Answer :
To analyze the relationship between the functions [tex]\( f(x) = \sqrt{16} x \)[/tex] and [tex]\( g(x) = \sqrt[3]{64} \)[/tex], let's start by understanding each function separately.
1. Function [tex]\( f(x) = \sqrt{16} x \)[/tex]:
- Simplify the expression [tex]\(\sqrt{16}\)[/tex]:
[tex]\[ \sqrt{16} = 4 \][/tex]
- Therefore, the function can be written as:
[tex]\[ f(x) = 4x \][/tex]
- This is a linear function with a slope of 4 and a y-intercept of 0. Its graph is a straight line passing through the origin (0,0) with a slope that indicates it rises 4 units for every 1 unit it moves to the right.
2. Function [tex]\( g(x) = \sqrt[3]{64} \)[/tex]:
- Simplify the expression [tex]\(\sqrt[3]{64}\)[/tex]:
[tex]\[ \sqrt[3]{64} = 4 \][/tex]
- Therefore, the function can be written as:
[tex]\[ g(x) = 4 \][/tex]
- This is a constant function. No matter the value of [tex]\( x \)[/tex], [tex]\( g(x) \)[/tex] always equals 4. Therefore, its graph is a horizontal line at [tex]\( y = 4 \)[/tex].
### Comparing the functions:
1. Initial Values:
- The initial value of [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex] is:
[tex]\[ f(0) = 4 \times 0 = 0 \][/tex]
- The initial value of [tex]\( g(x) \)[/tex] at [tex]\( x = 0 \)[/tex] is:
[tex]\[ g(0) = 4 \][/tex]
- Therefore, [tex]\( g(x) \)[/tex] has a greater initial value compared to [tex]\( f(x) \)[/tex].
2. Rate of Change:
- [tex]\( f(x) \)[/tex] increases linearly with a slope of 4. This means it increases at a constant rate.
- [tex]\( g(x) \)[/tex] is a constant function, which means it has no rate of change; it remains the same for all values of [tex]\( x \)[/tex].
- Since [tex]\( g(x) \)[/tex] does not change, it cannot be said to increase or decrease at any rate.
3. Equivalence and Other Properties:
- The functions are not equivalent in general since [tex]\( f(x) = 4x \)[/tex] changes with [tex]\( x \)[/tex], whereas [tex]\( g(x) = 4 \)[/tex] remains constant.
### Conclusion:
The correct inference from the analysis is that the function [tex]\( g(x) \)[/tex] has a greater initial value.
Therefore, the correct choice is:
- The function [tex]\( g(x) \)[/tex] has a greater initial value.
1. Function [tex]\( f(x) = \sqrt{16} x \)[/tex]:
- Simplify the expression [tex]\(\sqrt{16}\)[/tex]:
[tex]\[ \sqrt{16} = 4 \][/tex]
- Therefore, the function can be written as:
[tex]\[ f(x) = 4x \][/tex]
- This is a linear function with a slope of 4 and a y-intercept of 0. Its graph is a straight line passing through the origin (0,0) with a slope that indicates it rises 4 units for every 1 unit it moves to the right.
2. Function [tex]\( g(x) = \sqrt[3]{64} \)[/tex]:
- Simplify the expression [tex]\(\sqrt[3]{64}\)[/tex]:
[tex]\[ \sqrt[3]{64} = 4 \][/tex]
- Therefore, the function can be written as:
[tex]\[ g(x) = 4 \][/tex]
- This is a constant function. No matter the value of [tex]\( x \)[/tex], [tex]\( g(x) \)[/tex] always equals 4. Therefore, its graph is a horizontal line at [tex]\( y = 4 \)[/tex].
### Comparing the functions:
1. Initial Values:
- The initial value of [tex]\( f(x) \)[/tex] at [tex]\( x = 0 \)[/tex] is:
[tex]\[ f(0) = 4 \times 0 = 0 \][/tex]
- The initial value of [tex]\( g(x) \)[/tex] at [tex]\( x = 0 \)[/tex] is:
[tex]\[ g(0) = 4 \][/tex]
- Therefore, [tex]\( g(x) \)[/tex] has a greater initial value compared to [tex]\( f(x) \)[/tex].
2. Rate of Change:
- [tex]\( f(x) \)[/tex] increases linearly with a slope of 4. This means it increases at a constant rate.
- [tex]\( g(x) \)[/tex] is a constant function, which means it has no rate of change; it remains the same for all values of [tex]\( x \)[/tex].
- Since [tex]\( g(x) \)[/tex] does not change, it cannot be said to increase or decrease at any rate.
3. Equivalence and Other Properties:
- The functions are not equivalent in general since [tex]\( f(x) = 4x \)[/tex] changes with [tex]\( x \)[/tex], whereas [tex]\( g(x) = 4 \)[/tex] remains constant.
### Conclusion:
The correct inference from the analysis is that the function [tex]\( g(x) \)[/tex] has a greater initial value.
Therefore, the correct choice is:
- The function [tex]\( g(x) \)[/tex] has a greater initial value.