Answer :
Let's examine each of the given statements about the function [tex]\( f \)[/tex].
The function [tex]\( f \)[/tex] is defined piecewise as follows:
[tex]\[ f(x) = \begin{cases} 2^x & \text{if } x<0, \\ -x^2 - 4x + 1 & \text{if } 02. \end{cases} \][/tex]
### Option A: The function is increasing over its entire domain.
To determine if the function is increasing over its entire domain, we should consider each piece of the function separately:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex]. Since [tex]\( 2^x \)[/tex] decreases as [tex]\( x \)[/tex] decreases, this part is not increasing.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. The quadratic function [tex]\( -x^2 - 4x + 1 \)[/tex] opens downward (because the coefficient of [tex]\( x^2 \)[/tex] is negative), implying that it is not increasing.
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. This linear function has a positive slope ([tex]\(\frac{1}{2}\)[/tex]), so it is increasing in this interval.
Since the function is not increasing in the intervals [tex]\( x < 0 \)[/tex] and [tex]\( 0 < x < 2 \)[/tex], we conclude that:
Statement A is not true.
### Option B: The domain is all real numbers.
To verify the domain, we need to see if the function is defined for all values of [tex]\(x\)[/tex]:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex] is defined for all [tex]\( x < 0 \)[/tex].
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex] is defined for [tex]\( 0 < x < 2 \)[/tex].
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex] is defined for all [tex]\( x > 2 \)[/tex].
Since each piece covers a part of the real number line and there are no gaps between these intervals, the function is defined for all real numbers:
Statement B is true.
### Option C: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
Let's consider the behavior of the function as [tex]\( x \)[/tex] approaches positive infinity:
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] increases without bound, this linear function also increases without bound.
Therefore, [tex]\( f(x) \)[/tex] approaches positive infinity as [tex]\( x \)[/tex] approaches positive infinity:
Statement C is true.
### Option D: The function is continuous.
To determine if the function is continuous, we must check for potential discontinuities at the boundaries of the intervals:
- The function pieces join at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].
- For [tex]\( f(x) \)[/tex] to be continuous at these points, the limit from the left must equal the function value and the limit from the right.
At [tex]\( x = 0 \)[/tex]:
[tex]\[ \lim_{x \to 0^-} f(x) = 2^0 = 1, \][/tex]
but [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex].
At [tex]\( x = 2 \)[/tex]:
[tex]\[ \lim_{x \to 2^-} f(x) = -2^2 - 4(2) + 1 = -4 - 8 + 1 = -11, \][/tex]
[tex]\[ \lim_{x \to 2^+} f(x) = \frac{1}{2}(2) + 3 = 1 + 3 = 4. \][/tex]
These limits do not match, indicating a discontinuity.
Since there are discontinuities at the boundaries:
Statement D is not true.
Based on our evaluations, the only correct statement is:
B. The domain is all real numbers.
The function [tex]\( f \)[/tex] is defined piecewise as follows:
[tex]\[ f(x) = \begin{cases} 2^x & \text{if } x<0, \\ -x^2 - 4x + 1 & \text{if } 02. \end{cases} \][/tex]
### Option A: The function is increasing over its entire domain.
To determine if the function is increasing over its entire domain, we should consider each piece of the function separately:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex]. Since [tex]\( 2^x \)[/tex] decreases as [tex]\( x \)[/tex] decreases, this part is not increasing.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. The quadratic function [tex]\( -x^2 - 4x + 1 \)[/tex] opens downward (because the coefficient of [tex]\( x^2 \)[/tex] is negative), implying that it is not increasing.
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. This linear function has a positive slope ([tex]\(\frac{1}{2}\)[/tex]), so it is increasing in this interval.
Since the function is not increasing in the intervals [tex]\( x < 0 \)[/tex] and [tex]\( 0 < x < 2 \)[/tex], we conclude that:
Statement A is not true.
### Option B: The domain is all real numbers.
To verify the domain, we need to see if the function is defined for all values of [tex]\(x\)[/tex]:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex] is defined for all [tex]\( x < 0 \)[/tex].
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex] is defined for [tex]\( 0 < x < 2 \)[/tex].
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex] is defined for all [tex]\( x > 2 \)[/tex].
Since each piece covers a part of the real number line and there are no gaps between these intervals, the function is defined for all real numbers:
Statement B is true.
### Option C: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
Let's consider the behavior of the function as [tex]\( x \)[/tex] approaches positive infinity:
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] increases without bound, this linear function also increases without bound.
Therefore, [tex]\( f(x) \)[/tex] approaches positive infinity as [tex]\( x \)[/tex] approaches positive infinity:
Statement C is true.
### Option D: The function is continuous.
To determine if the function is continuous, we must check for potential discontinuities at the boundaries of the intervals:
- The function pieces join at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].
- For [tex]\( f(x) \)[/tex] to be continuous at these points, the limit from the left must equal the function value and the limit from the right.
At [tex]\( x = 0 \)[/tex]:
[tex]\[ \lim_{x \to 0^-} f(x) = 2^0 = 1, \][/tex]
but [tex]\( f(x) \)[/tex] is not defined at [tex]\( x = 0 \)[/tex].
At [tex]\( x = 2 \)[/tex]:
[tex]\[ \lim_{x \to 2^-} f(x) = -2^2 - 4(2) + 1 = -4 - 8 + 1 = -11, \][/tex]
[tex]\[ \lim_{x \to 2^+} f(x) = \frac{1}{2}(2) + 3 = 1 + 3 = 4. \][/tex]
These limits do not match, indicating a discontinuity.
Since there are discontinuities at the boundaries:
Statement D is not true.
Based on our evaluations, the only correct statement is:
B. The domain is all real numbers.
