The quantity of a drug Q mg present in the body t hours after an injection of the drug is given is : Q = f (t) = 70 te-0.6tFind f (1), f ' (1), f (6), and f ' (6). give units and interpret the answer. Round your answer to two decimal placesf(1) = f '(1)= f(6)= f '(6)= Interpretation of f (1) and f ' (1):One hour after the drug was administered, the quantity of drug in the body is approximately Interpretation of f (6) and f ' (6) :Six hours after the drug was administered, the quantity of drug in the body is approximately

A) f(1) =38.43mg

B) f'(1) = -15.37 mg/hour

C) f(6) = 11.47mg

D) f'(6) = -4.97mg/hour

F) 23.06mg

H) 0

Step-by-step explanation:

Given the function:

Q = 70te^-0.6t(1)

Where Q is measured in mg and t is in hours

(A) f(1) is the quantity Q in mg after 1 hour

f(1) = 70(1)e^-0.6(1) = 70e^-0.6

= 70×0.5488

= 38.43 mg

(B) f'(1) = dQ/dt, at t=1

= 70t × -0.6e^-0.6t +70e^-0.6t

= -42te^-0.6t + 70e^-0.6t

= -42(1) × 0.5488 + 70 × 0.5488

=28 × 0.5488

= -15.37mg/hour

(C) f(6) = 70te^-0.6t

= 70(6)e^-0.6(6)

= 420e^-3.6

= 420 × 0.0273

= 11.47mg

(D) f'(6) = dQ/dt, at t= 6

f' = -42te^-0.6t + 70e^-0.6t

f'(6) = -42(6)e^-3.6 + 70e^-3.6

= -252 × 0.0273 + 70 × 0.0273

= -6.8796 + 1.911

= -4.9686

= -4.97mg/hour

(E) f(1) is the quantity of the drug present in the body 1 hour after injection.

f'(1) is the rate of decrease of the drug 1 hour after injection or it is the rate at which the drug is decreasing 1 hour after injection.

(F) The quantity Q is given by

Q= 38.43 - 15.37

= 23.06mg

(G) f(6) is the quantity of the drug present in the body 6 hours after injection

f'(6) is the rate at which the drug is removed or decreased from the body 6 hours after injection.

(H) The quantity is given by

Q = 11.47 - 4.97×6

Q= 11.47-29.82

Q= -18.35mg

This is not possible, hence after 6 hours of injection the quantity of drug remaining is zero.

their is no graph showing

step-by-step explanation:

hope this !

let me know