A real estate firm estimates that the value of a certain property will be given by the function V (t) = 2300+ 125 - 517 where V(t) is the value of the property (in thousands of dollars), and t is the time (in years). Determine how fast the value of the property is changing when t = 9 years.
A real estate firm estimates that the value of a certain property will be given by the function V (t) = 2300+ 125 - 517 where V(t) is the value of the property (in thousands of dollars), and t is the time (in years). Determine how fast the value of the property is changing when t = 9 years.
Chapter2: Functions And Their Graphs
Section2.4: A Library Of Parent Functions
Problem 47E: During a nine-hour snowstorm, it snows at a rate of 1 inch per hour for the first 2 hours, at a rate...
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![## Estimating Property Value Growth
A real estate firm estimates that the value of a certain property will be given by the function:
\[ V(t) = 2300 + \frac{125}{t} - \frac{517}{t^2} \]
where \( V(t) \) is the value of the property (in thousands of dollars), and \( t \) is the time (in years).
### Problem Statement
Determine how fast the value of the property is changing when \( t = 9 \) years.
### Instructions
1. Find the derivative of \( V(t) \) to determine the rate of change of the property value.
2. Evaluate the derivative at \( t = 9 \) years.
3. Indicate whether the value is increasing or decreasing and include the correct units.
### Solution
The derivative of \( V(t) = 2300 + \frac{125}{t} - \frac{517}{t^2} \) with respect to \( t \) is:
\[ V'(t) = -\frac{125}{t^2} + \frac{1034}{t^3} \]
Evaluating the derivative at \( t = 9 \) years:
\[ V'(9) = -\frac{125}{9^2} + \frac{1034}{9^3} = -\frac{125}{81} + \frac{1034}{729} \]
Converting these fractions to a common denominator:
\[ V'(9) = -\frac{1125}{729} + \frac{1034}{729} = \frac{-1125 + 1034}{729} = -\frac{91}{729} \approx -0.125 \]
Thus, \( V'(9) \approx -0.125 \). This negative value indicates that the property value is decreasing at a rate of approximately 0.125 thousand dollars per year (or $125 per year) when \( t = 9 \) years.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbc3f4f4-d793-4827-8e3d-2f51535d2e99%2F6c482bc8-9668-41c9-bd5e-1c8af03974df%2Flkzws5g_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Estimating Property Value Growth
A real estate firm estimates that the value of a certain property will be given by the function:
\[ V(t) = 2300 + \frac{125}{t} - \frac{517}{t^2} \]
where \( V(t) \) is the value of the property (in thousands of dollars), and \( t \) is the time (in years).
### Problem Statement
Determine how fast the value of the property is changing when \( t = 9 \) years.
### Instructions
1. Find the derivative of \( V(t) \) to determine the rate of change of the property value.
2. Evaluate the derivative at \( t = 9 \) years.
3. Indicate whether the value is increasing or decreasing and include the correct units.
### Solution
The derivative of \( V(t) = 2300 + \frac{125}{t} - \frac{517}{t^2} \) with respect to \( t \) is:
\[ V'(t) = -\frac{125}{t^2} + \frac{1034}{t^3} \]
Evaluating the derivative at \( t = 9 \) years:
\[ V'(9) = -\frac{125}{9^2} + \frac{1034}{9^3} = -\frac{125}{81} + \frac{1034}{729} \]
Converting these fractions to a common denominator:
\[ V'(9) = -\frac{1125}{729} + \frac{1034}{729} = \frac{-1125 + 1034}{729} = -\frac{91}{729} \approx -0.125 \]
Thus, \( V'(9) \approx -0.125 \). This negative value indicates that the property value is decreasing at a rate of approximately 0.125 thousand dollars per year (or $125 per year) when \( t = 9 \) years.
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