Answer :
Let's analyze the piecewise function given to determine the validity of the statements:
[tex]\[ f(x)=\begin{cases} 2^x, & \text{if } x<0 \\ -x^2-4x+1, & \text{if } 02 \end{cases} \][/tex]
### Statement A: The function is continuous.
To check if the function is continuous, we need to ensure that:
1. The left-hand limit equals the right-hand limit and the function value at [tex]\( x = 0 \)[/tex].
2. The left-hand limit equals the right-hand limit and the function value at [tex]\( x = 2 \)[/tex].
At [tex]\( x = 0 \)[/tex]:
- Left-hand limit, as [tex]\( x \to 0^- \)[/tex]: [tex]\( \lim_{x \to 0^-} 2^x = 1 \)[/tex].
- Right-hand limit, as [tex]\( x \to 0^+ \)[/tex]: [tex]\(( -0^2 - 4(0) + 1 = 1 )\)[/tex].
Since the left-hand limit and right-hand limit match at [tex]\( x = 0 \)[/tex], the function is continuous at [tex]\( x = 0 \)[/tex].
At [tex]\( x = 2 \)[/tex]:
- Left-hand limit, as [tex]\( x \to 2^- \)[/tex]: [tex]\((-2^2 - 4(2) + 1 = -11 )\)[/tex].
- Right-hand limit, as [tex]\( x \to 2^+ \)[/tex]: [tex]\(( \frac{1}{2}(2) + 3 = 4 )\)[/tex].
Since the left-hand limit and right-hand limit do not match at [tex]\( x = 2 \)[/tex], the function is not continuous at [tex]\( x = 2 \)[/tex].
Therefore, statement A is False.
### Statement B: The function is increasing over its entire domain.
To determine if the function is increasing over its entire domain, we need to check each piece:
- [tex]\( 2^x \)[/tex] for [tex]\( x < 0 \)[/tex]: This function is increasing because the exponential function [tex]\( 2^x \)[/tex] increases as [tex]\( x \)[/tex] increases.
- [tex]\( -x^2 - 4x + 1 \)[/tex] for [tex]\( 0 < x < 2 \)[/tex]: This quadratic function opens downwards (the coefficient of [tex]\( x^2 \)[/tex] is negative), meaning it is not increasing over this interval.
- [tex]\( \frac{1}{2}x + 3 \)[/tex] for [tex]\( x > 2 \)[/tex]: This linear function is increasing because the coefficient of [tex]\( x \)[/tex] is positive.
Since [tex]\( -x^2 - 4x + 1 \)[/tex] is not increasing over [tex]\( 0 < x < 2 \)[/tex], the function [tex]\( f \)[/tex] is not increasing over its entire domain.
Therefore, statement B is False.
### Statement C: The domain is all real numbers.
By examining the piecewise definition of [tex]\( f(x) \)[/tex], each piece is defined for their respective intervals, and there are no gaps in the domain.
Therefore, statement C is True.
### Statement D: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 2 \)[/tex], the function is [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( \frac{1}{2}x + 3 \)[/tex] also approaches positive infinity.
Therefore, statement D is True.
Given the analysis, the correct evaluations for each statement about the function [tex]\( f \)[/tex] are:
- A. False
- B. False
- C. True
- D. True
[tex]\[ f(x)=\begin{cases} 2^x, & \text{if } x<0 \\ -x^2-4x+1, & \text{if } 02 \end{cases} \][/tex]
### Statement A: The function is continuous.
To check if the function is continuous, we need to ensure that:
1. The left-hand limit equals the right-hand limit and the function value at [tex]\( x = 0 \)[/tex].
2. The left-hand limit equals the right-hand limit and the function value at [tex]\( x = 2 \)[/tex].
At [tex]\( x = 0 \)[/tex]:
- Left-hand limit, as [tex]\( x \to 0^- \)[/tex]: [tex]\( \lim_{x \to 0^-} 2^x = 1 \)[/tex].
- Right-hand limit, as [tex]\( x \to 0^+ \)[/tex]: [tex]\(( -0^2 - 4(0) + 1 = 1 )\)[/tex].
Since the left-hand limit and right-hand limit match at [tex]\( x = 0 \)[/tex], the function is continuous at [tex]\( x = 0 \)[/tex].
At [tex]\( x = 2 \)[/tex]:
- Left-hand limit, as [tex]\( x \to 2^- \)[/tex]: [tex]\((-2^2 - 4(2) + 1 = -11 )\)[/tex].
- Right-hand limit, as [tex]\( x \to 2^+ \)[/tex]: [tex]\(( \frac{1}{2}(2) + 3 = 4 )\)[/tex].
Since the left-hand limit and right-hand limit do not match at [tex]\( x = 2 \)[/tex], the function is not continuous at [tex]\( x = 2 \)[/tex].
Therefore, statement A is False.
### Statement B: The function is increasing over its entire domain.
To determine if the function is increasing over its entire domain, we need to check each piece:
- [tex]\( 2^x \)[/tex] for [tex]\( x < 0 \)[/tex]: This function is increasing because the exponential function [tex]\( 2^x \)[/tex] increases as [tex]\( x \)[/tex] increases.
- [tex]\( -x^2 - 4x + 1 \)[/tex] for [tex]\( 0 < x < 2 \)[/tex]: This quadratic function opens downwards (the coefficient of [tex]\( x^2 \)[/tex] is negative), meaning it is not increasing over this interval.
- [tex]\( \frac{1}{2}x + 3 \)[/tex] for [tex]\( x > 2 \)[/tex]: This linear function is increasing because the coefficient of [tex]\( x \)[/tex] is positive.
Since [tex]\( -x^2 - 4x + 1 \)[/tex] is not increasing over [tex]\( 0 < x < 2 \)[/tex], the function [tex]\( f \)[/tex] is not increasing over its entire domain.
Therefore, statement B is False.
### Statement C: The domain is all real numbers.
By examining the piecewise definition of [tex]\( f(x) \)[/tex], each piece is defined for their respective intervals, and there are no gaps in the domain.
Therefore, statement C is True.
### Statement D: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 2 \)[/tex], the function is [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( \frac{1}{2}x + 3 \)[/tex] also approaches positive infinity.
Therefore, statement D is True.
Given the analysis, the correct evaluations for each statement about the function [tex]\( f \)[/tex] are:
- A. False
- B. False
- C. True
- D. True
