Step 1
Given
f(0) = 1 is the normal level of oxygen.
Step 2
Write the function
[tex]f(t)\text{ = }\frac{t^2-t+1}{t^2+1}[/tex]
a) Graph the function
b)
[tex]\begin{gathered} \text{ In the long run, t = }\infty \\ f(t)\text{ = }\frac{t^2-t+1}{t^2+1} \\ \lim_{t\to\infty}f(t)=\text{ }\frac{\frac{t^2}{t^2}-\frac{t}{t^2}+\frac{1}{t^2}}{\frac{t^2}{t^2}+\frac{1}{t^2}} \\ \lim_{t\to\infty}f(t)\text{ = }\frac{1\text{ - }\frac{1}{t}+\frac{1}{t^2}}{1\text{ + }\frac{1}{t^2}} \\ =\text{ }\frac{1\text{ - }\frac{1}{\infty\text{ }}+\text{ }\frac{1}{\infty}}{1\text{ + }\frac{1}{\infty}} \\ =\text{ }\frac{1}{1} \\ =\text{ 1} \end{gathered}[/tex]
The oxygen level will eventually returns to its normal in the long run.
c)
70% of it original level = 0.7
[tex]\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}[/tex]
Next, solve for t.
[tex]\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 = }\frac{1}{3} \end{gathered}[/tex]
3 weeks
