The percent of concentration of a drug in the bloodstream t hours after the drug is administered is given by K(t) =5t/(t^(2)+1) On what time intervals is the concentration of the drug increasing?b. On what intervals is it decreasing?c. Does the percent concentration have a maximum value?d. Find and interpret the limt→+[infinity]K(t).e. What are the maximum and minimum percentages of the concentration? Are theserelative extrema or absolute extrema?f. Give a rough sketch of the functionK(t).

Answer:   a) (0, 1)   b) (1, ∞)   c) yes   d) The concentration tends toward zero as time increases.   e) The percent concentration has a maximum value of 2.5 at t=1. It has a minimum of 0 at t=0, and approaching 0 as t → ∞. These are absolute extrema.Step-by-step explanation:The derivative is ...   K'(t) = 5/(t^2 +1) -10t/(t^2 +1)^2This is zero when ...   5(t^2 +1)-10t^2 = 0 . . . . . numerator of the derivative function   -5(t^2 -1) = 0 . . . . . . . . . . simplify   t = 1 . . . . . . . . . . . . . . . . . positive solutionThe denominator of the derivative function is positive everywhere (and is monotonically increasing), so the sign of the numerator determines the sign of the derivative. a) The first derivative is positive for x < 1, so the drug concentration is increasing for the first hour.__b) For x > 1, the sign of the first derivative is negative, so the percent concentration is decreasing after the first hour.__c) The concentration has a maximum value at the point where it stops increasing and starts decreasing.__d) The ratio of the highest-degree terms of the rational function is ...   5t/t^2 = 5/tThis has a limit of zero as t goes to infinity. The blood concentration decays over time.__e) K(1) = 5/2 = 2.5 . . . percent, the maximum concentration   K(0) = 0 . . . . the minimum concentrationThese are absolute extrema. (A graph of the function helps you see this.)