- At the same horizontal coordinate, distinguish the open point from the filled point. - Read the vertical coordinate of the point that is included in the function.

1. At \(t=2\), the open point is at \(72\), but an open point is not included in the function. 2. The filled point is at \(30\), so it gives the actual function value at the breakpoint.

\(30\)

- At \(t=4\), the second branch should still equal \(200\). - A \(10\%\) decrease leaves \(90\%\) of the previous value. - The exponent should count time elapsed since day \(4\).

At the moment the second branch begins, \(t=4\), the decay factor must have exponent \(0\) so the branch starts at \(200\). Only \(200(0.9)^{t-4}\) has that property. Also, a \(10\%\) decrease means a remaining factor of \(0.9\).

b) \(200(0.9)^{t-4}\)

- Check the strict and non-strict inequalities at the boundary. - Which exponent do you get when the time equals the start of the second phase?

1. The first rule excludes \(t=3\), while the second includes it. 2. Use the second rule: \(R(3)=50(0.8)^0=50\).

The second rule; \(R(3)=50\).

- What value does the first part approach at the switching time? - How can you choose the starting value of the second part so that it connects to the first?

1. At \(t=2\), the first rule approaches \(2^2=4\). 2. The second rule starts at \(c\left(\frac{1}{2}\right)^0=c\). 3. The two values agree when \(c=4\).

\(c=4\)

- Compare each input time with the breakpoint \(t=4\). - After choosing the correct row, substitute the time only into that rule. - At the breakpoint, what is the value of the exponent \(t-4\)?

1. Since \(2<4\), use the first rule for \(Q(2)\). Since \(6\ge4\), use the second rule for \(Q(6)\). 2. \(Q(2)=500(1.2)^2=720\). 3. \(Q(6)=500(1.2)^4(0.9)^2=839.808\). 4. At \(t=4\), the second rule gives \(Q(4)=500(1.2)^4(0.9)^0=1036.8\). The exponent \(t-4\) is \(0\), so the new factor is \(1\) exactly at the breakpoint.

a) Use the first rule for \(Q(2)\) and the second rule for \(Q(6)\). b) \(Q(2)=720\); \(Q(6)=839.808\) c) \(Q(4)=1036.8\); at \(t=4\), \((0.9)^{t-4}=1\).

- Compare the two branch values at the breakpoint. - A growth factor greater than \(1\) indicates increase. - Convert each growth factor to a percent growth rate before comparing them.

a) At \(t=3\), the first branch gives \(100(1.10)^3\). The second branch approaches \(100(1.10)^3(1.04)^0=100(1.10)^3\), so the model is continuous. b) Both growth factors, \(1.10\) and \(1.04\), are greater than \(1\), so the quantity increases on both intervals. c) The first interval grows by \(10\%\) per time unit, while the second grows by \(4\%\). Therefore, the first interval has the larger percent growth.

a) Yes. b) Increasing on both intervals. c) The first interval; it grows by \(10\%\) per time unit instead of \(4\%\).

- How many hours of decrease have passed when the filter turns on? - Which request rate starts the second phase: the rate before or after the sudden drop? - The second exponent should count time since which moment?

1. Before the filter, the rate is \(R(t)=80\left(\frac{1}{2}\right)^t\) for \(0\le t<2\). Just before the switch it approaches \(80\left(\frac{1}{2}\right)^2=20\) requests per minute. 2. The filter leaves \(75\%\), so the new starting rate is \(20\cdot0.75=15\). 3. A \(50\%\) hourly increase means a factor of \(1.5\). Thus the new rule is \(R(t)=15(1.5)^{t-2}\) for \(t\ge2\).

\(R(t)=80\left(\frac{1}{2}\right)^t\) for \(0\le t<2\), and \(R(t)=15(1.5)^{t-2}\) for \(t\ge2\).

- Use the initial amount and doubling time for the first rule. - Find the amount when the change occurs. - In the second rule, count time since the switch rather than since week zero.

1. In the first phase, \(N(t)=30\cdot2^t\) for \(0\le t<2\). 2. At week \(2\), the model reaches \(30\cdot2^2=120\) organisms. 3. After the switch, \(N(t)=120\left(\frac{1}{2}\right)^{t-2}\) for \(t\ge2\).

\(N(t)=30\cdot2^t\) for \(0\le t<2\), and \(N(t)=120\left(\frac{1}{2}\right)^{t-2}\) for \(t\ge2\).

- Find the multiplication factor before the filter is applied. - Is the starting amount of the second phase the value immediately before or after the filter takes effect? - Write the second exponent using time since day \(2\).

1. Before day \(2\), requests double daily from \(80\), giving \(V(t)=80\cdot2^t\) for \(0\le t<2\). 2. The second phase begins with \(160\) requests per minute at day \(2\) and halves daily. 3. Its rule is \(V(t)=160\left(\frac{1}{2}\right)^{t-2}\) for \(t\ge2\).

\(V(t)=80\cdot2^t\) for \(0\le t<2\), and \(V(t)=160\left(\frac{1}{2}\right)^{t-2}\) for \(t\ge2\).

- Look at the graph just after the sudden drop. - Compare the heights at two consecutive whole-day marks on the second curve. - Is the change described by adding the same amount or multiplying by the same factor?

1. The second phase begins at day \(2\) with \(8\) requests per second. 2. At day \(3\) it has \(4\) requests per second, and at day \(4\) it has \(2\). 3. The daily change factor is \(\frac{1}{2}\). A matching second rule is \(R(t)=8\left(\frac{1}{2}\right)^{t-2}\) for \(t\ge2\).

Second-phase starting value: \(8\) requests per second. Daily change factor: \(\frac{1}{2}\).

- Look for the time at which the graph changes from growth to decay. - Compare the marked values before the breakpoint to identify a doubling interval. - Compare the marked values after the breakpoint to identify a halving interval. - Continuity at the breakpoint requires the two branch values to agree there.

1. The graph shows \(P(0)=100\), a breakpoint at \(t=4\), and \(P(4)=200\). 2. From \(t=0\) to \(t=4\), the value doubles from \(100\) to \(200\), so the first branch is \(P(t)=100(2)^{t/4}\) for \(0\le t<4\). 3. After the breakpoint, the marked values fall from \(200\) at \(t=4\) to \(100\) at \(t=6\) and \(50\) at \(t=8\). The half-life is \(2\) time units, so the second branch is \(P(t)=200(0.5)^{(t-4)/2}\) for \(t\ge4\). 4. At \(t=4\), the first formula gives \(100(2)^1=200\), and the second gives \(200(0.5)^0=200\). The values agree, so the model is continuous at the breakpoint.

a) Initial value \(100\); breakpoint \(t=4\); breakpoint value \(200\) b) \(P(t)=100(2)^{t/4}\) for \(0\le t<4\), and \(P(t)=200(0.5)^{(t-4)/2}\) for \(t\ge4\) c) Both branches give \(200\) at \(t=4\), so the model is continuous there.

- Use the first weekly growth factor to find the value exactly at the change time. - For continuity, the second branch must begin from the value already reached at week \(5\). - In the second branch, count elapsed time from the breakpoint rather than from week \(0\).

1. For the first \(5\) weeks, \(S(t)=120(1.08)^t\). Thus, \(S(5)=120(1.08)^5\approx176.32\,\text{cm}^2\). 2. To keep the model continuous, use the week-5 value as the starting value of the second branch. 3. Therefore, \(S(t)=120(1.08)^t\) for \(0\le t\le5\), and \(S(t)=120(1.08)^5(1.03)^{t-5}\) for \(t>5\). 4. At week \(9\), \(S(9)=120(1.08)^5(1.03)^4\approx198.45\,\text{cm}^2\).

a) About \(176.32\,\text{cm}^2\) b) \(S(t)=120(1.08)^t\) for \(0\le t\le5\), and \(S(t)=120(1.08)^5(1.03)^{t-5}\) for \(t>5\) c) About \(198.45\,\text{cm}^2\)

- Which growth or decay rule applies before and after the filter activates? - Does the rate start below or above the target? Could it cross before the filter? - Work out the value at the filter change to start the second branch. - On which side of each crossing does the graph lie below the target?

1. Just before the filter activates, the rate is \(400(1.5)^3=1350\) requests per minute. It instantly falls by half to \(R(3)=675\). 2. The first branch is \(R(t)=400(1.5)^t\) for \(0\le t<3\). The second branch is \(R(t)=675(0.9)^{t-3}\) for \(t\ge3\). 3. Before the filter, solve \(400(1.5)^t=500\). Then \(t=\frac{\ln(1.25)}{\ln(1.5)}\approx0.55\) hours. 4. After the filter, solve \(675(0.9)^{t-3}=500\). Then \(t=3+\frac{\ln(500/675)}{\ln(0.9)}\approx5.85\) hours. 5. The first branch is increasing from \(400\), so \(R(t)<500\) for \(0\le t<\frac{\ln(1.25)}{\ln(1.5)}\). The second branch decreases below \(500\) after \(t=3+\frac{\ln(500/675)}{\ln(0.9)}\). Between these intervals, including the filter jump, the rate stays above \(500\).

a) Just before: \(1350\) requests per minute; immediately after: \(675\) requests per minute b) \(R(t)=400(1.5)^t\) for \(0\le t<3\), and \(R(t)=675(0.9)^{t-3}\) for \(t\ge3\) c) The rate equals \(500\) at about \(0.55\) hours and again at about \(5.85\) hours. It is below \(500\) from \(t=0\) until the first crossing, and after the second crossing (not including the crossing times).

- Compare heights one hour apart on each side of the switch. - Which point at the same time is filled, and which one is open? - For the second part, how can the exponent measure time since the switch?

1. Before hour \(2\), the plotted population goes from \(9\) to \(3\) in one hour, so it is divided by \(3\) each hour: \(N(t)=9\left(\frac{1}{3}\right)^t\) for \(0\le t<2\). 2. The open circle at \((2, 1)\) shows that the first rule approaches \(1\) but does not apply at hour \(2\). The filled dot at \((2, 3)\) shows \(N(2)=3\). 3. In the second phase, the plotted population goes from \(3\) to \(6\) in one hour. Thus \(N(t)=3\cdot2^{t-2}\) for \(2\le t\le4\). 4. The modeled number of players jumps upward from the approaching value \(1\) to \(3\) at hour \(2\).

\(N(t)=9\left(\frac{1}{3}\right)^t\) for \(0\le t<2\). \(N(t)=3\cdot2^{t-2}\) for \(2\le t\le4\). \(N(2)=3\) players; the model jumps up from \(1\) to \(3\) at the switch.

- Which term in the single-campaign model approaches zero as the campaign ages? - Which campaigns contribute before day fifty? - How many days has Campaign 2 been active at a global time \(t\) after its start? - Add both campaigns' contributions only after the second has begun.

1. As \(u\) increases, \(e^{-0.1u}\) approaches \(0\). Thus, \(S_k(u)\) approaches \(10k\) visits per day. 2. Before day \(50\), only Campaign 1 contributes: \(G(t)=1000+200e^{-0.1t}\) for \(0\le t<50\). 3. For \(t\ge50\), Campaign 2 has been running for \(t-50\) days. Its contribution is \(S_{150}(t-50)=1500+300e^{-0.1(t-50)}\). 4. Therefore, for \(t\ge50\), \(G(t)=2500+200e^{-0.1t}+300e^{-0.1(t-50)}\). 5. At \(t=60\), \(G(60)=2500+200e^{-6}+300e^{-1}\approx2610.86\), or \(2611\) visits after rounding. 6. In the long term both exponential terms approach \(0\), leaving \(2500\) daily visits.

a) \(10k\) daily visits b) \(G(t)=1000+200e^{-0.1t}\) for \(0\le t<50\), and \(G(t)=2500+200e^{-0.1t}+300e^{-0.1(t-50)}\) for \(t\ge50\) c) About \(2611\) daily visits d) \(2500\) daily visits

- Read the marked rate-change point against the horizontal axis before writing the second branch. - The second branch of \(B\) must start from the value already reached at the breakpoint. - The crossover after the breakpoint comes from setting that second branch equal to \(A\). - For the final comparison, distinguish the accumulated value at the breakpoint from the growth factor used afterward.

1. The graph marks the rate change at \(t=4\). 2. The single-rate model is \(A(t)=100(1.12)^t\). For \(B\), use \(B(t)=100(1.05)^t\) for \(0\le t\le4\), and \(B(t)=100(1.05)^4(1.25)^{t-4}\) for \(t>4\). 3. For the crossover after year \(4\), solve \(100(1.12)^t=100(1.05)^4(1.25)^{t-4}\). 4. Taking logarithms gives \(t\ln(1.12)=4\ln(1.05)+(t-4)\ln(1.25)\), so \(t=\frac{4(\ln(1.25)-\ln(1.05))}{\ln(1.25)-\ln(1.12)}\approx6.351\). 5. At year \(10\), \(A(10)=100(1.12)^{10}\approx310.58\), while \(B(10)=100(1.05)^4(1.25)^6\approx463.68\). 6. Although \(A\) is larger at the rate-change time, afterward \(B\) has the larger annual growth factor, so it eventually catches and then exceeds \(A\).

a) Year \(4\) b) \(A(t)=100(1.12)^t\); \(B(t)=100(1.05)^t\) for \(0\le t\le4\), and \(B(t)=100(1.05)^4(1.25)^{t-4}\) for \(t>4\). The crossover is \(t=\frac{4(\ln(1.25)-\ln(1.05))}{\ln(1.25)-\ln(1.12)}\approx6.351\,\text{years}\). c) \(B\) is larger at year \(10\): \(A(10)\approx310.58\) and \(B(10)\approx463.68\). After the rate change, \(B\) has the larger yearly growth factor.

- Identify when the factor \(0.8\) first begins to act. - At a later time, count only the time spent in the second regime. - Compare what each proposed second branch would do immediately after the breakpoint. - At day \(8\), compare how many decay factors the two models apply.

1. The factor \(0.8\) begins acting only after day \(6\). At time \(t\), the number of days spent in the decay regime is \(t-6\), not \(t\). Jordan's exponent incorrectly applies six extra days of decay. 2. The correct model is \(P(t)=50(1.1)^t\) for \(0\le t\le6\), and \(P(t)=50(1.1)^6(0.8)^{t-6}\) for \(t>6\). 3. At day \(8\), the correct model gives \(50(1.1)^6(0.8)^2\approx56.69\). 4. The proposed model gives \(50(1.1)^6(0.8)^8\approx14.86\). 5. The incorrect model has multiplied by six additional factors of \(0.8\), so its prediction is only \((0.8)^6\approx0.262\) times the correct prediction.

a) The decay exponent must count time since day \(6\), so it should be \(t-6\), not \(t\). b) \(P(t)=50(1.1)^t\) for \(0\le t\le6\), and \(P(t)=50(1.1)^6(0.8)^{t-6}\) for \(t>6\) c) Correct: about \(56.69\); proposed: about \(14.86\). The proposed model applies six extra daily decay factors.

- Treat each change time as a new elapsed-time origin for the factor that begins there. - At day \(4\), distinguish the value just before the intervention from the value immediately after it. - At day \(10\), use the value already reached by the second branch as the starting value of the third branch. - Continuity depends on whether the branch values agree at the breakpoint, not merely on whether both branches are exponential.

1. Before the first breakpoint, \(Q(t)=500(1.10)^t\) for \(0\le t<4\). 2. Just before day \(4\), the value is \(500(1.10)^4\). The intervention leaves \(80\%\) of that value, so the second branch begins at \(0.8\cdot500(1.10)^4\). 3. Therefore, \(Q(t)=0.8\cdot500(1.10)^4(0.95)^{t-4}\) for \(4\le t<10\). 4. At day \(10\), there is no jump. The second branch reaches \(0.8\cdot500(1.10)^4(0.95)^6\), so the third branch is \(Q(t)=0.8\cdot500(1.10)^4(0.95)^6(1.03)^{t-10}\) for \(t\ge10\). 5. At \(t=4\), the left-hand branch approaches \(500(1.10)^4\), while the actual branch value is \(80\%\) of that amount. The model is discontinuous there. 6. At \(t=10\), the second-branch value and third-branch starting value are identical, so the model is continuous there. 7. \(Q(12)=0.8\cdot500(1.10)^4(0.95)^6(1.03)^2\approx456.7\).

a) \(Q(t)=500(1.10)^t\) for \(0\le t<4\); \(Q(t)=0.8\cdot500(1.10)^4(0.95)^{t-4}\) for \(4\le t<10\); \(Q(t)=0.8\cdot500(1.10)^4(0.95)^6(1.03)^{t-10}\) for \(t\ge10\) b) Discontinuous at \(t=4\) because of the instantaneous \(20\%\) drop; continuous at \(t=10\) because the third branch starts from the second branch's day-10 value. c) \(Q(12)\approx456.7\)