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Sagot :
Let's solve the question step-by-step as a math teacher.
### Part a) State an appropriate domain and range for [tex]\( f(x) \)[/tex]
To determine the domain for [tex]\( f(x) \)[/tex], we observe the intervals given in the table. The intervals for [tex]\( x \)[/tex] are [tex]\( 0 \leq x < 1 \)[/tex], [tex]\( 1 \leq x < 2 \)[/tex], [tex]\( 2 \leq x < 3 \)[/tex], [tex]\( 3 \leq x < 4 \)[/tex], and [tex]\( 4 \leq x < 5 \)[/tex]. Collectively, these intervals cover [tex]\( 0 \leq x < 5 \)[/tex]. Hence, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ \text{Domain of } f(x): \quad 0 \leq x < 5 \][/tex]
To determine the range for [tex]\( f(x) \)[/tex], we need to examine how [tex]\( f(x) \)[/tex] changes across these intervals.
- Let's assume [tex]\( f(0) = 0 \)[/tex] for simplicity.
- From [tex]\( 0 \leq x < 1 \)[/tex], [tex]\( f(x) \)[/tex] increases by 3.1, so [tex]\( f(1) = 3.1 \)[/tex].
- From [tex]\( 1 \leq x < 2 \)[/tex], [tex]\( f(x) \)[/tex] increases by 3.8, so [tex]\( f(2) = 3.1 + 3.8 = 6.9 \)[/tex].
- From [tex]\( 2 \leq x < 3 \)[/tex], [tex]\( f(x) \)[/tex] increases by 4.0, so [tex]\( f(3) = 6.9 + 4.0 = 10.9 \)[/tex].
- From [tex]\( 3 \leq x < 4 \)[/tex], [tex]\( f(x) \)[/tex] increases by 4.2, so [tex]\( f(4) = 10.9 + 4.2 = 15.1 \)[/tex].
- From [tex]\( 4 \leq x < 5 \)[/tex], [tex]\( f(x) \)[/tex] increases by 4.4, so [tex]\( f(5) = 15.1 + 4.4 = 19.5 \)[/tex].
Therefore, the range of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] varies from [tex]\( 0 \leq x < 5 \)[/tex] is:
[tex]\[ \text{Range of } f(x): \quad 0 \leq f(x) \leq 19.5 \][/tex]
### Part b) Determine the rate of change for [tex]\( f(x) \)[/tex] on the sub-interval [tex]\( 2 \leq x < 5 \)[/tex]
To calculate the rate of change for [tex]\( f(x) \)[/tex] on the sub-interval [tex]\( 2 \leq x < 5 \)[/tex], we consider the intervals [tex]\( 2 \leq x < 3 \)[/tex], [tex]\( 3 \leq x < 4 \)[/tex], and [tex]\( 4 \leq x < 5 \)[/tex].
The changes in [tex]\( f(x) \)[/tex] in these intervals are:
- For [tex]\( 2 \leq x < 3 \)[/tex], the change is 4.0.
- For [tex]\( 3 \leq x < 4 \)[/tex], the change is 4.2.
- For [tex]\( 4 \leq x < 5 \)[/tex], the change is 4.4.
The average rate of change in these intervals:
[tex]\[ \text{Rate of Change} = \frac{4.0 + 4.2 + 4.4}{3} = \frac{12.6}{3} = 4.2 \][/tex]
So, the rate of change for [tex]\( f(x) \)[/tex] on the interval [tex]\( 2 \leq x < 5 \)[/tex] is:
[tex]\[ 4.2 \][/tex]
### Part c) Which of the functions is best modeled by a piecewise-linear function with two linear segments with different slopes?
To decide which function is best modeled by a piecewise-linear function with two linear segments with different slopes, we need to analyze the changes in each function over the intervals.
- [tex]\( f(x) \)[/tex] shows consistent positive changes (3.1, 3.8, 4.0, 4.2, 4.4). This indicates a single increasing slope with slight variation but not distinctively two different slopes.
- [tex]\( g(x) \)[/tex] changes as (2.1, 1.9, -4.0, -4.1, -4.1). Initially, there is a positive slope (increasing), followed by a negative slope (decreasing).
- [tex]\( h(x) \)[/tex] changes as (3.3, 3.1, 2.9, 2.7, 2.5). This shows a consistently decreasing rate but not distinctly two linear segments with different slopes.
Thus, [tex]\( g(x) \)[/tex] is best modeled by a piecewise-linear function with two linear segments with different slopes because it shows a distinct increase initially and then a consistent decrease.
In conclusion:
- Domain of [tex]\( f(x) \)[/tex]: [tex]\( 0 \leq x < 5 \)[/tex]
- Range of [tex]\( f(x) \)[/tex]: [tex]\( 0 \leq f(x) \leq 19.5 \)[/tex]
- Rate of change for [tex]\( f(x) \)[/tex] on [tex]\( 2 \leq x < 5 \)[/tex]: [tex]\( 4.2 \)[/tex]
- Function best modeled by a piecewise-linear function with two linear segments with different slopes: [tex]\( g(x) \)[/tex]
### Part a) State an appropriate domain and range for [tex]\( f(x) \)[/tex]
To determine the domain for [tex]\( f(x) \)[/tex], we observe the intervals given in the table. The intervals for [tex]\( x \)[/tex] are [tex]\( 0 \leq x < 1 \)[/tex], [tex]\( 1 \leq x < 2 \)[/tex], [tex]\( 2 \leq x < 3 \)[/tex], [tex]\( 3 \leq x < 4 \)[/tex], and [tex]\( 4 \leq x < 5 \)[/tex]. Collectively, these intervals cover [tex]\( 0 \leq x < 5 \)[/tex]. Hence, the domain of [tex]\( f(x) \)[/tex] is:
[tex]\[ \text{Domain of } f(x): \quad 0 \leq x < 5 \][/tex]
To determine the range for [tex]\( f(x) \)[/tex], we need to examine how [tex]\( f(x) \)[/tex] changes across these intervals.
- Let's assume [tex]\( f(0) = 0 \)[/tex] for simplicity.
- From [tex]\( 0 \leq x < 1 \)[/tex], [tex]\( f(x) \)[/tex] increases by 3.1, so [tex]\( f(1) = 3.1 \)[/tex].
- From [tex]\( 1 \leq x < 2 \)[/tex], [tex]\( f(x) \)[/tex] increases by 3.8, so [tex]\( f(2) = 3.1 + 3.8 = 6.9 \)[/tex].
- From [tex]\( 2 \leq x < 3 \)[/tex], [tex]\( f(x) \)[/tex] increases by 4.0, so [tex]\( f(3) = 6.9 + 4.0 = 10.9 \)[/tex].
- From [tex]\( 3 \leq x < 4 \)[/tex], [tex]\( f(x) \)[/tex] increases by 4.2, so [tex]\( f(4) = 10.9 + 4.2 = 15.1 \)[/tex].
- From [tex]\( 4 \leq x < 5 \)[/tex], [tex]\( f(x) \)[/tex] increases by 4.4, so [tex]\( f(5) = 15.1 + 4.4 = 19.5 \)[/tex].
Therefore, the range of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] varies from [tex]\( 0 \leq x < 5 \)[/tex] is:
[tex]\[ \text{Range of } f(x): \quad 0 \leq f(x) \leq 19.5 \][/tex]
### Part b) Determine the rate of change for [tex]\( f(x) \)[/tex] on the sub-interval [tex]\( 2 \leq x < 5 \)[/tex]
To calculate the rate of change for [tex]\( f(x) \)[/tex] on the sub-interval [tex]\( 2 \leq x < 5 \)[/tex], we consider the intervals [tex]\( 2 \leq x < 3 \)[/tex], [tex]\( 3 \leq x < 4 \)[/tex], and [tex]\( 4 \leq x < 5 \)[/tex].
The changes in [tex]\( f(x) \)[/tex] in these intervals are:
- For [tex]\( 2 \leq x < 3 \)[/tex], the change is 4.0.
- For [tex]\( 3 \leq x < 4 \)[/tex], the change is 4.2.
- For [tex]\( 4 \leq x < 5 \)[/tex], the change is 4.4.
The average rate of change in these intervals:
[tex]\[ \text{Rate of Change} = \frac{4.0 + 4.2 + 4.4}{3} = \frac{12.6}{3} = 4.2 \][/tex]
So, the rate of change for [tex]\( f(x) \)[/tex] on the interval [tex]\( 2 \leq x < 5 \)[/tex] is:
[tex]\[ 4.2 \][/tex]
### Part c) Which of the functions is best modeled by a piecewise-linear function with two linear segments with different slopes?
To decide which function is best modeled by a piecewise-linear function with two linear segments with different slopes, we need to analyze the changes in each function over the intervals.
- [tex]\( f(x) \)[/tex] shows consistent positive changes (3.1, 3.8, 4.0, 4.2, 4.4). This indicates a single increasing slope with slight variation but not distinctively two different slopes.
- [tex]\( g(x) \)[/tex] changes as (2.1, 1.9, -4.0, -4.1, -4.1). Initially, there is a positive slope (increasing), followed by a negative slope (decreasing).
- [tex]\( h(x) \)[/tex] changes as (3.3, 3.1, 2.9, 2.7, 2.5). This shows a consistently decreasing rate but not distinctly two linear segments with different slopes.
Thus, [tex]\( g(x) \)[/tex] is best modeled by a piecewise-linear function with two linear segments with different slopes because it shows a distinct increase initially and then a consistent decrease.
In conclusion:
- Domain of [tex]\( f(x) \)[/tex]: [tex]\( 0 \leq x < 5 \)[/tex]
- Range of [tex]\( f(x) \)[/tex]: [tex]\( 0 \leq f(x) \leq 19.5 \)[/tex]
- Rate of change for [tex]\( f(x) \)[/tex] on [tex]\( 2 \leq x < 5 \)[/tex]: [tex]\( 4.2 \)[/tex]
- Function best modeled by a piecewise-linear function with two linear segments with different slopes: [tex]\( g(x) \)[/tex]
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How many atoms lie in a straight line in the molecule 2-butyne, CH₃C≡CCH₃?
a. 10
b. 8
c. 6
d. 4
e. 2
