Answer :
Let's go through each question step-by-step with detailed solutions.
### 1.
The population of a city, in thousands, is given by the function [tex]\( P(t) = 50 + 80(0.95)^t \)[/tex], where [tex]\( t \)[/tex] represents the time, in years, after the initial population was recorded.
#### a. Find [tex]\( P(4) \)[/tex] and interpret your answer in the context of this problem.
To find [tex]\( P(4) \)[/tex], we substitute [tex]\( t = 4 \)[/tex] into the function:
[tex]\[ P(4) = 50 + 80(0.95)^4 \][/tex]
After calculating, we get:
[tex]\[ P(4) = 50 + 80(0.95)^4 \approx 115.1605 \][/tex]
Interpretation:
After 4 years, the population of the city is approximately 115.16 thousand people.
#### b. What is an appropriate domain for this situation? Explain.
The domain refers to the set of possible values for [tex]\( t \)[/tex]. Since [tex]\( t \)[/tex] represents time in years, it must be non-negative. Therefore, the domain is:
[tex]\[ (0, \infty) \][/tex]
This means [tex]\( t \)[/tex] can be any non-negative value (including zero) representing time.
#### c. What is an appropriate range for this situation? Explain.
The range refers to the set of possible values for [tex]\( P(t) \)[/tex]. Given the function [tex]\( P(t) = 50 + 80(0.95)^t \)[/tex]:
- As [tex]\( t \to 0 \)[/tex], [tex]\( P(t) \)[/tex] approaches 130 (because [tex]\( 0.95^0 = 1 \)[/tex])
- As [tex]\( t \to \infty \)[/tex], [tex]\( P(t) \)[/tex] approaches 50 (because [tex]\( 0.95^t \to 0 \)[/tex])
Thus, the range is:
[tex]\[ (50, 130) \][/tex]
This means the population will never go below 50 thousand and can be as much as 130 thousand initially.
### 2.
Let [tex]\( f(x) = \sqrt{81 - x^2} \)[/tex].
#### a. What is the domain of the function?
The domain of this function consists of all [tex]\( x \)[/tex] values that make the expression under the square root non-negative:
[tex]\[ 81 - x^2 \geq 0 \][/tex]
Solving this inequality:
[tex]\[ 81 \geq x^2 \implies -9 \leq x \leq 9 \][/tex]
So, the domain is:
[tex]\[ [-9, 9] \][/tex]
#### b. What is the range of the function?
The range consists of the possible [tex]\( y \)[/tex] values that [tex]\( f(x) \)[/tex] can take. The minimum value of [tex]\( f(x) \)[/tex]:
[tex]\[ \sqrt{81 - x^2} \geq 0 \][/tex]
The maximum value occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{81 - 0} = 9 \][/tex]
So, the range is:
[tex]\[ [0, 9] \][/tex]
### 3.
Let [tex]\( f(x) = \frac{1}{2x - 10} \)[/tex].
#### a. Identify the domain of [tex]\( f \)[/tex].
The domain consists of all [tex]\( x \)[/tex] values for which the function is defined. The denominator cannot be zero:
[tex]\[ 2x - 10 \neq 0 \implies x \neq 5 \][/tex]
Thus, the domain is:
[tex]\[ (-\infty, 5) \cup (5, \infty) \][/tex]
#### b. Identify the range of [tex]\( f \)[/tex].
The range consists of all possible [tex]\( y \)[/tex] values that [tex]\( f(x) \)[/tex] can take. Let's analyze the behavior:
[tex]\[ f(x) = \frac{1}{2x - 10} \][/tex]
For any value of [tex]\( x \)[/tex] other than 5, the function is defined. As [tex]\( x \)[/tex] approaches 5, the function values increase or decrease without bound. The function will never cross [tex]\( y = 0 \)[/tex], because the function's output is a fraction with 1 in the numerator, meaning it never results in 0.
So, the range:
[tex]\[ (-\infty, 0) \cup (0, \infty) \][/tex]
### Final Answer
1.
a. [tex]\( P(4) \approx 115.16 \)[/tex] thousand.
b. Domain: [tex]\( (0, \infty) \)[/tex]
c. Range: [tex]\( (50, 130) \)[/tex]
2.
a. Domain: [tex]\( [-9, 9] \)[/tex]
b. Range: [tex]\( [0, 9] \)[/tex]
3.
a. Domain: [tex]\( (-\infty, 5) \cup (5, \infty) \)[/tex]
b. Range: [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex]#
### 1.
The population of a city, in thousands, is given by the function [tex]\( P(t) = 50 + 80(0.95)^t \)[/tex], where [tex]\( t \)[/tex] represents the time, in years, after the initial population was recorded.
#### a. Find [tex]\( P(4) \)[/tex] and interpret your answer in the context of this problem.
To find [tex]\( P(4) \)[/tex], we substitute [tex]\( t = 4 \)[/tex] into the function:
[tex]\[ P(4) = 50 + 80(0.95)^4 \][/tex]
After calculating, we get:
[tex]\[ P(4) = 50 + 80(0.95)^4 \approx 115.1605 \][/tex]
Interpretation:
After 4 years, the population of the city is approximately 115.16 thousand people.
#### b. What is an appropriate domain for this situation? Explain.
The domain refers to the set of possible values for [tex]\( t \)[/tex]. Since [tex]\( t \)[/tex] represents time in years, it must be non-negative. Therefore, the domain is:
[tex]\[ (0, \infty) \][/tex]
This means [tex]\( t \)[/tex] can be any non-negative value (including zero) representing time.
#### c. What is an appropriate range for this situation? Explain.
The range refers to the set of possible values for [tex]\( P(t) \)[/tex]. Given the function [tex]\( P(t) = 50 + 80(0.95)^t \)[/tex]:
- As [tex]\( t \to 0 \)[/tex], [tex]\( P(t) \)[/tex] approaches 130 (because [tex]\( 0.95^0 = 1 \)[/tex])
- As [tex]\( t \to \infty \)[/tex], [tex]\( P(t) \)[/tex] approaches 50 (because [tex]\( 0.95^t \to 0 \)[/tex])
Thus, the range is:
[tex]\[ (50, 130) \][/tex]
This means the population will never go below 50 thousand and can be as much as 130 thousand initially.
### 2.
Let [tex]\( f(x) = \sqrt{81 - x^2} \)[/tex].
#### a. What is the domain of the function?
The domain of this function consists of all [tex]\( x \)[/tex] values that make the expression under the square root non-negative:
[tex]\[ 81 - x^2 \geq 0 \][/tex]
Solving this inequality:
[tex]\[ 81 \geq x^2 \implies -9 \leq x \leq 9 \][/tex]
So, the domain is:
[tex]\[ [-9, 9] \][/tex]
#### b. What is the range of the function?
The range consists of the possible [tex]\( y \)[/tex] values that [tex]\( f(x) \)[/tex] can take. The minimum value of [tex]\( f(x) \)[/tex]:
[tex]\[ \sqrt{81 - x^2} \geq 0 \][/tex]
The maximum value occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{81 - 0} = 9 \][/tex]
So, the range is:
[tex]\[ [0, 9] \][/tex]
### 3.
Let [tex]\( f(x) = \frac{1}{2x - 10} \)[/tex].
#### a. Identify the domain of [tex]\( f \)[/tex].
The domain consists of all [tex]\( x \)[/tex] values for which the function is defined. The denominator cannot be zero:
[tex]\[ 2x - 10 \neq 0 \implies x \neq 5 \][/tex]
Thus, the domain is:
[tex]\[ (-\infty, 5) \cup (5, \infty) \][/tex]
#### b. Identify the range of [tex]\( f \)[/tex].
The range consists of all possible [tex]\( y \)[/tex] values that [tex]\( f(x) \)[/tex] can take. Let's analyze the behavior:
[tex]\[ f(x) = \frac{1}{2x - 10} \][/tex]
For any value of [tex]\( x \)[/tex] other than 5, the function is defined. As [tex]\( x \)[/tex] approaches 5, the function values increase or decrease without bound. The function will never cross [tex]\( y = 0 \)[/tex], because the function's output is a fraction with 1 in the numerator, meaning it never results in 0.
So, the range:
[tex]\[ (-\infty, 0) \cup (0, \infty) \][/tex]
### Final Answer
1.
a. [tex]\( P(4) \approx 115.16 \)[/tex] thousand.
b. Domain: [tex]\( (0, \infty) \)[/tex]
c. Range: [tex]\( (50, 130) \)[/tex]
2.
a. Domain: [tex]\( [-9, 9] \)[/tex]
b. Range: [tex]\( [0, 9] \)[/tex]
3.
a. Domain: [tex]\( (-\infty, 5) \cup (5, \infty) \)[/tex]
b. Range: [tex]\( (-\infty, 0) \cup (0, \infty) \)[/tex]#
