Answer :
To determine which statement is true about the given piecewise function [tex]\( f \)[/tex], we need to examine several aspects of the function.
The function is given by:
[tex]\[ f(x) = \begin{cases} 2^x & \text{if } x < 0 \\ -x^2 - 4x + 1 & \text{if } 0 < x < 2 \\ \frac{1}{2}x + 3 & \text{if } x > 2 \end{cases} \][/tex]
Let's analyze each statement:
Statement A: The function is continuous.
For the function to be continuous at points where the pieces connect (i.e., [tex]\(x = 0\)[/tex] and [tex]\(x = 2\)[/tex]), the left-hand limit and right-hand limit at these points must equal the value of the function at those points.
1. At [tex]\( x = 0 \)[/tex]:
- The left-hand limit as [tex]\( x \)[/tex] approaches 0 from the left is [tex]\( \lim_{x \to 0^-} 2^x = 2^0 = 1 \)[/tex].
- The right-hand limit as [tex]\( x \)[/tex] approaches 0 from the right is [tex]\( \lim_{x \to 0^+} (-x^2 - 4x + 1) = -(0)^2 - 4(0) + 1 = 1 \)[/tex].
- Since [tex]\( f \)[/tex] is not defined at [tex]\( x = 0 \)[/tex], [tex]\( f(0) \)[/tex] does not exist, thus [tex]\( f \)[/tex] is not continuous at [tex]\( x = 0 \)[/tex].
Because [tex]\( f \)[/tex] is not defined at [tex]\( x = 0 \)[/tex], [tex]\( f \)[/tex] is discontinuous at [tex]\( x = 0 \)[/tex].
Statement B: The domain is all real numbers.
By inspecting the given function pieces:
- The first piece, [tex]\( 2^x \)[/tex] for [tex]\( x < 0 \)[/tex], is defined for all [tex]\( x < 0 \)[/tex].
- The second piece, [tex]\( -x^2 - 4x + 1 \)[/tex] for [tex]\( 0 < x < 2 \)[/tex], is defined for [tex]\( 0 < x < 2 \)[/tex].
- The third piece, [tex]\( \frac{1}{2}x + 3 \)[/tex] for [tex]\( x > 2 \)[/tex], is defined for all [tex]\( x > 2 \)[/tex].
The function [tex]\( f \)[/tex] is not defined at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], so [tex]\( f \)[/tex] does not have all real numbers as its domain. The domain is [tex]\( (-\infty, 0) \cup (0, 2) \cup (2, \infty) \)[/tex] and is not all real numbers.
Statement C: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
Let's inspect the behavior of [tex]\( f \)[/tex] for [tex]\( x > 2 \)[/tex]:
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( \frac{1}{2}x + 3 \to \infty \)[/tex].
Therefore, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] does indeed approach positive infinity.
Statement D: The function is increasing over its entire domain.
We need to check if each piece is increasing over the relevant intervals and if the transitions between pieces maintain increasing behavior:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex] is increasing because the exponential function with base 2 is strictly increasing.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. To determine if this is increasing or decreasing, we find its derivative: [tex]\( f'(x) = -2x - 4 \)[/tex]. For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f'(x) < 0 \)[/tex], indicating it is decreasing over this interval.
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex] is increasing because the slope [tex]\( \frac{1}{2} \)[/tex] is positive.
Because the function is decreasing in the interval [tex]\( 0 < x < 2 \)[/tex], it cannot be increasing over its entire domain.
Conclusion:
The correct statement is:
C. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
The function is given by:
[tex]\[ f(x) = \begin{cases} 2^x & \text{if } x < 0 \\ -x^2 - 4x + 1 & \text{if } 0 < x < 2 \\ \frac{1}{2}x + 3 & \text{if } x > 2 \end{cases} \][/tex]
Let's analyze each statement:
Statement A: The function is continuous.
For the function to be continuous at points where the pieces connect (i.e., [tex]\(x = 0\)[/tex] and [tex]\(x = 2\)[/tex]), the left-hand limit and right-hand limit at these points must equal the value of the function at those points.
1. At [tex]\( x = 0 \)[/tex]:
- The left-hand limit as [tex]\( x \)[/tex] approaches 0 from the left is [tex]\( \lim_{x \to 0^-} 2^x = 2^0 = 1 \)[/tex].
- The right-hand limit as [tex]\( x \)[/tex] approaches 0 from the right is [tex]\( \lim_{x \to 0^+} (-x^2 - 4x + 1) = -(0)^2 - 4(0) + 1 = 1 \)[/tex].
- Since [tex]\( f \)[/tex] is not defined at [tex]\( x = 0 \)[/tex], [tex]\( f(0) \)[/tex] does not exist, thus [tex]\( f \)[/tex] is not continuous at [tex]\( x = 0 \)[/tex].
Because [tex]\( f \)[/tex] is not defined at [tex]\( x = 0 \)[/tex], [tex]\( f \)[/tex] is discontinuous at [tex]\( x = 0 \)[/tex].
Statement B: The domain is all real numbers.
By inspecting the given function pieces:
- The first piece, [tex]\( 2^x \)[/tex] for [tex]\( x < 0 \)[/tex], is defined for all [tex]\( x < 0 \)[/tex].
- The second piece, [tex]\( -x^2 - 4x + 1 \)[/tex] for [tex]\( 0 < x < 2 \)[/tex], is defined for [tex]\( 0 < x < 2 \)[/tex].
- The third piece, [tex]\( \frac{1}{2}x + 3 \)[/tex] for [tex]\( x > 2 \)[/tex], is defined for all [tex]\( x > 2 \)[/tex].
The function [tex]\( f \)[/tex] is not defined at [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex], so [tex]\( f \)[/tex] does not have all real numbers as its domain. The domain is [tex]\( (-\infty, 0) \cup (0, 2) \cup (2, \infty) \)[/tex] and is not all real numbers.
Statement C: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
Let's inspect the behavior of [tex]\( f \)[/tex] for [tex]\( x > 2 \)[/tex]:
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex].
- As [tex]\( x \to \infty \)[/tex], [tex]\( \frac{1}{2}x + 3 \to \infty \)[/tex].
Therefore, as [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] does indeed approach positive infinity.
Statement D: The function is increasing over its entire domain.
We need to check if each piece is increasing over the relevant intervals and if the transitions between pieces maintain increasing behavior:
- For [tex]\( x < 0 \)[/tex], [tex]\( f(x) = 2^x \)[/tex] is increasing because the exponential function with base 2 is strictly increasing.
- For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. To determine if this is increasing or decreasing, we find its derivative: [tex]\( f'(x) = -2x - 4 \)[/tex]. For [tex]\( 0 < x < 2 \)[/tex], [tex]\( f'(x) < 0 \)[/tex], indicating it is decreasing over this interval.
- For [tex]\( x > 2 \)[/tex], [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex] is increasing because the slope [tex]\( \frac{1}{2} \)[/tex] is positive.
Because the function is decreasing in the interval [tex]\( 0 < x < 2 \)[/tex], it cannot be increasing over its entire domain.
Conclusion:
The correct statement is:
C. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
