Question

C(h)= 2h^2+5h/h^3+8

1.Find domain
2. Find vertical and horizontal asymptotes
3. All intercepts
4. How many hours after injection does a maximum concentration of the drug occur in the bloodstream? Round to nearest hundredth
5. 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 8am, what time should the next injection be given?

PLEASE SHOW ALL WORK/STEPS

Answer 1

Explanation:

Given is a function

`c(h) = (2h^2+5h)/(h^3+8)`

1) When denominator =0 i.e. when

`h^3=-8 \\h=-2`

makes the function undefined

So domain is R-{-2}

2) Vertical asymptote is h=-2

Since degree of numerator is smaller than degree of denominator we get

C=0 is a horizontal asymptote

3) X intercepts are when

`2h^2+5h =0\\h=0, -5/2` are x intercepts

4) Derivative test will give the result

I derivative =
`((h^3+8)(4h+5)-(2h^2+5h)3h^2)/((h^3+8)^2)`

Equate I derivative to 0.

`4h^4+40+32h+5h^3-(6h^4+15h^3)=0\\\\-2h^4-10h^3+32h+40=0\\Solving we find that maximum at h = 1.945\\i.e. after 1.95 hours\\`

5) WE find that when x =0.656, y =0.5

Hence after 0.656 hours or 39.36 minutes or at8.40 a.m. apprxy the next injection.

Also after 5.544 hours i.e. 1.30 p.m.

Answer 2

1. 0 ≤ h
2. The only asymptote in the domain is the horizontal asymptote c = 0.
3. The only intercept in the domain is (x, y) = (0, 0).
4. Maximum concentration occurs at h ≈ 1.95.
5. About 1:33 pm

Explanation:

1. For functions involving time, it usually makes no sense to evaluate them for negative values of time. At some point in the future, the value of the function is sufficiently small as to be "don't care", so that value of time is a good upper limit on the domain. Here, that might be about 100 hours, where the concentration is about 0.02%.

A reasonable domain might be 0 ≤ h ≤ 100.

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2. In the domain, the only asymptote is horizontal. As h gets large, the value of the function becomes 2/h, so approaches zero.

The horizontal asymptote is C=0.

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3. The only intercept is the origin, where C(0) = 0.

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4. A graph shows the function maximum occurs at about h = 1.95.

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5. A graph shows the value of C(5.544) ≈ 0.5. If we assume that C(h) represents the concentration in %, then the next injection should be given about 5.544 hours after 8 am, or at 1:33 pm. (My calculator converts hours to hours:minutes, so there is no "work" to show.)