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Sagot :
Let's analyze each statement about the function [tex]\( f(x) \)[/tex]:
### Statement A: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
To evaluate this statement, we look at the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \to \infty \)[/tex]. The piecewise function specifies that for [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] becomes very large (approaches infinity), the term [tex]\( \frac{1}{2}x \)[/tex] dominates, and [tex]\( f(x) \)[/tex] will increase without bound. Therefore, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] also approaches positive infinity.
Thus, statement A is true.
### Statement B: The function is continuous.
To determine if [tex]\( f(x) \)[/tex] is continuous, we need to check its continuity at the boundaries [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].
- At [tex]\( x = 0 \)[/tex]:
- From the left, as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex], [tex]\( f(x) \)[/tex] is described by [tex]\( 2^x \)[/tex]. The limit from the left is [tex]\( 2^0 = 1 \)[/tex].
- From the right, [tex]\( f(x) \)[/tex] is evaluated for [tex]\( 0 < x < 2 \)[/tex], hence there is no limit defined from the right at [tex]\( x = 0 \)[/tex].
- Since the function is not defined at [tex]\( x = 0 \)[/tex], it is not continuous here.
- At [tex]\( x = 2 \)[/tex]:
- From the right, as [tex]\( x \)[/tex] approaches [tex]\( 2 \)[/tex], [tex]\( f(x) \)[/tex] is described by [tex]\( \frac{1}{2}x + 3 \)[/tex]. The limit from the right is [tex]\( \frac{1}{2}(2) + 3 = 4 \)[/tex].
- However, [tex]\( f(x) \)[/tex] is not defined for [tex]\( x = 2 \)[/tex], hence there is no limit defined directly at [tex]\( x = 2 \)[/tex].
Since the function is not continuous at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], statement B is false.
### Statement C: The function is increasing over its entire domain.
To determine if [tex]\( f(x) \)[/tex] is increasing over its entire domain, we need to examine each interval separately where the function is defined.
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex] which is an exponential function. This function is increasing as [tex]\( x \)[/tex] becomes less negative.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex], which is a downward-opening parabola (decreasing in nature).
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex], which is a linear function with a positive slope, hence it is increasing.
Since [tex]\( f(x) \)[/tex] decreases in the interval [tex]\( 0 < x < 2 \)[/tex], the function is not increasing over its entire domain. Thus, statement C is false.
### Statement D: The domain is all real numbers.
To verify the domain, we examine where the function [tex]\( f(x) \)[/tex] is defined. The piecewise definition of [tex]\( f(x) \)[/tex] is given for:
- [tex]\( x < 0 \)[/tex],
- [tex]\( 0 < x < 2 \)[/tex],
- [tex]\( x > 2 \)[/tex].
However, [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex]. Thus, the function's domain does not cover all real numbers.
Therefore, statement D is false.
### Conclusion:
After evaluating all the statements, the correct answer is:
A. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
### Statement A: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
To evaluate this statement, we look at the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \to \infty \)[/tex]. The piecewise function specifies that for [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] becomes very large (approaches infinity), the term [tex]\( \frac{1}{2}x \)[/tex] dominates, and [tex]\( f(x) \)[/tex] will increase without bound. Therefore, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] also approaches positive infinity.
Thus, statement A is true.
### Statement B: The function is continuous.
To determine if [tex]\( f(x) \)[/tex] is continuous, we need to check its continuity at the boundaries [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].
- At [tex]\( x = 0 \)[/tex]:
- From the left, as [tex]\( x \)[/tex] approaches [tex]\( 0 \)[/tex], [tex]\( f(x) \)[/tex] is described by [tex]\( 2^x \)[/tex]. The limit from the left is [tex]\( 2^0 = 1 \)[/tex].
- From the right, [tex]\( f(x) \)[/tex] is evaluated for [tex]\( 0 < x < 2 \)[/tex], hence there is no limit defined from the right at [tex]\( x = 0 \)[/tex].
- Since the function is not defined at [tex]\( x = 0 \)[/tex], it is not continuous here.
- At [tex]\( x = 2 \)[/tex]:
- From the right, as [tex]\( x \)[/tex] approaches [tex]\( 2 \)[/tex], [tex]\( f(x) \)[/tex] is described by [tex]\( \frac{1}{2}x + 3 \)[/tex]. The limit from the right is [tex]\( \frac{1}{2}(2) + 3 = 4 \)[/tex].
- However, [tex]\( f(x) \)[/tex] is not defined for [tex]\( x = 2 \)[/tex], hence there is no limit defined directly at [tex]\( x = 2 \)[/tex].
Since the function is not continuous at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], statement B is false.
### Statement C: The function is increasing over its entire domain.
To determine if [tex]\( f(x) \)[/tex] is increasing over its entire domain, we need to examine each interval separately where the function is defined.
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex] which is an exponential function. This function is increasing as [tex]\( x \)[/tex] becomes less negative.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex], which is a downward-opening parabola (decreasing in nature).
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex], which is a linear function with a positive slope, hence it is increasing.
Since [tex]\( f(x) \)[/tex] decreases in the interval [tex]\( 0 < x < 2 \)[/tex], the function is not increasing over its entire domain. Thus, statement C is false.
### Statement D: The domain is all real numbers.
To verify the domain, we examine where the function [tex]\( f(x) \)[/tex] is defined. The piecewise definition of [tex]\( f(x) \)[/tex] is given for:
- [tex]\( x < 0 \)[/tex],
- [tex]\( 0 < x < 2 \)[/tex],
- [tex]\( x > 2 \)[/tex].
However, [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex]. Thus, the function's domain does not cover all real numbers.
Therefore, statement D is false.
### Conclusion:
After evaluating all the statements, the correct answer is:
A. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
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