Answer :
Let's analyze each statement step by step:
1. Function [tex]\( f \)[/tex] is continuous.
To determine if the function [tex]\( f \)[/tex] is continuous at [tex]\( x = 1 \)[/tex], we must check if the left-hand limit equals the right-hand limit at this point.
For [tex]\( x \leq 1 \)[/tex]:
[tex]\[ f(x) = \left(\frac{1}{3}\right)^x \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
For [tex]\( x > 1 \)[/tex]:
[tex]\[ f(x) = -x^2 + 2x - 1 \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -(1)^2 + 2(1) - 1 = -1 + 2 - 1 = 0 \][/tex]
Since [tex]\(\frac{1}{3} \neq 0\)[/tex], the left-hand limit does not equal the right-hand limit at [tex]\( x = 1 \)[/tex]. Therefore, the function [tex]\( f \)[/tex] is not continuous.
2. The domain of function [tex]\( f \)[/tex] is all real numbers.
The function is defined for all [tex]\( x \leq 1 \)[/tex] by [tex]\( \left( \frac{1}{3} \right)^x \)[/tex] and for all [tex]\( x > 1 \)[/tex] by [tex]\( -x^2 + 2x - 1 \)[/tex]. Both parts are defined for their respective domains without any restrictions. Therefore, the domain of function [tex]\( f \)[/tex] is all real numbers.
3. Function [tex]\( f \)[/tex] is decreasing over the entire domain.
- For [tex]\( x \leq 1 \)[/tex], [tex]\( f(x) = \left( \frac{1}{3} \right)^x \)[/tex] is decreasing because [tex]\( \frac{1}{3} < 1 \)[/tex].
- For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. The derivative is:
[tex]\[ f'(x) = -2x + 2 \][/tex]
At [tex]\( x = 1 \)[/tex], [tex]\( f'(1) = -2(1) + 2 = 0 \)[/tex]. For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) > 0 \)[/tex] indicating an increasing function, and for [tex]\( x > 1 \)[/tex], [tex]\( f'(x) < 0 \)[/tex], indicating a decreasing function. However, since it’s not consistently decreasing over [tex]\( x \leq 1 \)[/tex] and [tex]\( x > 1 \)[/tex] domains, [tex]\( f \)[/tex] is not decreasing over the entire domain.
4. The value of [tex]\( f(1) \)[/tex] is 0.
At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 0 \][/tex]
This is based on the right-hand expression, even though the left-hand expression yields [tex]\(\frac{1}{3}\)[/tex]. From either part, it has previously been confirmed that the value of [tex]\( f(1) \)[/tex] directly here is 0.
5. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. As [tex]\( x \to +\infty \)[/tex]:
[tex]\[ -x^2 \text{ term dominates, leading to } f(x) \to -\infty \][/tex]
Therefore, as [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \)[/tex] does not approach positive infinity, but rather negative infinity.
Based on this detailed analysis:
Correct selections:
- The domain of function [tex]\( f \)[/tex] is all real numbers.
- The value of [tex]\( f(1) \)[/tex] is 0.
1. Function [tex]\( f \)[/tex] is continuous.
To determine if the function [tex]\( f \)[/tex] is continuous at [tex]\( x = 1 \)[/tex], we must check if the left-hand limit equals the right-hand limit at this point.
For [tex]\( x \leq 1 \)[/tex]:
[tex]\[ f(x) = \left(\frac{1}{3}\right)^x \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \left(\frac{1}{3}\right)^1 = \frac{1}{3} \][/tex]
For [tex]\( x > 1 \)[/tex]:
[tex]\[ f(x) = -x^2 + 2x - 1 \][/tex]
When [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = -(1)^2 + 2(1) - 1 = -1 + 2 - 1 = 0 \][/tex]
Since [tex]\(\frac{1}{3} \neq 0\)[/tex], the left-hand limit does not equal the right-hand limit at [tex]\( x = 1 \)[/tex]. Therefore, the function [tex]\( f \)[/tex] is not continuous.
2. The domain of function [tex]\( f \)[/tex] is all real numbers.
The function is defined for all [tex]\( x \leq 1 \)[/tex] by [tex]\( \left( \frac{1}{3} \right)^x \)[/tex] and for all [tex]\( x > 1 \)[/tex] by [tex]\( -x^2 + 2x - 1 \)[/tex]. Both parts are defined for their respective domains without any restrictions. Therefore, the domain of function [tex]\( f \)[/tex] is all real numbers.
3. Function [tex]\( f \)[/tex] is decreasing over the entire domain.
- For [tex]\( x \leq 1 \)[/tex], [tex]\( f(x) = \left( \frac{1}{3} \right)^x \)[/tex] is decreasing because [tex]\( \frac{1}{3} < 1 \)[/tex].
- For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. The derivative is:
[tex]\[ f'(x) = -2x + 2 \][/tex]
At [tex]\( x = 1 \)[/tex], [tex]\( f'(1) = -2(1) + 2 = 0 \)[/tex]. For [tex]\( x < 1 \)[/tex], [tex]\( f'(x) > 0 \)[/tex] indicating an increasing function, and for [tex]\( x > 1 \)[/tex], [tex]\( f'(x) < 0 \)[/tex], indicating a decreasing function. However, since it’s not consistently decreasing over [tex]\( x \leq 1 \)[/tex] and [tex]\( x > 1 \)[/tex] domains, [tex]\( f \)[/tex] is not decreasing over the entire domain.
4. The value of [tex]\( f(1) \)[/tex] is 0.
At [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = 0 \][/tex]
This is based on the right-hand expression, even though the left-hand expression yields [tex]\(\frac{1}{3}\)[/tex]. From either part, it has previously been confirmed that the value of [tex]\( f(1) \)[/tex] directly here is 0.
5. As [tex]\( x \)[/tex] approaches positive infinity, [tex]\( f(x) \)[/tex] approaches positive infinity.
For [tex]\( x > 1 \)[/tex], [tex]\( f(x) = -x^2 + 2x - 1 \)[/tex]. As [tex]\( x \to +\infty \)[/tex]:
[tex]\[ -x^2 \text{ term dominates, leading to } f(x) \to -\infty \][/tex]
Therefore, as [tex]\( x \to +\infty \)[/tex], [tex]\( f(x) \)[/tex] does not approach positive infinity, but rather negative infinity.
Based on this detailed analysis:
Correct selections:
- The domain of function [tex]\( f \)[/tex] is all real numbers.
- The value of [tex]\( f(1) \)[/tex] is 0.
