A patient receives a 150 mg injection of a drug every 4 hours. The graph shows the amount f(t) of the drug in the bloodstream after t hours.The t f(t) coordinate plane is given.The curve begins at the open point (0, 150), goes down and right to the approximate closed point (4, 75). The curve starts again at the approximate open point (4, 225), goes down and right to the approximate closed point (8, 125). The curve starts again at the approximate open point (8, 275), goes down and right to the approximate closed point (12, 150). The curve starts again at the approximate open point (12, 300), goes down and right to the approximate closed point (16, 175). The curve starts again at the approximate open point (16, 325), goes down and right, and ends at the approximate closed point (20, 225).Find lim t → 4− f(t)   and   lim t → 4+ f(t).lim t → 4− f(t)= mglim t → 4+ f(t)= mg

The graph of d(t) = Ae-k tsin(kt), where A and k are positive real constants, can be used to describe drug absorption in a patient's bloodstream, using units mg/litre per unit of time in minutes.Consider the special case where A = 1 and k = 1, and discuss this with respect to a dose of a drug taken at t = 0.

When a certain medical drug is administered to a patient, the number of milligrams remaining in the patient's bloodstream after t hours is modeled byD(t) = 40e−0.4t.

The graph of a function g is shown.The x y-coordinate plane is given. A function labeled y = g(x) with 4 parts is graphed.The first part is a curve, enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 2.5, and ends at the open point (2, 3).The second part is a curve begins again at the open point (2, 1), goes up and right becoming less steep, and ends at the open point (5, 2).The third part is the closed approximate point (5, 1.2).The fourth part is a curve, begins at the open point (5, 2) goes down and right becoming more steep, and exits the window in the first quadrant.Use it to state the values (if they exist) of the following:(a)  lim x → 2− g(x)(b)  lim x → 2+ g(x)(c)  lim x → 2 g(x)(d)  lim x → 5− g(x)(e)  lim x → 5+ g(x)(f)  lim x → 5 g(x)SolutionLooking at the graph we see that the values of g(x) approach as x approaches 2 from the left, but they approach as x approaches 2 from the right.Therefore (a) lim x → 2− g(x) =     and    (b) lim x → 2+ g(x) = .Since the left and right limits are different, we conclude that (c) the limit as x approaches 2 of g(x) does not exist.The graph also shows that (d) lim x → 5− g(x) =     and    (e) lim x → 5+ g(x) = .This time, the left and right limits are the same and so, by this theorem, we have (f) lim x → 5 g(x) = Despite this fact, notice that g(5) ≠ 2.

what is peak drug x level attained at steady state if 100mg is given intravenous injection every 6hrs. the patient ke=0.35/hr and Vd = 20L

What is the peak drug Z level attained at steady state if 150 mg is given by IV injection every 8 hr, the half-life of drug Z in patient being 6 hr and volume of distribution = 0.3 L/kg? (NB pt wt =75 Kg).