Question 1025358
{{{c(t) =(5t)/(t^2+1) }}}

a.  {{{graph( 300, 200, 0, 8, -8, 8,(5x)/(x^2+1))}}}

b.  c'(t) = {{{(5t - 5t^2)/(t^2+1)^2}}}.  Setting this derivative to zero to get the extreme values, we get t = 0, 1.  Incidentally an absolute maximum exists at t = 1.  ( c(1) = 2.5 mg/L.)

c.  As {{{t->infinity}}}, {{{c(t)->0}}}, and so the drug concentration disappears over a long period of time.

d. You have to find the solution to the equation {{{(5t)/(t^2+1) = 0.3}}}, or equivalently, {{{3t^2 - 50t+3 = 0}}}.  (Use the quadratic formula.)