sequence is constant when \(a_1-200=0\), so \(a_1=200\). 5. Because \(1.5^{n-1}\) grows without bound, \(a_n\to\infty\) when \(a_1>200\), and \(a_n\to-\infty\) when \(a_1<200\).

a) \(a_n=200+(a_1-200)1.5^{n-1}\) b) \(a_1=200\) c) If \(a_1>200\), then \(a_n\to\infty\). If \(a_1<200\), then \(a_n\to-\infty\).

- Substitute \(B_0\), \(B_1\), and \(G\) into the model to solve for \(k\). - Use the unrounded population from one week in the next calculation. - Examine the factor \(G-B_n\) when the population is close to capacity.

1. Substitute the given values: \(528=400+k\cdot400(2000-400)\). Therefore, \(128=640{,}000k\), so \(k=0.0002\). 2. For week \(2\), \(B_2=528+0.0002\cdot528(2000-528)=683.4432\approx683\). 3. Using the unrounded value of \(B_2\), \(B_3=683.4432+0.0002\cdot683.4432(2000-683.4432)\approx863.4016\approx863\). 4. As \(B_n\) approaches \(G\), the factor \(G-B_n\) approaches \(0\). Therefore, the weekly increase approaches \(0\).

1. \(k=0.0002\) 2. \(B_2\approx683\); \(B_3\approx863\) 3. The weekly increase approaches \(0\).

- Use the first-year increase to determine the unknown growth parameter. - Compare the harvest rate with the natural per-capita growth rate when the population is very small. - At the critical harvest fraction, simplify the recurrence and examine the multiplier applied to a positive population. - To justify approach to zero, consider what would happen if the population stayed bounded away from zero by some positive amount.

1. Substitute \(B_0=800\), \(B_1=912\), and \(K=5000\): \(112=r\cdot800\cdot\frac{4200}{5000}=672r\). Hence \(r=\frac16\approx0.1667\). 2. The population change is \(B_{n+1}-B_n=B_n\left(r\frac{K-B_n}{K}-q\right)\). If \(q<r\), then for sufficiently small positive \(B_n\), the expression in parentheses is positive. Therefore no harvest fraction below \(r\) can make every positive initial population decrease toward \(0\). 3. Take \(q=r=\frac16\). Then \(B_{n+1}=B_n\left(1-\frac{B_n}{6K}\right)\). For \(0<B_n\le K\), the multiplier lies between \(\frac56\) and \(1\), so \(0<B_{n+1}<B_n\). 4. To justify that the decreasing population tends to \(0\), suppose instead that it stayed at least \(\varepsilon>0\). Then \(B_{n+1}\le B_n\left(1-\frac{\varepsilon}{6K}\right)\). The factor in parentheses is a fixed number less than \(1\), so repeated application would force \(B_n\) toward \(0\), contradicting \(B_n\ge\varepsilon\). Therefore \(B_n\) approaches \(0\). 5. Thus the minimum harvest fraction is \(q=\frac16\), or approximately \(16.67\%\).

1. \(r=\frac16\approx0.1667\). 2. The minimum harvest fraction is \(q=\frac16\), or approximately \(16.67\%\). At this value every positive population up to \(K\) stays positive, decreases, and approaches \(0\).

- Rearrange each fixed-point equation back to the polynomial while keeping track of any denominator restriction. - Use each unrounded output as the next input. - Compare whether the iterates stabilize inside the target interval, leave it, or head toward a different value. - Separate what four computed steps show from a claim about every later step.

1. Form (1) gives \(5x=x^3+1\), form (2) gives \(x^3=5x-1\), and form (3) gives \(x(5-x^2)=1\). Each rearranges to \(x^3-5x+1=0\). Values with \(5-x^2=0\) are outside the domain of form (3) and are not roots of the polynomial. 2. From \(x_0=0\), form (1) gives \(0.2\), \(0.2016\), \(0.2016387\), \(0.2016397\). Form (2) gives \(-1\), \(-1.8171206\), \(-2.1605648\), \(-2.2768197\). Form (3) gives \(0.2\), \(0.2016129\), \(0.2016392\), \(0.2016397\). Thus forms (1) and (3) appear to stabilize near the small positive solution, while form (2) moves away from that interval. 3. From \(x_0=2\), form (1) gives \(1.8\), \(1.3664\), \(0.7102271\), \(0.2716509\). Form (2) gives \(2.0800838\), \(2.1104857\), \(2.1218007\), \(2.1259814\). Form (3) gives \(1\), \(0.25\), \(0.2025316\), \(0.2016543\). Only form (2) appears to move toward the solution in \([2,3]\). 4. A finite list of iterates can show numerical behavior over the computed steps, but it does not by itself prove what all later iterates do. Therefore the conclusions here are stated as numerical evidence, not proofs of convergence.

a) All three fixed-point equations rearrange to \(x^3-5x+1=0\) on their domains. b) Forms (1) and (3) appear to move toward the solution in \([0,1]\); form (2) does not. c) From \(x_0=2\), only form (2) appears to move toward the solution in \([2,3]\). The computed terms are numerical evidence, not a convergence proof.

- A fixed point occurs where the graph of the recurrence function intersects the line \(y=x\). - Follow each cobweb step vertically to the function and horizontally to the line to see how the next input is produced. - For the exact value, set the next value equal to the current value and check any solutions against the original square-root equation.

a) The two graphs intersect near \(x=2\), so the fixed point is approximately \(2\). b) Iterating gives \(x_1\approx1.581\), \(x_2\approx1.892\), \(x_3\approx1.973\), and \(x_4\approx1.993\). c) A fixed point satisfies \(x=\sqrt{x+2}\). Squaring gives \(x^2-x-2=0\), so \((x-2)(x+1)=0\). Because \(\sqrt{x+2}\) is nonnegative, the fixed point shown is \(x=2\). The cobweb steps and numerical iterates approach \(2\), so the displayed iteration appears to converge to \(2\).

a) Approximately \(2\). b) \(x_1\approx1.581\), \(x_2\approx1.892\), \(x_3\approx1.973\), \(x_4\approx1.993\). c) The exact fixed point is \(2\), and the displayed iterates appear to converge to it.