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Certainly! Let's analyze the function [tex]\( R(x) = \frac{2100 x^2}{x^2 + 4} \)[/tex] to solve for the required parts:
### Part (a): Compute [tex]\( R(1) \)[/tex]
We need to find the revenue at 1 year after publication. Substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[ R(1) = \frac{2100 \cdot 1^2}{1^2 + 4} \][/tex]
Calculating the numerator and the denominator:
[tex]\[ R(1) = \frac{2100 \cdot 1}{1 + 4} = \frac{2100}{5} = 420 \][/tex]
Therefore, the revenue 1 year after publication is [tex]\( \boxed{420} \)[/tex] million dollars.
### Part (b): Compute [tex]\( R(10) \)[/tex]
Next, we need to find the revenue 10 years after publication. Substitute [tex]\( x = 10 \)[/tex] into the function:
[tex]\[ R(10) = \frac{2100 \cdot 10^2}{10^2 + 4} \][/tex]
Calculating the numerator and the denominator:
[tex]\[ 10^2 = 100 \][/tex]
[tex]\[ 100 + 4 = 104 \][/tex]
So,
[tex]\[ R(10) = \frac{2100 \cdot 100}{104} = \frac{210000}{104} = \frac{52500}{26} = \frac{26250}{13} \][/tex]
Thus, the revenue 10 years after publication is [tex]\( \boxed{\frac{26250}{13}} \)[/tex] million dollars.
### Part (c): Compute [tex]\( R(100) \)[/tex]
Now, we will find the revenue 100 years after publication. Substitute [tex]\( x = 100 \)[/tex] into the function:
[tex]\[ R(100) = \frac{2100 \cdot 100^2}{100^2 + 4} \][/tex]
Calculating the numerator and the denominator:
[tex]\[ 100^2 = 10000 \][/tex]
[tex]\[ 10000 + 4 = 10004 \][/tex]
So,
[tex]\[ R(100) = \frac{2100 \cdot 10000}{10004} = \frac{21000000}{10004} = \frac{5250000}{2501} \][/tex]
Therefore, the revenue 100 years after publication is [tex]\( \boxed{\frac{5250000}{2501}} \)[/tex] million dollars.
### Part (d): Find the limit of [tex]\( R(x) \)[/tex] as [tex]\( x \)[/tex] approaches infinity:
Lastly, we need to determine the limit of [tex]\( R(x) \)[/tex] as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} R(x) = \lim_{x \to \infty} \frac{2100 x^2}{x^2 + 4} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the [tex]\( +4 \)[/tex] in the denominator becomes negligible compared to [tex]\( x^2 \)[/tex]. Hence, the function simplifies to:
[tex]\[ \lim_{x \to \infty} \frac{2100 x^2}{x^2} = \lim_{x \to \infty} 2100 \cdot \frac{x^2}{x^2} = 2100 \cdot 1 = 2100 \][/tex]
Thus, the limit as [tex]\( x \)[/tex] approaches infinity is [tex]\( \boxed{2100} \)[/tex] million dollars.
In conclusion, after performing these calculations, we have:
[tex]\[ \begin{align*} &\text{(a) } R(1) = 420 \\ &\text{(b) } R(10) = \frac{26250}{13} \\ &\text{(c) } R(100) = \frac{5250000}{2501} \\ &\text{(d) } \lim_{x \to \infty} R(x) = 2100 \\ \end{align*} \][/tex]
These solutions describe the total revenue in millions of dollars at different times since publication.
### Part (a): Compute [tex]\( R(1) \)[/tex]
We need to find the revenue at 1 year after publication. Substitute [tex]\( x = 1 \)[/tex] into the function:
[tex]\[ R(1) = \frac{2100 \cdot 1^2}{1^2 + 4} \][/tex]
Calculating the numerator and the denominator:
[tex]\[ R(1) = \frac{2100 \cdot 1}{1 + 4} = \frac{2100}{5} = 420 \][/tex]
Therefore, the revenue 1 year after publication is [tex]\( \boxed{420} \)[/tex] million dollars.
### Part (b): Compute [tex]\( R(10) \)[/tex]
Next, we need to find the revenue 10 years after publication. Substitute [tex]\( x = 10 \)[/tex] into the function:
[tex]\[ R(10) = \frac{2100 \cdot 10^2}{10^2 + 4} \][/tex]
Calculating the numerator and the denominator:
[tex]\[ 10^2 = 100 \][/tex]
[tex]\[ 100 + 4 = 104 \][/tex]
So,
[tex]\[ R(10) = \frac{2100 \cdot 100}{104} = \frac{210000}{104} = \frac{52500}{26} = \frac{26250}{13} \][/tex]
Thus, the revenue 10 years after publication is [tex]\( \boxed{\frac{26250}{13}} \)[/tex] million dollars.
### Part (c): Compute [tex]\( R(100) \)[/tex]
Now, we will find the revenue 100 years after publication. Substitute [tex]\( x = 100 \)[/tex] into the function:
[tex]\[ R(100) = \frac{2100 \cdot 100^2}{100^2 + 4} \][/tex]
Calculating the numerator and the denominator:
[tex]\[ 100^2 = 10000 \][/tex]
[tex]\[ 10000 + 4 = 10004 \][/tex]
So,
[tex]\[ R(100) = \frac{2100 \cdot 10000}{10004} = \frac{21000000}{10004} = \frac{5250000}{2501} \][/tex]
Therefore, the revenue 100 years after publication is [tex]\( \boxed{\frac{5250000}{2501}} \)[/tex] million dollars.
### Part (d): Find the limit of [tex]\( R(x) \)[/tex] as [tex]\( x \)[/tex] approaches infinity:
Lastly, we need to determine the limit of [tex]\( R(x) \)[/tex] as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} R(x) = \lim_{x \to \infty} \frac{2100 x^2}{x^2 + 4} \][/tex]
As [tex]\( x \)[/tex] becomes very large, the [tex]\( +4 \)[/tex] in the denominator becomes negligible compared to [tex]\( x^2 \)[/tex]. Hence, the function simplifies to:
[tex]\[ \lim_{x \to \infty} \frac{2100 x^2}{x^2} = \lim_{x \to \infty} 2100 \cdot \frac{x^2}{x^2} = 2100 \cdot 1 = 2100 \][/tex]
Thus, the limit as [tex]\( x \)[/tex] approaches infinity is [tex]\( \boxed{2100} \)[/tex] million dollars.
In conclusion, after performing these calculations, we have:
[tex]\[ \begin{align*} &\text{(a) } R(1) = 420 \\ &\text{(b) } R(10) = \frac{26250}{13} \\ &\text{(c) } R(100) = \frac{5250000}{2501} \\ &\text{(d) } \lim_{x \to \infty} R(x) = 2100 \\ \end{align*} \][/tex]
These solutions describe the total revenue in millions of dollars at different times since publication.
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