Problem 1
Answer: Choice C) 1/2
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Explanation:
Let's plug 8 into f(x)
[tex]f(x) = \frac{x+3}{x^2-4x-21}\\\\f(8) = \frac{8+3}{8^2-4(8)-21}\\\\f(8) = \frac{8+3}{64-32-21}\\\\f(8) = \frac{11}{11}\\\\f(8) = 1\\\\[/tex]
Repeat for g(x)
[tex]g(x) = \log_{4}(x)\\\\g(8) = \log_{4}(8)\\\\g(8) = \frac{\log(8)}{\log(4)}\\\\g(8) = \frac{\log(2^3)}{\log(2^2)}\\\\g(8) = \frac{3*\log(2)}{2*\log(2)}\\\\g(8) = \frac{3}{2}\\\\[/tex]
I used the change of base formula in the third step.
Lastly, we subtract the results like so
[tex](g-f)(x) = g(x) - f(x)\\\\(g-f)(8) = g(8) - f(8)\\\\(g-f)(8) = \frac{3}{2} - 1\\\\(g-f)(8) = \frac{3}{2} - \frac{2}{2}\\\\(g-f)(8) = \frac{3-2}{2}\\\\(g-f)(8) = \frac{1}{2}\\\\[/tex]
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Problem 2
Answer: Choice A) 8
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Explanation:
Through trial and error (I recommend the rational root theorem), you should find that the roots of R(x) are: -1, 3, and 8
We'll only focus on positive x values. So we focus on the roots 3 and 8.
It turns out that the R(x) curve dips below the x axis when x is between 3 and 8. When x > 8 is when R(x) is positive and goes upward forever. This represents when the company is making money.
The company does make money when x is between 0 and 3; however, this is a short timespan compared to the other interval. So it's much more effective and realistic to say that the company makes money after 8 days.
In other words, yes the company makes money between day 0 and day 3, but from day 3 to day 8 is when the company loses money. Ultimately the company is profitable again when it's beyond day 8.
