Answer :
To solve this problem, we need to analyze each of the given functions to see which one meets specific conditions: it must be always negative to the left of the y-axis (x < 0), always positive to the right of the y-axis (x > 0), and must be a decreasing function over every interval in its domain. Let's examine each option one by one.
### A) [tex]\(\frac{1}{x}\)[/tex]
1. Left of the y-axis (x < 0):
- When [tex]\(x\)[/tex] is negative, [tex]\(\frac{1}{x}\)[/tex] is negative. For example, if [tex]\(x = -2\)[/tex], [tex]\(\frac{1}{-2} = -0.5\)[/tex].
2. Right of the y-axis (x > 0):
- When [tex]\(x\)[/tex] is positive, [tex]\(\frac{1}{x}\)[/tex] is positive. For example, if [tex]\(x = 2\)[/tex], [tex]\(\frac{1}{2} = 0.5\)[/tex].
3. Decreasing function:
- As [tex]\(x\)[/tex] increases, [tex]\( \frac{1}{x} \)[/tex] decreases. For instance, moving from [tex]\(x = 1\)[/tex] to [tex]\(x = 2\)[/tex], the function goes from 1 to 0.5, showing a decrease. Similarly, moving from [tex]\(x = -1\)[/tex] to [tex]\(x = -2\)[/tex], the function goes from -1 to -0.5, showing a decrease in absolute value.
### B) [tex]\(\frac{1}{1+e^{-x}}\)[/tex]
1. Left of the y-axis (x < 0):
- When [tex]\(x\)[/tex] is negative, [tex]\(\frac{1}{1+e^{-x}}\)[/tex] is between 0 and 0.5. However, it is always positive because [tex]\(e^{-x}\)[/tex] is positive for all real [tex]\(x\)[/tex].
2. Right of the y-axis (x > 0):
- When [tex]\(x\)[/tex] is positive, [tex]\(\frac{1}{1+e^{-x}}\)[/tex] is between 0.5 and 1. It is always positive.
3. Decreasing function:
- This function is not always decreasing. For example, as [tex]\(x\)[/tex] increases from -1 to 1, [tex]\(\frac{1}{1+e^{-x}}\)[/tex] moves from approximately 0.2689 to 0.731. Thus, the function increases instead of decreases.
### C) [tex]\(x^3\)[/tex]
1. Left of the y-axis (x < 0):
- When [tex]\(x\)[/tex] is negative, [tex]\(x^3\)[/tex] is negative. For example, if [tex]\(x = -2\)[/tex], then [tex]\((-2)^3 = -8\)[/tex].
2. Right of the y-axis (x > 0):
- When [tex]\(x\)[/tex] is positive, [tex]\(x^3\)[/tex] is positive. For example, if [tex]\(x = 2\)[/tex], then [tex]\(2^3 = 8\)[/tex].
3. Decreasing function:
- [tex]\(x^3\)[/tex] is an increasing function. As [tex]\(x\)[/tex] increases, [tex]\(x^3\)[/tex] also increases. Therefore, this function does not meet the condition of being decreasing over every interval.
### D) [tex]\(\text{int}(x)\)[/tex]
1. Left of the y-axis (x < 0):
- The integer part of [tex]\(x\)[/tex] for negative [tex]\(x\)[/tex] is negative but it is a step function and not continuously negative.
2. Right of the y-axis (x > 0):
- For positive [tex]\(x\)[/tex], int([tex]\(x\)[/tex]) is positive but, again, it is a discrete step function.
3. Decreasing function:
- int([tex]\(x\)[/tex]) is not a continuously decreasing function. It is a step function which is constant over intervals and increases in jumps at integer values.
### Conclusion
After examining all the options, we see that option A, [tex]\(\frac{1}{x}\)[/tex], is the only function that is always negative to the left of the y-axis, always positive to the right of the y-axis, and decreases over every interval in its domain. Thus, the correct answer is:
A) [tex]\(\frac{1}{x}\)[/tex]
### A) [tex]\(\frac{1}{x}\)[/tex]
1. Left of the y-axis (x < 0):
- When [tex]\(x\)[/tex] is negative, [tex]\(\frac{1}{x}\)[/tex] is negative. For example, if [tex]\(x = -2\)[/tex], [tex]\(\frac{1}{-2} = -0.5\)[/tex].
2. Right of the y-axis (x > 0):
- When [tex]\(x\)[/tex] is positive, [tex]\(\frac{1}{x}\)[/tex] is positive. For example, if [tex]\(x = 2\)[/tex], [tex]\(\frac{1}{2} = 0.5\)[/tex].
3. Decreasing function:
- As [tex]\(x\)[/tex] increases, [tex]\( \frac{1}{x} \)[/tex] decreases. For instance, moving from [tex]\(x = 1\)[/tex] to [tex]\(x = 2\)[/tex], the function goes from 1 to 0.5, showing a decrease. Similarly, moving from [tex]\(x = -1\)[/tex] to [tex]\(x = -2\)[/tex], the function goes from -1 to -0.5, showing a decrease in absolute value.
### B) [tex]\(\frac{1}{1+e^{-x}}\)[/tex]
1. Left of the y-axis (x < 0):
- When [tex]\(x\)[/tex] is negative, [tex]\(\frac{1}{1+e^{-x}}\)[/tex] is between 0 and 0.5. However, it is always positive because [tex]\(e^{-x}\)[/tex] is positive for all real [tex]\(x\)[/tex].
2. Right of the y-axis (x > 0):
- When [tex]\(x\)[/tex] is positive, [tex]\(\frac{1}{1+e^{-x}}\)[/tex] is between 0.5 and 1. It is always positive.
3. Decreasing function:
- This function is not always decreasing. For example, as [tex]\(x\)[/tex] increases from -1 to 1, [tex]\(\frac{1}{1+e^{-x}}\)[/tex] moves from approximately 0.2689 to 0.731. Thus, the function increases instead of decreases.
### C) [tex]\(x^3\)[/tex]
1. Left of the y-axis (x < 0):
- When [tex]\(x\)[/tex] is negative, [tex]\(x^3\)[/tex] is negative. For example, if [tex]\(x = -2\)[/tex], then [tex]\((-2)^3 = -8\)[/tex].
2. Right of the y-axis (x > 0):
- When [tex]\(x\)[/tex] is positive, [tex]\(x^3\)[/tex] is positive. For example, if [tex]\(x = 2\)[/tex], then [tex]\(2^3 = 8\)[/tex].
3. Decreasing function:
- [tex]\(x^3\)[/tex] is an increasing function. As [tex]\(x\)[/tex] increases, [tex]\(x^3\)[/tex] also increases. Therefore, this function does not meet the condition of being decreasing over every interval.
### D) [tex]\(\text{int}(x)\)[/tex]
1. Left of the y-axis (x < 0):
- The integer part of [tex]\(x\)[/tex] for negative [tex]\(x\)[/tex] is negative but it is a step function and not continuously negative.
2. Right of the y-axis (x > 0):
- For positive [tex]\(x\)[/tex], int([tex]\(x\)[/tex]) is positive but, again, it is a discrete step function.
3. Decreasing function:
- int([tex]\(x\)[/tex]) is not a continuously decreasing function. It is a step function which is constant over intervals and increases in jumps at integer values.
### Conclusion
After examining all the options, we see that option A, [tex]\(\frac{1}{x}\)[/tex], is the only function that is always negative to the left of the y-axis, always positive to the right of the y-axis, and decreases over every interval in its domain. Thus, the correct answer is:
A) [tex]\(\frac{1}{x}\)[/tex]
