Question 1169364
c(t)=5t/(t^2+1)
at t=0, c(t)=0
at t=1, c(t)=2.5
at t=10, c(t)=50/101
as t goes to infinity, c(t) goes to 0, consistent with the drug's gradually disappearing.
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highest concentration: this is 5t(t^2+1)^(-1)
maximum is found by setting the derivative=0.
-10t^2/(t^2+1)^2+5/(t^2+1)=0
move the term on the left to the right and divide by 5
(2t^2)/(t^2+1)^2=1/(t^2+1)
t=1, c(t)=2.5 as above.
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5t/(t^2+1)=0.3
5t=0.3t^2+0.3
0=3t^2-50t+3
t=(1/6)(50+/-sqrt(2464)); sqrt (2464)=49.64
 roots are 0.602 hours and  16.606 hours and 0.062 hours.  The first is on the ascending phase right after the injection, and it is 16.61 hours before the concentration falls below 0.3 mg/L., or 16.607 hours to three decimal places.
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{{{graph(300,300,-0.3,10,-3,4,(5x)/(x^2+1))}}}