Question

Please help me solve this question. I’m not sure how

Answer 1

Step 1

Given

f(0) = 1 is the normal level of oxygen.

Step 2

Write the function

`f(t)\text{ = }(t^2-t+1)/(t^2+1)`

a) Graph the function

b)

`\begin{gathered} \text{ In the long run, t = }\infty \\ f(t)\text{ = }(t^2-t+1)/(t^2+1) \\ \lim_(t\to\infty)f(t)=\text{ }((t^2)/(t^2)-(t)/(t^2)+(1)/(t^2))/((t^2)/(t^2)+(1)/(t^2)) \\ \lim_(t\to\infty)f(t)\text{ = }\frac{1\text{ - }(1)/(t)+(1)/(t^2)}{1\text{ + }(1)/(t^2)} \\ =\text{ }\frac{1\text{ - }\frac{1}{\infty\text{ }}+\text{ }(1)/(\infty)}{1\text{ + }(1)/(\infty)} \\ =\text{ }(1)/(1) \\ =\text{ 1} \end{gathered}`

The oxygen level will eventually returns to its normal in the long run.

c)

70% of it original level = 0.7

`\begin{gathered} 0.7\text{ = }\frac{t^2-\text{ t + 1}}{t^2+1} \\ t^2\text{ - t + 1 = 0.7t}^2\text{ + 0.7} \\ t^2-0.7t^2\text{ - t + 1 - 0.7 = 0} \\ 0.3t^2\text{ - t + 0.3 = 0} \end{gathered}`

Next, solve for t.

`\begin{gathered} 0.3t^2-t+0.3=0 \\ \mathrm{Multiply\:both\:sides\:by\:}10 \\ 0.3t^2\cdot \:10-t\cdot \:10+0.3\cdot \:10=0\cdot \:10 \\ \mathrm{Refine} \\ 3t^2-10t+3=0 \\ 3t^2-9t-t+3\text{ = 0} \\ 3t(t\text{ - 3\rparen -1\lparen t - 3\rparen = 0} \\ (3t\text{ - 1\rparen\lparen t - 3\rparen = 0} \\ \text{t = 3, t = }(1)/(3) \end{gathered}`

3 weeks