Answer :
Let's analyze each statement given about the function [tex]\( f(x) \)[/tex].
### Statement A: The function is increasing over its entire domain.
To determine if the function is increasing, we need to look at the behavior in each interval:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex]. The function [tex]\( 2^x \)[/tex] is decreasing because the exponent is negative.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. This is a downward parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative), which means the function is not increasing in this interval.
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. This is a linear equation with a positive slope, so it is increasing.
Since [tex]\( f(x) \)[/tex] is not increasing in the entire domain (it is decreasing for [tex]\( x < 0 \)[/tex] and [tex]\( 0 < x < 2 \)[/tex]), statement A is false.
### Statement B: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] becomes very large (approaches positive infinity), the term [tex]\(\frac{1}{2}x + 3\)[/tex] also becomes very large. Thus, [tex]\( f(x) \)[/tex] does approach positive infinity as [tex]\( x \)[/tex] approaches positive infinity.
So, statement B is true.
### Statement C: The domain is all real numbers.
Let's examine the domain of each part of the function:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex]. This is defined for all [tex]\( x < 0 \)[/tex].
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. This is defined for [tex]\( 0 < x < 2 \)[/tex].
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. This is defined for all [tex]\( x > 2 \)[/tex].
However, there are no definitions for [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex] based on the piecewise function representation. Thus, [tex]\( f(x) \)[/tex] is not defined for all real numbers.
So, statement C is false.
### Statement D: The function is continuous.
To test for continuity at potential points of discontinuity ([tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex]):
- At [tex]\( x = 0 \)[/tex]:
- Left-hand limit as [tex]\( x \)[/tex] approaches 0 from the left: [tex]\( \lim_{x \to 0^-} 2^x = 2^0 = 1 \)[/tex].
- Right-hand limit as [tex]\( x \)[/tex] approaches 0 from the right: [tex]\( \lim_{x \to 0^+} (-x^2 - 4x + 1) = 1 \)[/tex].
- However, [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex].
- At [tex]\( x = 2 \)[/tex]:
- Left-hand limit as [tex]\( x \)[/tex] approaches 2 from the left: [tex]\( \lim_{x \to 2^-} (-2^2 - 4(2) + 1) = -15 \)[/tex].
- Right-hand limit as [tex]\( x \)[/tex] approaches 2 from the right: [tex]\( \lim_{x \to 2^+} \left(\frac{1}{2}(2) + 3 \right) = 4 \)[/tex].
- However, [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 2 \)[/tex].
Since [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], it cannot be continuous over its entire domain.
So, statement D is false.
### Conclusion
The correct statement about the function [tex]\( f(x) \)[/tex] is:
B. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
### Statement A: The function is increasing over its entire domain.
To determine if the function is increasing, we need to look at the behavior in each interval:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex]. The function [tex]\( 2^x \)[/tex] is decreasing because the exponent is negative.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. This is a downward parabola (since the coefficient of [tex]\( x^2 \)[/tex] is negative), which means the function is not increasing in this interval.
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. This is a linear equation with a positive slope, so it is increasing.
Since [tex]\( f(x) \)[/tex] is not increasing in the entire domain (it is decreasing for [tex]\( x < 0 \)[/tex] and [tex]\( 0 < x < 2 \)[/tex]), statement A is false.
### Statement B: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] becomes very large (approaches positive infinity), the term [tex]\(\frac{1}{2}x + 3\)[/tex] also becomes very large. Thus, [tex]\( f(x) \)[/tex] does approach positive infinity as [tex]\( x \)[/tex] approaches positive infinity.
So, statement B is true.
### Statement C: The domain is all real numbers.
Let's examine the domain of each part of the function:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex]. This is defined for all [tex]\( x < 0 \)[/tex].
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. This is defined for [tex]\( 0 < x < 2 \)[/tex].
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. This is defined for all [tex]\( x > 2 \)[/tex].
However, there are no definitions for [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex] based on the piecewise function representation. Thus, [tex]\( f(x) \)[/tex] is not defined for all real numbers.
So, statement C is false.
### Statement D: The function is continuous.
To test for continuity at potential points of discontinuity ([tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex]):
- At [tex]\( x = 0 \)[/tex]:
- Left-hand limit as [tex]\( x \)[/tex] approaches 0 from the left: [tex]\( \lim_{x \to 0^-} 2^x = 2^0 = 1 \)[/tex].
- Right-hand limit as [tex]\( x \)[/tex] approaches 0 from the right: [tex]\( \lim_{x \to 0^+} (-x^2 - 4x + 1) = 1 \)[/tex].
- However, [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex].
- At [tex]\( x = 2 \)[/tex]:
- Left-hand limit as [tex]\( x \)[/tex] approaches 2 from the left: [tex]\( \lim_{x \to 2^-} (-2^2 - 4(2) + 1) = -15 \)[/tex].
- Right-hand limit as [tex]\( x \)[/tex] approaches 2 from the right: [tex]\( \lim_{x \to 2^+} \left(\frac{1}{2}(2) + 3 \right) = 4 \)[/tex].
- However, [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 2 \)[/tex].
Since [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], it cannot be continuous over its entire domain.
So, statement D is false.
### Conclusion
The correct statement about the function [tex]\( f(x) \)[/tex] is:
B. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
