SOLUTION: A patient is being treated for a chronic illness. The concentration C (x) (in g/mL) of a certain medication in her bloodstream x weeks from now is approximated by the following equ

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Question 1190276: A patient is being treated for a chronic illness. The concentration C (x) (in g/mL) of a certain medication in her bloodstream x weeks from now is approximated by the following equation.
C(x)=(12x² - 33x+44) / (4x² - 12x + 12)
Complete the following.
(a) Find the value of x that maximizes the concentration. Then give the maximum concentration. Round your answers to the nearest hundredth.
Value of x that maximizes concentration:
Maximum concentration:
(b) Complete the following sentence.
For very large x, the concentration appears to
(Choose one)
Increase without bound.
Decrease without bound.
Approach a finite value.
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Part (a)

x = number of weeks
C(x) = concentration of the medication in the bloodstream

C(x) = f(x)/g(x)
f(x) = 12x^2-33x+44
g(x) = 4x^2-12x+12

Apply derivatives
f ' (x) = 24x-33
g ' (x) = 8x-12
So,
C(x) = f(x)/g(x)
C ' (x) = [ f ' (x)*g(x) - f(x)*g ' (x) ]/[ g(x) ]^2 ... quotient rule
C ' (x) = [ (24x-33)*(4x^2-12x+12) - (12x^2-33x+44)*(8x-12) ]/[ 4x^2-12x+12 ]^2
C ' (x) = [ 96x^3-420x^2+684x-396 -96x^3+408x^2-748x+528 ]/[ 4x^2-12x+12 ]^2
C ' (x) = [ -12x^2-64x+132 ]/[ 4x^2-12x+12 ]^2

Set C'(x) equal to zero and solve for x
C'(x) = 0
[ -12x^2-64x+132 ]/[ 4x^2-12x+12 ]^2 = 0
-12x^2-64x+132 = 0

Next you'll need to use the quadratic formula
I'll skip those steps, but you should get these roots:
x = -6.9224, x = 1.5890

x represents the number of weeks, so the proper domain is x > 0. Meaning we'll ignore the negative root.

The only practical root here is roughly x = 1.5890

Now to conduct the first derivative test.
Pick something to the left of this to plug into C'(x)
If you plugged in x = 1, you should find that C'(1) = 3.5
The value itself doesn't matter. All we care about is the sign of C'(x). In this case, it's positive.

Now try something to the right of the root found. Let's do x = 2.
You should find that C'(2) = -2.75 so C'(x) is negative when x > 1.5890

The transition from positive to negative means that there is a local max at around x = 1.5890
This rounds to 1.59

The last thing to do is plug the root found into the original C(x) function to find the corresponding concentration
Doing so should get you roughly C(x) = 7.2112 and this rounds to 7.21

Answers:
Value of x that maximizes concentration: 1.59
Maximum Concentration: 7.21
============================================================
Part (b)

For very large x, the concentration appears to Approach a finite value

If you were to use a graph, then you may be able to see that the curve slowly would approach the horizontal asymptote y = 3. I recommend GeoGebra or Desmos as two options you can use to make the graph. They are both free.

A non-graph approach would be to compare the leading terms from the numerator and denominator (12x^2 and 4x^2 respectively). Dividing said terms leads to the ratio 3
(12x^2)/(4x^2) = 3. This trick of dividing the leading terms only works when the degree of each is the same.

This works because as x gets really big, the only terms that will determine the size of numerator and denominator will be those leading terms. If x is big, then x^2 is much bigger. I recommend looking at a table of values to see what I mean.
So at large x values, the other terms won't really affect things too much. They play a role, just a very small part.
This explains why we don't arrive at y = 3 itself when x is really really large.

To summarize what's going on with the C(x) function, we have:
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