Section II Part A
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An increasing, continuous function $$g(x)$$ has outputs shown in the table:A function $$f(x)$$ is shown graphically.(A) (i) Estimate the value of $$g(f^{-1}(1))$$ if it is defined.(ii) If $$h(x)=\\frac{g(x)}{f(x)}$$, identify any discontinuities for $$x$$>$$0$$. State their type and location.(B) Suppose $$ j(x)=\\frac{1}{4.998}\\ln(x)\\cdot e^{2\\sin(\\sqrt{x})}$$(i) Determine the end behavior of $$j(x)$$ as $$x\\to\\infty$$. Express your answer using limit notation.(ii) Determine the vertical asymptote of $$j(x)$$ as $$x\\to0^+$$ using limit notation.(C) Determine whether $$g(x)$$ is best modeled by a linear, quadratic, exponential, or logarithmic function. Justify your answer using the pattern in the table.⌄
(A)(i) From the graph, estimate the value where:$$f(x)=1.$$This occurs at approximately:$$x\approx2.$$So:$$f^{-1}(1)\approx2.$$Now evaluate:$$g(2)=1.$$Answer: $$1$$(A)(ii) Discontinuities of:$$h(x)=\frac{g(x)}{f(x)}$$ occur where:$$f(x)=0.$$From the graph, $$f(x)=0$$ near:$$x\approx1.$$Since the denominator is zero and the numerator is finite, this is a vertical asymptote.Answer: vertical asymptote at $$x\approx1$$(B)(i) As $$x\to\infty$$:$$\ln(x)\to\infty.$$$$\sin(\sqrt{x})$$ oscillates between $$-1$$ and $$1,$$so:$$e^{2\sin(\sqrt{x})}$$ remains bounded.Thus:$$j(x)\to\infty.$$Answer: $$\lim_{x\to\infty}j(x)=\infty$$(B)(ii) As $$x\to0^+$$:$$\ln(x)\to-\infty.$$The exponential term stays finite.So:$$j(x)\to-\infty.$$Answer:$$\lim_{x\to0^+}j(x)=-\infty$$Vertical asymptote at $$x=0$$(C) Check differences in $$g(x).$$The values increase slowly and the rate of increase decreases.This behavior matches a logarithmic function.Answer: logarithmic model
A tank contains a certain number of liters of water. A small leak causes water to drain continuously, and the rate at which the water leaves is proportional to the amount remaining. Measurements show that after $$1$$ minute, $$200$$ liters remain, and after $$3$$ minutes, $$128$$ liters remain. The situation can be modeled by an exponential function of the form $$W(t)=Ae^{kt}$$ where $$W(t)$$ is the amount of water in liters at time $$t$$ minutes.(A) Find an exponential function that models this situation.(B)(i) Find the average rate of change of the amount of water from $$t=0$$ to $$t=4$$.(ii) Interpret the meaning of your answer in context.(C) Will the average rate of change from $$t=4$$ to $$t=t_a,$$ where $$t_a$$>$$4,$$ be greater than or less than the average rate of change from $$t=0$$ to $$t=4$$? Justify your answer.(D) Estimate the time when the tank will have $$50$$ liters remaining.⌄
(A) Start with:$$W(t)=Ae^{kt}.$$Use the given data.At $$t=1$$:$$200=Ae^k.$$At $$t=3$$:$$128=Ae^{3k}.$$Divide the second equation by the first:$$\frac{128}{200}=e^{2k}.$$$$\frac{16}{25}=e^{2k}.$$Take the natural logarithm:$$2k=\ln\left(\frac{16}{25}\right).$$$$k=\frac{1}{2}\ln\left(\frac{16}{25}\right).$$Now solve for $$A$$ using $$200=Ae^k$$:$$A=\frac{200}{e^k}.$$Since:$$e^k=\left(\frac{16}{25}\right)^{1/2}=\frac{4}{5},$$we get:$$A=\frac{200}{4/5}=250.$$So the model is:$$W(t)=250e^{\left(\frac{1}{2}\ln(16/25)\right)t}.$$This can also be written as:$$W(t)=250\left(\frac{4}{5}\right)^t.$$(B)(i) The average rate of change from $$t=0$$ to $$t=4$$ is:$$\frac{W(4)-W(0)}{4-0}.$$Compute the values:$$W(0)=250\left(\frac{4}{5}\right)^0=250.$$$$W(4)=250\left(\frac{4}{5}\right)^4=250\cdot\frac{256}{625}=102.4.$$So:$$\frac{W(4)-W(0)}{4}=\frac{102.4-250}{4}.$$$$=\frac{-147.6}{4}=-36.9.$$Final answer for (B)(i): $$-36.9$$ liters per minute.(B)(ii) Interpretation:From $$t=0$$ to $$t=4,$$ the amount of water in the tank decreases by an average of $$36.9$$ liters per minute.(C) Since: $$W(t)=250\left(\frac{4}{5}\right)^t,$$ the function is exponential decay.As time increases, the graph decreases but becomes less steep.That means the rate of decrease gets smaller in magnitude over time.So, the average rate of change from $$t=4$$ to $$t=t_a,$$ where $$t_a$$>$$4,$$ will be greater than the average rate of change from $$t=0$$ to $$t=4.$$It will still be negative, but it will be closer to $$0.$$(D) Set the amount equal to $$50$$ liters:$$250\left(\frac{4}{5}\right)^t=50.$$Divide by $$250$$:$$\left(\frac{4}{5}\right)^t=\frac{1}{5}.$$Take the natural logarithm:$$t\ln\left(\frac{4}{5}\right)=\ln\left(\frac{1}{5}\right).$$So:$$t=\frac{\ln(1/5)}{\ln(4/5)}.$$Approximate:$$t\approx\frac{-1.6094}{-0.2231}\approx7.21.$$Final answer for (D): $$7.21$$ minutes.