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B.) Would this concern a medical professional? Explain. C.) Find the equation of

ID: 3018433 • Letter: B

Question

B.) Would this concern a medical professional? Explain.

C.) Find the equation of the horizontal asymptote. What does this mean in this context?

D.) Find all of the intercepts of C and interpret their meaning in the context of this problem.

E.) How many hours after injection does a maximum concentration of the drug occur in the bloodstream? Round answer to nearest hundredth.

F.) Suppose you need to re-administer this injection at the point when the concentration is less than 0.5%. If the first injection was given at 8:00 am, what time should the next injection be given?

h3+8

Explanation / Answer

given C(h)=(2h2+5h)/(h3+8)

=>C(h)=h(2h+5)/(h3+23)

=>C(h)=h(2h+5)/((h+2)(h2-2h+4))

a)

for vertical asymptote denominator =0

=>(h+2)(h2-2h+4)=0

=>h+2=0

=>h=-2

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B)

this Wouldnot concern a medical professional. since h is negative. actually time is always positive.

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C)

horizontal symptote ,y= lim[h->] C(h)

=>y=lim[h->](2h2+5h)/(h3+8)

=>y=lim[h->]h2(2+(5/h))/(h3(1+(8/h3)))

=>y=lim[h->](2+(5/h))/(h(1+(8/h3)))

=>y=0

horizontal asymptote is y=0

after a long time concentration of medicine in bloodstream nears zero percent

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D)

for horizontal intercept , C(h)=0

=>(2h2+5h)/(h3+8)=0

=>(2h2+5h)=0

=>h(2h+5)=0

=>h=0, h=-2.5

h=0 => concentration of medicine in bloodstream is zero when h=0. h=-2.5 has no significance

for vertical intercept , C(0)=(2*02+5*0)/(03+8)

=>C(0)=0

initial concentration of medicine in bloodstream is zero

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E)

C(h)=(2h2+5h)/(h3+8)

C'(h)=-2(h4+5h3-16h-20)/(h3+8)2

for local extremum , C'(h)=0

=>-2(h4+5h3-16h-20)/(h3+8)2=0

=>h=-4.41, h=1.95

h=-4.41 has no significance

maximum concentration of the drug occur in the bloodstream after 1.95 hours

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F)

C(h)=(2h2+5h)/(h3+8)=0.5

=>(2h2+5h)=0.5(h3+8)

=>(4h2+10h)=(h3+8)

=>h3-4h2-10h+8=0

=>h=0.65607,5.5436

injection given at h=0,i.e, 8:00am

cocentration will be less than 0.5% after 5.5436 hours

=>injection should be given at 8+5+(60*.5436)

=>injection should be given at 13:33

=>injection should be given at 1:33 pm