Answer :
Let's analyze the function [tex]\( f(x) \)[/tex] given as a piecewise function and evaluate the truth of each statement provided.
### Function Definition:
[tex]\[ f(x)=\left\{\begin{array}{ll} 2^x, & x<0 \\ -x^2-4x+1, & 02 \end{array}\right. \][/tex]
### Statement (A): The domain is all real numbers.
- The function [tex]\( f(x) \)[/tex] provides a piecewise definition for all ranges of [tex]\( x \)[/tex]: [tex]\( x < 0 \)[/tex], [tex]\( 0 < x < 2 \)[/tex], and [tex]\( x > 2 \)[/tex].
- There is no restriction on the values of [tex]\( x \)[/tex] provided in any of these pieces.
Conclusion: The domain of [tex]\( f(x) \)[/tex] is all real numbers [tex]\( \mathbb{R} \)[/tex].
### 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] approaches positive infinity, the term [tex]\( \frac{1}{2}x \)[/tex] will also approach positive infinity.
Conclusion: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] indeed approaches positive infinity.
### Statement (C): The function is increasing over its entire domain.
- For [tex]\( x < 0 \)[/tex]: [tex]\( f(x) = 2^x \)[/tex]. The derivative [tex]\( f'(x) = 2^x \ln 2 \)[/tex]. Since [tex]\( 2^x \ln 2 > 0 \)[/tex], this part is increasing.
- For [tex]\( 0 < x < 2 \)[/tex]: [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. The derivative [tex]\( f'(x) = -2x - 4 \)[/tex]. Since [tex]\( -2x - 4 < 0 \)[/tex] for [tex]\( 0 < x < 2 \)[/tex], this part is decreasing.
- For [tex]\( x > 2 \)[/tex]: [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. The derivative [tex]\( f'(x) = \frac{1}{2} > 0 \)[/tex], indicating that this part is increasing.
Conclusion: The function is not increasing over its entire domain.
### Statement (D): The function is continuous.
- We need to check the continuity at the boundary points [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].
- At [tex]\( x = 0 \)[/tex]: Left-hand limit as [tex]\( x \to 0^- \)[/tex] from [tex]\( 2^x \)[/tex] and right-hand limit as [tex]\( x \to 0^+ \)[/tex] from [tex]\( -x^2 - 4x + 1 \)[/tex].
[tex]\[ \lim_{x \to 0^-} 2^x = 2^0 = 1 \][/tex]
[tex]\[ \lim_{x \to 0^+} (-x^2 - 4x + 1) = -0^2 - 4 \cdot 0 + 1 = 1 \][/tex]
Both are equal, and the function value at [tex]\( x = 0 \)[/tex] is also defined to match.
- At [tex]\( x = 2 \)[/tex]: Left-hand limit as [tex]\( x \to 2^- \)[/tex] from [tex]\( -x^2 - 4x + 1 \)[/tex] and right-hand limit as [tex]\( x \to 2^+ \)[/tex] from [tex]\( \frac{1}{2} x + 3 \)[/tex].
[tex]\[ \lim_{x \to 2^-} (-x^2 - 4x + 1) = -2^2 - 4 \cdot 2 + 1 = -4 - 8 + 1 = -11 \][/tex]
[tex]\[ \lim_{x \to 2^+} \left(\frac{1}{2} x + 3\right) = \frac{1}{2} \cdot 2 + 3 = 1 + 3 = 4 \][/tex]
The limits do not match and the function values at [tex]\( x = 2 \)[/tex].
Conclusion: The function is not continuous at [tex]\( x = 2 \)[/tex].
### Final Answer:
Based on our analysis:
- (A) True
- (B) True
- (C) False
- (D) False
Multiple statements are true but based on wording that suggests a single correct choice, the best answer is:
B. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
### Function Definition:
[tex]\[ f(x)=\left\{\begin{array}{ll} 2^x, & x<0 \\ -x^2-4x+1, & 02 \end{array}\right. \][/tex]
### Statement (A): The domain is all real numbers.
- The function [tex]\( f(x) \)[/tex] provides a piecewise definition for all ranges of [tex]\( x \)[/tex]: [tex]\( x < 0 \)[/tex], [tex]\( 0 < x < 2 \)[/tex], and [tex]\( x > 2 \)[/tex].
- There is no restriction on the values of [tex]\( x \)[/tex] provided in any of these pieces.
Conclusion: The domain of [tex]\( f(x) \)[/tex] is all real numbers [tex]\( \mathbb{R} \)[/tex].
### 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] approaches positive infinity, the term [tex]\( \frac{1}{2}x \)[/tex] will also approach positive infinity.
Conclusion: As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] indeed approaches positive infinity.
### Statement (C): The function is increasing over its entire domain.
- For [tex]\( x < 0 \)[/tex]: [tex]\( f(x) = 2^x \)[/tex]. The derivative [tex]\( f'(x) = 2^x \ln 2 \)[/tex]. Since [tex]\( 2^x \ln 2 > 0 \)[/tex], this part is increasing.
- For [tex]\( 0 < x < 2 \)[/tex]: [tex]\( f(x) = -x^2 - 4x + 1 \)[/tex]. The derivative [tex]\( f'(x) = -2x - 4 \)[/tex]. Since [tex]\( -2x - 4 < 0 \)[/tex] for [tex]\( 0 < x < 2 \)[/tex], this part is decreasing.
- For [tex]\( x > 2 \)[/tex]: [tex]\( f(x) = \frac{1}{2}x + 3 \)[/tex]. The derivative [tex]\( f'(x) = \frac{1}{2} > 0 \)[/tex], indicating that this part is increasing.
Conclusion: The function is not increasing over its entire domain.
### Statement (D): The function is continuous.
- We need to check the continuity at the boundary points [tex]\( x = 0 \)[/tex] and [tex]\( x = 2 \)[/tex].
- At [tex]\( x = 0 \)[/tex]: Left-hand limit as [tex]\( x \to 0^- \)[/tex] from [tex]\( 2^x \)[/tex] and right-hand limit as [tex]\( x \to 0^+ \)[/tex] from [tex]\( -x^2 - 4x + 1 \)[/tex].
[tex]\[ \lim_{x \to 0^-} 2^x = 2^0 = 1 \][/tex]
[tex]\[ \lim_{x \to 0^+} (-x^2 - 4x + 1) = -0^2 - 4 \cdot 0 + 1 = 1 \][/tex]
Both are equal, and the function value at [tex]\( x = 0 \)[/tex] is also defined to match.
- At [tex]\( x = 2 \)[/tex]: Left-hand limit as [tex]\( x \to 2^- \)[/tex] from [tex]\( -x^2 - 4x + 1 \)[/tex] and right-hand limit as [tex]\( x \to 2^+ \)[/tex] from [tex]\( \frac{1}{2} x + 3 \)[/tex].
[tex]\[ \lim_{x \to 2^-} (-x^2 - 4x + 1) = -2^2 - 4 \cdot 2 + 1 = -4 - 8 + 1 = -11 \][/tex]
[tex]\[ \lim_{x \to 2^+} \left(\frac{1}{2} x + 3\right) = \frac{1}{2} \cdot 2 + 3 = 1 + 3 = 4 \][/tex]
The limits do not match and the function values at [tex]\( x = 2 \)[/tex].
Conclusion: The function is not continuous at [tex]\( x = 2 \)[/tex].
### Final Answer:
Based on our analysis:
- (A) True
- (B) True
- (C) False
- (D) False
Multiple statements are true but based on wording that suggests a single correct choice, the best answer is:
B. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
